Re: [PHP-DB] Please help a newbie
Hello,
May be try something like this:
$query1=
CREATE TABLE contacts(
id int(16) NOT NULL auto_increment,
phone varchar(15) NOT NULL,
name varchar(15) NOT NULL,
address varchar(15) NOT NULL,
PRIMARY KEY (id)
);
$query2 = INSERT INTO contacts VALUES ('NULL','$phone', '$name',
'$address');
P:S
id incremented automatically by MYSQL now
GR
mrfroasty
Rij wrote:
Hello,
I am new to the world of PHP and MySQL. My objective is to create a
table, insert values in it and read it back.
Here's the partial code to create a table from a PHP file:
if (!$table_exists) {
$query=CREATE TABLE contacts (id int(20) NOT NULL, name
varchar(15) NOT NULL, address varchar(15),PRIMARY KEY(id)
if (mysql_query($query, $con)) echo Table contacts created;
else die('Unable to create table : '.mysql_error());
}
I input the values from a HTML form. Here is the partial code.
$phone = $_POST['phone'];
$name = $_POST['name'];
$address = $_POST['address'];
$query = INSERT INTO contacts VALUES ('$phone', '$name', '$address');
if (mysql_query($query, $con)) echo Values inserted;
else die('Unable to create table : '.mysql_error());
Now the problem that I am facing is that when I make my first insert,
the id field shows a garbage value and not the number that I entered.
Subsequent entries into the table show up just fine. It's only the
first one.
What am I doing wrong?
Thanks, Rij
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Re: [PHP-DB] Please help a newbie
Rij wrote:
I input the values from a HTML form. Here is the partial code.
$phone = $_POST['phone'];
$name = $_POST['name'];
$address = $_POST['address'];
$query = INSERT INTO contacts VALUES ('$phone', '$name', '$address');
if (mysql_query($query, $con)) echo Values inserted;
else die('Unable to create table : '.mysql_error());
This is unsafe code. I suggest you lookup prepared statements and the
PDO library (which is part of PHP).
Daniel.
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Re: [PHP-DB] Please help a newbie: Fixed
Ok, figured it out.
I was using an int for the field but the value was too large for it. I
used bigint instead and life is suddenly so much better. Well it helps
that its bright and sunny here too :)
On Sun, Apr 19, 2009 at 10:37 AM, mrfroasty mrfroa...@gmail.com wrote:
Try using var_export() or var_dump() to debug to see why that 1st time is
having different values than what you want :-)
The idea from the sample code I have provided is not to use phone as id...
P:S
A wild guess would phone is some sort of string.
GR
mrfroasty
Rij wrote:
Ok. But that still doesn't tell me why I am getting the behavior from
my code. Where is the garbage value coming from? And why only the
first time I do an INSERT?
To Daniel Carrera, thanks for your tip. I sure will look up your
suggestion.
On Sun, Apr 19, 2009 at 2:32 AM, mrfroasty mrfroa...@gmail.com wrote:
Hello,
May be try something like this:
$query1=
CREATE TABLE contacts(
id int(16) NOT NULL auto_increment,
phone varchar(15) NOT NULL,
name varchar(15) NOT NULL,
address varchar(15) NOT NULL,
PRIMARY KEY (id)
);
$query2 = INSERT INTO contacts VALUES ('NULL','$phone', '$name',
'$address');
P:S
id incremented automatically by MYSQL now
GR
mrfroasty
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Re: [PHP-DB] Please help a newbie
mrfroasty wrote:
Hello,
May be try something like this:
$query1=
CREATE TABLE contacts(
id int(16) NOT NULL auto_increment,
phone varchar(15) NOT NULL,
name varchar(15) NOT NULL,
address varchar(15) NOT NULL,
PRIMARY KEY (id)
);
$query2 = INSERT INTO contacts VALUES ('NULL','$phone', '$name',
'$address');
P:S
id incremented automatically by MYSQL now
Maybe - but it's by accident. You're trying to insert the word NULL into
an int field (it's being treated as a word because of the single quotes
around it).
Don't specify the id field at all:
$query2 = insert into contacts(phone, name, address) values (' .
mysql_real_escape_string($_POST['phone']) . ', ' .
mysql_real_escape_string($_POST['name']) . ', ' .
mysql_real_escape_string($_POST['address']) . ');
You should always use the field names (as above) because if your table
gets reordered, your inserts will now break - if you put name before
phone, the data is now going into the wrong fields.
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Re: [PHP-DB] Please help
WHERE columname LIKE condition ^^^ example. On Friday 10 September 2004 10:22 am, Stuart Felenstein wrote: Okay, I was under the impression that where is implied inthe joins xx.xx = xx.xx . Is that not the case ? As a matter of fact, there isn't even a where or like in my query. Then again myabe that's why I can see *all* records but not run a search. Can you please provide an exmample of the column name bteween where and like ? Thank you. Stuart --- Micah Stevens [EMAIL PROTECTED] wrote: If you look at the query, there's no column name between 'WHERE' and 'LIKE' which is a syntax error. That's the problem. Looks like your sql generator has some issues, or you didn't specify the WHERE column properly. I'm not familiar with the system you're using, but keep in mind, that mysql_error will usually not steer you wrong. -Micah On Friday 10 September 2004 10:03 am, Stuart Felenstein wrote: I had not, my apologies. I think your post slipped by. Anyway, yes I have now inserted mysql_error() and got my return. Though I'm not entirely sure how to fix it. Here is the error. You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Like 'ACCFIN'' at line 1 SELECT `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit 0,1 -- --- Micah Stevens [EMAIL PROTECTED] wrote: did you make the change to the code I suggested? What does MySQL say the error is? -Micah On Friday 10 September 2004 07:49 am, Stuart Felenstein wrote: As I said this is a code generator (dbqwiksite). So, describing the process for creating the code is different. The $sql is fine, as far as typos or incorrect characterrs. I've gone through those statement very carefully. I've also tried to run a debug with no luck. But I do know that there is something wrong with the statement, since I can get a complete display of all records, but can't search using criteria. Below is the $sql, one I know works , though I've tried others. VendorJobs is the table I'm querying , and since it's made up of values from other static type tables I have all the joins in place. I've also taken the tick marks out, thinking maybe it was a style issue. I'm not sure where to go with it. Getting a response from the company is like pretty difficult. `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) Stuart --- Philip Thompson [EMAIL PROTECTED] wrote: I
Re: [PHP-DB] Please help
I think everyone knows that $sql is a statement. But what people are asking is: what is that statement?! Because if there are incorrect characters or 's in that statement, then that can break your code/statement. ~Philip On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote: Just getting back to this thing. There are 3 files involved, search.php, connections.php and functions.php. I searched through all and couldn't find the meaning of the $sql_ext. $sql is just the sql statement I inserted into the code. Great that I can't get a response from the company. If anyone thinks it would be useful to post the code I will. That is if anyone would want to see and look at it. The weird thing is searching on one table is great. The list (all records) works great. The problem starts when I insert all my joins. Now I've tested and retested the query and I know it's fine. Thank you , Stuart --- Steve Davies [EMAIL PROTECTED] wrote: What's contained in $sql and $sql_ext ??? Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
As I said this is a code generator (dbqwiksite). So, describing the process for creating the code is different. The $sql is fine, as far as typos or incorrect characterrs. I've gone through those statement very carefully. I've also tried to run a debug with no luck. But I do know that there is something wrong with the statement, since I can get a complete display of all records, but can't search using criteria. Below is the $sql, one I know works , though I've tried others. VendorJobs is the table I'm querying , and since it's made up of values from other static type tables I have all the joins in place. I've also taken the tick marks out, thinking maybe it was a style issue. I'm not sure where to go with it. Getting a response from the company is like pretty difficult. `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) Stuart --- Philip Thompson [EMAIL PROTECTED] wrote: I think everyone knows that $sql is a statement. But what people are asking is: what is that statement?! Because if there are incorrect characters or 's in that statement, then that can break your code/statement. ~Philip On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote: Just getting back to this thing. There are 3 files involved, search.php, connections.php and functions.php. I searched through all and couldn't find the meaning of the $sql_ext. $sql is just the sql statement I inserted into the code. Great that I can't get a response from the company. If anyone thinks it would be useful to post the code I will. That is if anyone would want to see and look at it. The weird thing is searching on one table is great. The list (all records) works great. The problem starts when I insert all my joins. Now I've tested and retested the query and I know it's fine. Thank you , Stuart --- Steve Davies [EMAIL PROTECTED] wrote: What's contained in $sql and $sql_ext ??? Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
did you make the change to the code I suggested? What does MySQL say the error is? -Micah On Friday 10 September 2004 07:49 am, Stuart Felenstein wrote: As I said this is a code generator (dbqwiksite). So, describing the process for creating the code is different. The $sql is fine, as far as typos or incorrect characterrs. I've gone through those statement very carefully. I've also tried to run a debug with no luck. But I do know that there is something wrong with the statement, since I can get a complete display of all records, but can't search using criteria. Below is the $sql, one I know works , though I've tried others. VendorJobs is the table I'm querying , and since it's made up of values from other static type tables I have all the joins in place. I've also taken the tick marks out, thinking maybe it was a style issue. I'm not sure where to go with it. Getting a response from the company is like pretty difficult. `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) Stuart --- Philip Thompson [EMAIL PROTECTED] wrote: I think everyone knows that $sql is a statement. But what people are asking is: what is that statement?! Because if there are incorrect characters or 's in that statement, then that can break your code/statement. ~Philip On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote: Just getting back to this thing. There are 3 files involved, search.php, connections.php and functions.php. I searched through all and couldn't find the meaning of the $sql_ext. $sql is just the sql statement I inserted into the code. Great that I can't get a response from the company. If anyone thinks it would be useful to post the code I will. That is if anyone would want to see and look at it. The weird thing is searching on one table is great. The list (all records) works great. The problem starts when I insert all my joins. Now I've tested and retested the query and I know it's fine. Thank you , Stuart --- Steve Davies [EMAIL PROTECTED] wrote: What's contained in $sql and $sql_ext ??? Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
I had not, my apologies. I think your post slipped by. Anyway, yes I have now inserted mysql_error() and got my return. Though I'm not entirely sure how to fix it. Here is the error. You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Like 'ACCFIN'' at line 1 SELECT `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit 0,1 -- --- Micah Stevens [EMAIL PROTECTED] wrote: did you make the change to the code I suggested? What does MySQL say the error is? -Micah On Friday 10 September 2004 07:49 am, Stuart Felenstein wrote: As I said this is a code generator (dbqwiksite). So, describing the process for creating the code is different. The $sql is fine, as far as typos or incorrect characterrs. I've gone through those statement very carefully. I've also tried to run a debug with no luck. But I do know that there is something wrong with the statement, since I can get a complete display of all records, but can't search using criteria. Below is the $sql, one I know works , though I've tried others. VendorJobs is the table I'm querying , and since it's made up of values from other static type tables I have all the joins in place. I've also taken the tick marks out, thinking maybe it was a style issue. I'm not sure where to go with it. Getting a response from the company is like pretty difficult. `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) Stuart --- Philip Thompson [EMAIL PROTECTED] wrote: I think everyone knows that $sql is a statement. But what people are asking is: what is that statement?! Because if there are incorrect characters or 's in that statement, then that can break your code/statement. ~Philip On Sep 9, 2004, at 5:49 PM, Stuart Felenstein wrote: Just getting back to this thing. There are 3 files involved, search.php, connections.php and functions.php. I searched through all and couldn't find the meaning of the $sql_ext. $sql is just the sql statement I inserted into the code. Great that I can't get a response from the company. If anyone thinks it would be useful to post the code I will. That is if anyone would want to see and look at it. The weird thing is searching on one table is great. The list (all records) works great. The problem starts when I insert all my joins. Now I've tested and retested the query and I know it's fine. Thank you , Stuart --- Steve Davies [EMAIL PROTECTED] wrote: What's contained in $sql and $sql_ext ??? Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or
Re: [PHP-DB] Please help
Okay, I was under the impression that where is implied inthe joins xx.xx = xx.xx . Is that not the case ? As a matter of fact, there isn't even a where or like in my query. Then again myabe that's why I can see *all* records but not run a search. Can you please provide an exmample of the column name bteween where and like ? Thank you. Stuart --- Micah Stevens [EMAIL PROTECTED] wrote: If you look at the query, there's no column name between 'WHERE' and 'LIKE' which is a syntax error. That's the problem. Looks like your sql generator has some issues, or you didn't specify the WHERE column properly. I'm not familiar with the system you're using, but keep in mind, that mysql_error will usually not steer you wrong. -Micah On Friday 10 September 2004 10:03 am, Stuart Felenstein wrote: I had not, my apologies. I think your post slipped by. Anyway, yes I have now inserted mysql_error() and got my return. Though I'm not entirely sure how to fix it. Here is the error. You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Like 'ACCFIN'' at line 1 SELECT `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit 0,1 -- --- Micah Stevens [EMAIL PROTECTED] wrote: did you make the change to the code I suggested? What does MySQL say the error is? -Micah On Friday 10 September 2004 07:49 am, Stuart Felenstein wrote: As I said this is a code generator (dbqwiksite). So, describing the process for creating the code is different. The $sql is fine, as far as typos or incorrect characterrs. I've gone through those statement very carefully. I've also tried to run a debug with no luck. But I do know that there is something wrong with the statement, since I can get a complete display of all records, but can't search using criteria. Below is the $sql, one I know works , though I've tried others. VendorJobs is the table I'm querying , and since it's made up of values from other static type tables I have all the joins in place. I've also taken the tick marks out, thinking maybe it was a style issue. I'm not sure where to go with it. Getting a response from the company is like pretty difficult. `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) Stuart --- Philip Thompson [EMAIL PROTECTED] wrote: I think everyone knows that $sql is a statement. But what people are asking is: what is that statement?! Because if there are incorrect characters or 's in that statement, then that can break your code/statement. ~Philip On Sep 9, 2004, at 5:49 PM, Stuart Felenstein === message truncated
Fwd: Re: [PHP-DB] Please help
Forgot to CC the list.. ---BeginMessage--- there is a WHERE on the last line of the statement you sent me. Where's are in the form of: WHERE value condition value Yours is in the form: WHERE condition value You're not providing anything to compare the latter value to. You can imply conditions in the join syntax for sure, but the fact of the matter remains you are in addition specifying a syntactically incorrect WHERE statement at the end. -Micah On Friday 10 September 2004 10:22 am, you wrote: Okay, I was under the impression that where is implied inthe joins xx.xx = xx.xx . Is that not the case ? As a matter of fact, there isn't even a where or like in my query. Then again myabe that's why I can see *all* records but not run a search. Can you please provide an exmample of the column name bteween where and like ? Thank you. Stuart --- Micah Stevens [EMAIL PROTECTED] wrote: If you look at the query, there's no column name between 'WHERE' and 'LIKE' which is a syntax error. That's the problem. Looks like your sql generator has some issues, or you didn't specify the WHERE column properly. I'm not familiar with the system you're using, but keep in mind, that mysql_error will usually not steer you wrong. -Micah On Friday 10 September 2004 10:03 am, Stuart Felenstein wrote: I had not, my apologies. I think your post slipped by. Anyway, yes I have now inserted mysql_error() and got my return. Though I'm not entirely sure how to fix it. Here is the error. You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near 'Like 'ACCFIN'' at line 1 SELECT `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`) LEFT OUTER JOIN `staTaxTerm` ON (`VendorJobs`.`TaxTerm` = `staTaxTerm`.`TaxTermID`) INNER JOIN `staTravelReq` ON (`VendorJobs`.`TravelReq` = `staTravelReq`.`TravelReqID`) INNER JOIN `VendorJobDetails` ON (`VendorJobs`.`JobID` = `VendorJobDetails`.`JobID`) where Like 'ACCFIN' limit 0,1 -- --- Micah Stevens [EMAIL PROTECTED] wrote: did you make the change to the code I suggested? What does MySQL say the error is? -Micah On Friday 10 September 2004 07:49 am, Stuart Felenstein wrote: As I said this is a code generator (dbqwiksite). So, describing the process for creating the code is different. The $sql is fine, as far as typos or incorrect characterrs. I've gone through those statement very carefully. I've also tried to run a debug with no luck. But I do know that there is something wrong with the statement, since I can get a complete display of all records, but can't search using criteria. Below is the $sql, one I know works , though I've tried others. VendorJobs is the table I'm querying , and since it's made up of values from other static type tables I have all the joins in place. I've also taken the tick marks out, thinking maybe it was a style issue. I'm not sure where to go with it. Getting a response from the company is like pretty difficult. `VendorJobs`.`JobID`, `VendorSignUp`.`CompanyName`, `StaIndTypes`.`CareerCategories`, `StaUSCities`.`City`, `USStates`.`States`, `VendorJobs`.`AreaCode`, `staTaxTerm`.`TaxTerm`, `VendorJobs`.`PayRate`, `staTravelReq`.`TravelReq`, `VendorJobDetails`.`Details`, `VendorJobs`.`PostStart`, `VendorJobs`.`JobTitle` FROM `VendorJobs` INNER JOIN `VendorSignUp` ON (`VendorJobs`.`VendorID` = `VendorSignUp`.`VendorID`) INNER JOIN `StaIndTypes` ON (`VendorJobs`.`Industry` = `StaIndTypes`.`CareerIDs`) LEFT OUTER JOIN `StaUSCities` ON (`VendorJobs`.`LocationCity` = `StaUSCities`.`CityID`) LEFT OUTER JOIN `USStates` ON (`VendorJobs`.`LocationState` = `USStates`.`StateID`)
Re: [PHP-DB] Please help
I think maybe there is an implied where in the
generated code, but not in my statement. I'm saying
when I chose ACCFIN (as in the referred error
message), the join should be looking at the referred
table , value ACCFIN . e.g. Code_Table.CodeID
(ACCFIN) = MAIN_Table.CodeID (ACCFIN) return label.
Here is the code, and I honestly am not trying to
argue , just understand. Apologies if I'm belabouring
the point.
?php
require('LFW3_connection.php');
require('qs_functions.php');
@session_start();
$err_string = ;
$strkeyword = ;
$sql = ;
$sql_ext = ;
$fields = array();
$fields[0] = JobID;
$fields[1] = CompanyName;
$fields[2] = CareerCategories;
$fields[3] = City;
$fields[4] = States;
$fields[5] = AreaCode;
$fields[6] = TaxTerm;
$fields[7] = PayRate;
$fields[8] = TravelReq;
$fields[9] = Details;
$fields[10] = PostStart;
$fields[11] = JobTitle;
$arryitemvalue = array();
$arryitemvalue[0] = ;
$arryitemvalue[1] = ;
$arryitemvalue[2] = ;
$arryitemvalue[3] = ;
$arryitemvalue[4] = ;
$arryitemvalue[5] = ;
$arryitemvalue[6] = ;
$arryitemvalue[7] = ;
$arryitemvalue[8] = ;
$arryitemvalue[9] = ;
$arryitemvalue[10] = ;
$arryitemvalue[11] = ;
$arryopt = array();
$arryopt[0] = ;
$arryopt[1] = ;
$arryopt[2] = ;
$arryopt[3] = ;
$arryopt[4] = ;
$arryopt[5] = ;
$arryopt[6] = ;
$arryopt[7] = ;
$arryopt[8] = ;
$arryopt[9] = ;
$arryopt[10] = ;
$arryopt[11] = ;
$sql .= SELECT ;
$sql .=VendorJobs.JobID,;
$sql .=VendorSignUp.CompanyName,;
$sql .=StaIndTypes.CareerCategories,;
$sql .=StaUSCities.City,;
$sql .=USStates.States,;
$sql .=VendorJobs.AreaCode,;
$sql .=staTaxTerm.TaxTerm,;
$sql .=VendorJobs.PayRate,;
$sql .=staTravelReq.TravelReq,;
$sql .=VendorJobDetails.Details,;
$sql .=VendorJobs.PostStart,;
$sql .=VendorJobs.JobTitle;
$sql .= FROM;
$sql .=VendorJobs;
$sql .=INNER JOIN VendorSignUp ON
(VendorJobs.VendorID = VendorSignUp.VendorID);
$sql .=INNER JOIN StaIndTypes ON
(VendorJobs.Industry = StaIndTypes.CareerIDs);
$sql .=LEFT OUTER JOIN StaUSCities ON
(VendorJobs.LocationCity = StaUSCities.CityID);
$sql .=LEFT OUTER JOIN USStates ON
(VendorJobs.LocationState = USStates.StateID);
$sql .=LEFT OUTER JOIN staTaxTerm ON
(VendorJobs.TaxTerm = staTaxTerm.TaxTermID);
$sql .=INNER JOIN staTravelReq ON
(VendorJobs.TravelReq = staTravelReq.TravelReqID);
$sql .=INNER JOIN VendorJobDetails ON
(VendorJobs.JobID = VendorJobDetails.JobID);
$result = mysql_query($sql . . $sql_ext . limit
0,1) or
die(mysql_error().br.$sql. .$sql_ext . limit
0,1br);
if (isset($_POST[QS_Submit])) {
$filter_string = ;
$qry_string = ;
$i = 0;
while ($i mysql_num_fields($result)) {
$meta = mysql_fetch_field($result);
$field_name = $meta-name;
$field_type = $meta-type;
if ((qsrequest(search_fd . $i) != )
(qsrequest(search_fd . $i) != *)) {
$idata = qsrequest(search_fd . $i);
if (strlen($idata) 1) {
if ($idata[strlen($idata) - 1] == *)
{
$idata = substr($idata, 0,
strlen($idata) - 1);
}
}
$idata = str_replace(*, %, $idata);
$irealdata = $idata;
if (qsrequest(search_optfd.$i) != ) {
$idata = qsrequest(search_optfd. $i) .
$idata ;
}
$iopt = substr($idata, 0, 2);
if (($iopt == =) || ($iopt == =)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif (($iopt == =) || ($iopt ==
=)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ==) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ) {
$irealdata = substr($idata, 2);
} elseif ($iopt == ^^) {
$iopt = *;
$idata = $iopt . $irealdata . $iopt; //
Contain
} elseif ($iopt == ^*) {
$iopt = *;
$idata = $irealdata . $iopt; // Start
With
} elseif ($iopt == *^) {
$iopt = *;
$idata = $iopt . $irealdata ; // End
With
} else {
$iopt = substr($idata, 0, 1);
if (($iopt == ) || ($iopt == )
|| ($iopt == =)) {
$irealdata = substr($idata,1);
} else {
$iopt = =;
}
}
if (!strcasecmp($idata,{current date and
time})) {
$idata = time();
} elseif (!strcasecmp($idata,{current
date})) {
$idata = time();
} elseif (!strcasecmp($idata,{current
time})) {
$idata = time();
}
if ($meta) {
if ((strtolower($field_type) ==
timestamp)
||(strtolower($field_type) ==
datetime)
||(strtolower($field_type) ==
smalldatetime)
||(strtolower($field_type)
Re: [PHP-DB] Please help
I understand your intent, but that is not really what is happening. The code
changes I suggested did two things:
1) Echo's the error statement that mysql produces
2) Echo's the actual SQL statement that is sent to the DB
you should be looking at #2 for answers, not your intended query. This
'actual' query is what I'm referring to.
-Micah
On Friday 10 September 2004 10:46 am, Stuart Felenstein wrote:
I think maybe there is an implied where in the
generated code, but not in my statement. I'm saying
when I chose ACCFIN (as in the referred error
message), the join should be looking at the referred
table , value ACCFIN . e.g. Code_Table.CodeID
(ACCFIN) = MAIN_Table.CodeID (ACCFIN) return label.
Here is the code, and I honestly am not trying to
argue , just understand. Apologies if I'm belabouring
the point.
?php
require('LFW3_connection.php');
require('qs_functions.php');
@session_start();
$err_string = ;
$strkeyword = ;
$sql = ;
$sql_ext = ;
$fields = array();
$fields[0] = JobID;
$fields[1] = CompanyName;
$fields[2] = CareerCategories;
$fields[3] = City;
$fields[4] = States;
$fields[5] = AreaCode;
$fields[6] = TaxTerm;
$fields[7] = PayRate;
$fields[8] = TravelReq;
$fields[9] = Details;
$fields[10] = PostStart;
$fields[11] = JobTitle;
$arryitemvalue = array();
$arryitemvalue[0] = ;
$arryitemvalue[1] = ;
$arryitemvalue[2] = ;
$arryitemvalue[3] = ;
$arryitemvalue[4] = ;
$arryitemvalue[5] = ;
$arryitemvalue[6] = ;
$arryitemvalue[7] = ;
$arryitemvalue[8] = ;
$arryitemvalue[9] = ;
$arryitemvalue[10] = ;
$arryitemvalue[11] = ;
$arryopt = array();
$arryopt[0] = ;
$arryopt[1] = ;
$arryopt[2] = ;
$arryopt[3] = ;
$arryopt[4] = ;
$arryopt[5] = ;
$arryopt[6] = ;
$arryopt[7] = ;
$arryopt[8] = ;
$arryopt[9] = ;
$arryopt[10] = ;
$arryopt[11] = ;
$sql .= SELECT ;
$sql .=VendorJobs.JobID,;
$sql .=VendorSignUp.CompanyName,;
$sql .=StaIndTypes.CareerCategories,;
$sql .=StaUSCities.City,;
$sql .=USStates.States,;
$sql .=VendorJobs.AreaCode,;
$sql .=staTaxTerm.TaxTerm,;
$sql .=VendorJobs.PayRate,;
$sql .=staTravelReq.TravelReq,;
$sql .=VendorJobDetails.Details,;
$sql .=VendorJobs.PostStart,;
$sql .=VendorJobs.JobTitle;
$sql .= FROM;
$sql .=VendorJobs;
$sql .=INNER JOIN VendorSignUp ON
(VendorJobs.VendorID = VendorSignUp.VendorID);
$sql .=INNER JOIN StaIndTypes ON
(VendorJobs.Industry = StaIndTypes.CareerIDs);
$sql .=LEFT OUTER JOIN StaUSCities ON
(VendorJobs.LocationCity = StaUSCities.CityID);
$sql .=LEFT OUTER JOIN USStates ON
(VendorJobs.LocationState = USStates.StateID);
$sql .=LEFT OUTER JOIN staTaxTerm ON
(VendorJobs.TaxTerm = staTaxTerm.TaxTermID);
$sql .=INNER JOIN staTravelReq ON
(VendorJobs.TravelReq = staTravelReq.TravelReqID);
$sql .=INNER JOIN VendorJobDetails ON
(VendorJobs.JobID = VendorJobDetails.JobID);
$result = mysql_query($sql . . $sql_ext . limit
0,1) or
die(mysql_error().br.$sql. .$sql_ext . limit
0,1br);
if (isset($_POST[QS_Submit])) {
$filter_string = ;
$qry_string = ;
$i = 0;
while ($i mysql_num_fields($result)) {
$meta = mysql_fetch_field($result);
$field_name = $meta-name;
$field_type = $meta-type;
if ((qsrequest(search_fd . $i) != )
(qsrequest(search_fd . $i) != *)) {
$idata = qsrequest(search_fd . $i);
if (strlen($idata) 1) {
if ($idata[strlen($idata) - 1] == *)
{
$idata = substr($idata, 0,
strlen($idata) - 1);
}
}
$idata = str_replace(*, %, $idata);
$irealdata = $idata;
if (qsrequest(search_optfd.$i) != ) {
$idata = qsrequest(search_optfd. $i) .
$idata ;
}
$iopt = substr($idata, 0, 2);
if (($iopt == =) || ($iopt == =)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif (($iopt == =) || ($iopt ==
=)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ==) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ) {
$irealdata = substr($idata, 2);
} elseif ($iopt == ^^) {
$iopt = *;
$idata = $iopt . $irealdata . $iopt; //
Contain
} elseif ($iopt == ^*) {
$iopt = *;
$idata = $irealdata . $iopt; // Start
With
} elseif ($iopt == *^) {
$iopt = *;
$idata = $iopt . $irealdata ; // End
With
} else {
$iopt = substr($idata, 0, 1);
if (($iopt == ) || ($iopt == )
|| ($iopt == =)) {
$irealdata = substr($idata,1);
} else {
$iopt
Re: [PHP-DB] Please help
Now that I'm not getting an invalid error message any
longer, how can I get it to echo the sql statement ?
Stuart
--- Micah Stevens [EMAIL PROTECTED] wrote:
I understand your intent, but that is not really
what is happening. The code
changes I suggested did two things:
1) Echo's the error statement that mysql produces
2) Echo's the actual SQL statement that is sent to
the DB
you should be looking at #2 for answers, not your
intended query. This
'actual' query is what I'm referring to.
-Micah
On Friday 10 September 2004 10:46 am, Stuart
Felenstein wrote:
I think maybe there is an implied where in the
generated code, but not in my statement. I'm
saying
when I chose ACCFIN (as in the referred error
message), the join should be looking at the
referred
table , value ACCFIN . e.g. Code_Table.CodeID
(ACCFIN) = MAIN_Table.CodeID (ACCFIN) return
label.
Here is the code, and I honestly am not trying to
argue , just understand. Apologies if I'm
belabouring
the point.
?php
require('LFW3_connection.php');
require('qs_functions.php');
@session_start();
$err_string = ;
$strkeyword = ;
$sql = ;
$sql_ext = ;
$fields = array();
$fields[0] = JobID;
$fields[1] = CompanyName;
$fields[2] = CareerCategories;
$fields[3] = City;
$fields[4] = States;
$fields[5] = AreaCode;
$fields[6] = TaxTerm;
$fields[7] = PayRate;
$fields[8] = TravelReq;
$fields[9] = Details;
$fields[10] = PostStart;
$fields[11] = JobTitle;
$arryitemvalue = array();
$arryitemvalue[0] = ;
$arryitemvalue[1] = ;
$arryitemvalue[2] = ;
$arryitemvalue[3] = ;
$arryitemvalue[4] = ;
$arryitemvalue[5] = ;
$arryitemvalue[6] = ;
$arryitemvalue[7] = ;
$arryitemvalue[8] = ;
$arryitemvalue[9] = ;
$arryitemvalue[10] = ;
$arryitemvalue[11] = ;
$arryopt = array();
$arryopt[0] = ;
$arryopt[1] = ;
$arryopt[2] = ;
$arryopt[3] = ;
$arryopt[4] = ;
$arryopt[5] = ;
$arryopt[6] = ;
$arryopt[7] = ;
$arryopt[8] = ;
$arryopt[9] = ;
$arryopt[10] = ;
$arryopt[11] = ;
$sql .= SELECT ;
$sql .=VendorJobs.JobID,;
$sql .=VendorSignUp.CompanyName,;
$sql .=StaIndTypes.CareerCategories,;
$sql .=StaUSCities.City,;
$sql .=USStates.States,;
$sql .=VendorJobs.AreaCode,;
$sql .=staTaxTerm.TaxTerm,;
$sql .=VendorJobs.PayRate,;
$sql .=staTravelReq.TravelReq,;
$sql .=VendorJobDetails.Details,;
$sql .=VendorJobs.PostStart,;
$sql .=VendorJobs.JobTitle;
$sql .= FROM;
$sql .=VendorJobs;
$sql .=INNER JOIN VendorSignUp ON
(VendorJobs.VendorID = VendorSignUp.VendorID);
$sql .=INNER JOIN StaIndTypes ON
(VendorJobs.Industry = StaIndTypes.CareerIDs);
$sql .=LEFT OUTER JOIN StaUSCities ON
(VendorJobs.LocationCity = StaUSCities.CityID);
$sql .=LEFT OUTER JOIN USStates ON
(VendorJobs.LocationState = USStates.StateID);
$sql .=LEFT OUTER JOIN staTaxTerm ON
(VendorJobs.TaxTerm = staTaxTerm.TaxTermID);
$sql .=INNER JOIN staTravelReq ON
(VendorJobs.TravelReq =
staTravelReq.TravelReqID);
$sql .=INNER JOIN VendorJobDetails ON
(VendorJobs.JobID = VendorJobDetails.JobID);
$result = mysql_query($sql . . $sql_ext .
limit
0,1) or
die(mysql_error().br.$sql. .$sql_ext .
limit
0,1br);
if (isset($_POST[QS_Submit])) {
$filter_string = ;
$qry_string = ;
$i = 0;
while ($i mysql_num_fields($result)) {
$meta = mysql_fetch_field($result);
$field_name = $meta-name;
$field_type = $meta-type;
if ((qsrequest(search_fd . $i) != )
(qsrequest(search_fd . $i) != *)) {
$idata = qsrequest(search_fd . $i);
if (strlen($idata) 1) {
if ($idata[strlen($idata) - 1] ==
*)
{
$idata = substr($idata, 0,
strlen($idata) - 1);
}
}
$idata = str_replace(*, %,
$idata);
$irealdata = $idata;
if (qsrequest(search_optfd.$i) !=
) {
$idata = qsrequest(search_optfd.
$i) .
$idata ;
}
$iopt = substr($idata, 0, 2);
if (($iopt == =) || ($iopt ==
=)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif (($iopt == =) || ($iopt ==
=)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ==) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ) {
$irealdata = substr($idata, 2);
} elseif ($iopt == ^^) {
$iopt = *;
$idata = $iopt . $irealdata . $iopt;
//
Contain
} elseif ($iopt == ^*) {
$iopt = *;
$idata = $irealdata . $iopt; //
Start
With
} elseif ($iopt ==
Re: [PHP-DB] Please help
echo $sql. .$sql_ext;
Read the docs:
http://www.php.net/echo
http://www.php.net/mysql_query
On Friday 10 September 2004 01:18 pm, you wrote:
Now that I'm not getting an invalid error message any
longer, how can I get it to echo the sql statement ?
Stuart
--- Micah Stevens [EMAIL PROTECTED] wrote:
I understand your intent, but that is not really
what is happening. The code
changes I suggested did two things:
1) Echo's the error statement that mysql produces
2) Echo's the actual SQL statement that is sent to
the DB
you should be looking at #2 for answers, not your
intended query. This
'actual' query is what I'm referring to.
-Micah
On Friday 10 September 2004 10:46 am, Stuart
Felenstein wrote:
I think maybe there is an implied where in the
generated code, but not in my statement. I'm
saying
when I chose ACCFIN (as in the referred error
message), the join should be looking at the
referred
table , value ACCFIN . e.g. Code_Table.CodeID
(ACCFIN) = MAIN_Table.CodeID (ACCFIN) return
label.
Here is the code, and I honestly am not trying to
argue , just understand. Apologies if I'm
belabouring
the point.
?php
require('LFW3_connection.php');
require('qs_functions.php');
@session_start();
$err_string = ;
$strkeyword = ;
$sql = ;
$sql_ext = ;
$fields = array();
$fields[0] = JobID;
$fields[1] = CompanyName;
$fields[2] = CareerCategories;
$fields[3] = City;
$fields[4] = States;
$fields[5] = AreaCode;
$fields[6] = TaxTerm;
$fields[7] = PayRate;
$fields[8] = TravelReq;
$fields[9] = Details;
$fields[10] = PostStart;
$fields[11] = JobTitle;
$arryitemvalue = array();
$arryitemvalue[0] = ;
$arryitemvalue[1] = ;
$arryitemvalue[2] = ;
$arryitemvalue[3] = ;
$arryitemvalue[4] = ;
$arryitemvalue[5] = ;
$arryitemvalue[6] = ;
$arryitemvalue[7] = ;
$arryitemvalue[8] = ;
$arryitemvalue[9] = ;
$arryitemvalue[10] = ;
$arryitemvalue[11] = ;
$arryopt = array();
$arryopt[0] = ;
$arryopt[1] = ;
$arryopt[2] = ;
$arryopt[3] = ;
$arryopt[4] = ;
$arryopt[5] = ;
$arryopt[6] = ;
$arryopt[7] = ;
$arryopt[8] = ;
$arryopt[9] = ;
$arryopt[10] = ;
$arryopt[11] = ;
$sql .= SELECT ;
$sql .=VendorJobs.JobID,;
$sql .=VendorSignUp.CompanyName,;
$sql .=StaIndTypes.CareerCategories,;
$sql .=StaUSCities.City,;
$sql .=USStates.States,;
$sql .=VendorJobs.AreaCode,;
$sql .=staTaxTerm.TaxTerm,;
$sql .=VendorJobs.PayRate,;
$sql .=staTravelReq.TravelReq,;
$sql .=VendorJobDetails.Details,;
$sql .=VendorJobs.PostStart,;
$sql .=VendorJobs.JobTitle;
$sql .= FROM;
$sql .=VendorJobs;
$sql .=INNER JOIN VendorSignUp ON
(VendorJobs.VendorID = VendorSignUp.VendorID);
$sql .=INNER JOIN StaIndTypes ON
(VendorJobs.Industry = StaIndTypes.CareerIDs);
$sql .=LEFT OUTER JOIN StaUSCities ON
(VendorJobs.LocationCity = StaUSCities.CityID);
$sql .=LEFT OUTER JOIN USStates ON
(VendorJobs.LocationState = USStates.StateID);
$sql .=LEFT OUTER JOIN staTaxTerm ON
(VendorJobs.TaxTerm = staTaxTerm.TaxTermID);
$sql .=INNER JOIN staTravelReq ON
(VendorJobs.TravelReq =
staTravelReq.TravelReqID);
$sql .=INNER JOIN VendorJobDetails ON
(VendorJobs.JobID = VendorJobDetails.JobID);
$result = mysql_query($sql . . $sql_ext .
limit
0,1) or
die(mysql_error().br.$sql. .$sql_ext .
limit
0,1br);
if (isset($_POST[QS_Submit])) {
$filter_string = ;
$qry_string = ;
$i = 0;
while ($i mysql_num_fields($result)) {
$meta = mysql_fetch_field($result);
$field_name = $meta-name;
$field_type = $meta-type;
if ((qsrequest(search_fd . $i) != )
(qsrequest(search_fd . $i) != *)) {
$idata = qsrequest(search_fd . $i);
if (strlen($idata) 1) {
if ($idata[strlen($idata) - 1] ==
*)
{
$idata = substr($idata, 0,
strlen($idata) - 1);
}
}
$idata = str_replace(*, %,
$idata);
$irealdata = $idata;
if (qsrequest(search_optfd.$i) !=
) {
$idata = qsrequest(search_optfd.
$i) .
$idata ;
}
$iopt = substr($idata, 0, 2);
if (($iopt == =) || ($iopt ==
=)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif (($iopt == =) || ($iopt ==
=)) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ==) {
$iopt = =;
$irealdata = substr($idata, 2);
} elseif ($iopt == ) {
$irealdata =
Re: [PHP-DB] Please help
Just getting back to this thing. There are 3 files involved, search.php, connections.php and functions.php. I searched through all and couldn't find the meaning of the $sql_ext. $sql is just the sql statement I inserted into the code. Great that I can't get a response from the company. If anyone thinks it would be useful to post the code I will. That is if anyone would want to see and look at it. The weird thing is searching on one table is great. The list (all records) works great. The problem starts when I insert all my joins. Now I've tested and retested the query and I know it's fine. Thank you , Stuart --- Steve Davies [EMAIL PROTECTED] wrote: What's contained in $sql and $sql_ext ??? Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
What are the variables $sql and $sql_ext set to? The query format is fine. It's probably in your variables. -- Peter Ellis - [EMAIL PROTECTED] Web Design and Development Consultant naturalaxis | http://www.naturalaxis.com/ On Wed, 2004-09-08 at 12:29 -0700, Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help
HI try $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query.mysql_error()); which will give more detail on the error. HTH Peter -Original Message- From: Stuart Felenstein [mailto:[EMAIL PROTECTED] Sent: 08 September 2004 20:29 To: [EMAIL PROTECTED] Subject: [PHP-DB] Please help I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
Not sure what the varialbe are set too. Going through the code and rebuilding the query. Post on hold to further notice :). Stuart --- Peter Ellis [EMAIL PROTECTED] wrote: What are the variables $sql and $sql_ext set to? The query format is fine. It's probably in your variables. -- Peter Ellis - [EMAIL PROTECTED] Web Design and Development Consultant naturalaxis | http://www.naturalaxis.com/ On Wed, 2004-09-08 at 12:29 -0700, Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
Change the code to this: $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(mysql_error().br.$sql. .$sql_ext . limit 0,1br); And you'll get the error from mysql and a copy of the actual query. Without that, it's pretty tough to help. -Micah On Wednesday 08 September 2004 12:29 pm, Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help
What exactly is in the $sql and $sql_ext?? Try to put it into a variable so you can print out what you're sending to the DB Such as: $query = $sql . . $sql_ext . limit 0,1; Echo $query . BR; If your $sql and $sql_ext combine to make a valid sql query, the is just putting a space inside the query, and it is valid Gary Every Sr. UNIX Administrator Ingram Entertainment Inc. 2 Ingram Blvd, La Vergne, TN 37089 Pay It Forward! -Original Message- From: Stuart Felenstein [mailto:[EMAIL PROTECTED] Sent: Wednesday, September 08, 2004 2:29 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Please help I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help
What's contained in $sql and $sql_ext ??? Stuart Felenstein wrote: I'm using a product called dbqwiksite pro. PHP generator for PHP - MySQL The code seems to be working fine except in my search page where I receive an invalid query $result = mysql_query($sql . . $sql_ext . limit 0,1) or die(Invalid query); This is the place I where the code is taking the die path. I'm guessing something belongs between the quotation marks , just not sure. Anyone ? Thank you Stuart -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help - Managing relational tables using PHP
You need to realize that phpMyAdmin is a database manangement tool, not a database. So I'll answer your questions based on the mySQL database ... 1. MySQL 3.23.43b introduced a table type of InnoDB which is the first to allow foreign key constraints. http://www.mysql.com/doc/en/SEC448.html ... However, you do not need to enforce constraints to have a relationship. I just assume that's what you're getting at. 2. Yes, if you enforce constraints with the CASCADE option it will. ..see same document: http://www.mysql.com/doc/en/SEC448.html ... Now the catch, my version of phpMyAdmin 2.3.0 does not allow me to choose the table type of InnoDB, even though my database is version 3.23.54. So I assume phpMyAdmin does not allow one to manage these attributes. Someone please correct me if I'm wrong. Ryan -Original Message- From: Phanivas Vemuri [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 02, 2003 12:50 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] Please help - Managing relational tables using PHP hi all, 1. How does phpMyAdmin manage tables with one to many or many to one relation ships. Is there any documentation about this. --or- 2. Does phpMyAdmin support editing data from more than one table that have a one to many or many to one relationship. Thanks and Wish you a happy new year, Phani. _ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help - Managing relational tables using PHP
Dear Mr. Ryan, May be I did not present my question clearly to you. My question in detail is ... -- Does phpMyAdmin supports / allow / interface, editing data from one to many or ,many to many related tables, using a single phpMyAdmin interface( a single HTML page) I have a (primary) table and one of its fields(category) has one or many values , all these values are also listed in another(secondary) table. I would like to know if phpMyAdmin provides any interface such that those values from the secondary table appears(listed) (for example : as a Form) in the field of the primary table. PrimaryTable(X,Y,category,Z) pk:X SecondaryTable(category)pk:category I am not intending to enforce referential integrity. My question is just about the interfacing. Thank you, Phani. From: Ryan Jameson (USA) [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Please help - Managing relational tables using PHP Date: Thu, 2 Jan 2003 13:32:15 -0700 You need to realize that phpMyAdmin is a database manangement tool, not a database. So I'll answer your questions based on the mySQL database ... 1. MySQL 3.23.43b introduced a table type of InnoDB which is the first to allow foreign key constraints. http://www.mysql.com/doc/en/SEC448.html ... However, you do not need to enforce constraints to have a relationship. I just assume that's what you're getting at. 2. Yes, if you enforce constraints with the CASCADE option it will. ..see same document: http://www.mysql.com/doc/en/SEC448.html ... Now the catch, my version of phpMyAdmin 2.3.0 does not allow me to choose the table type of InnoDB, even though my database is version 3.23.54. So I assume phpMyAdmin does not allow one to manage these attributes. Someone please correct me if I'm wrong. Ryan -Original Message- From: Phanivas Vemuri [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 02, 2003 12:50 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] Please help - Managing relational tables using PHP hi all, 1. How does phpMyAdmin manage tables with one to many or many to one relation ships. Is there any documentation about this. --or- 2. Does phpMyAdmin support editing data from more than one table that have a one to many or many to one relationship. Thanks and Wish you a happy new year, Phani. _ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php _ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help - Managing relational tables using PHP
Well, you can use the SQL tab to write a sql statement to do whatever you want, but no, the graphical interface doesn't appear to do that. Phanivas Vemuri [EMAIL PROTECTED]To: [EMAIL PROTECTED], [EMAIL PROTECTED] m cc: Subject: RE: [PHP-DB] Please help - Managing relational tables using PHP 01/02/03 04:34 PM Dear Mr. Ryan, May be I did not present my question clearly to you. My question in detail is ... -- Does phpMyAdmin supports / allow / interface, editing data from one to many or ,many to many related tables, using a single phpMyAdmin interface( a single HTML page) I have a (primary) table and one of its fields(category) has one or many values , all these values are also listed in another(secondary) table. I would like to know if phpMyAdmin provides any interface such that those values from the secondary table appears(listed) (for example : as a Form) in the field of the primary table. PrimaryTable(X,Y,category,Z) pk:X SecondaryTable(category)pk:category I am not intending to enforce referential integrity. My question is just about the interfacing. Thank you, Phani. From: Ryan Jameson (USA) [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Please help - Managing relational tables using PHP Date: Thu, 2 Jan 2003 13:32:15 -0700 You need to realize that phpMyAdmin is a database manangement tool, not a database. So I'll answer your questions based on the mySQL database ... 1. MySQL 3.23.43b introduced a table type of InnoDB which is the first to allow foreign key constraints. http://www.mysql.com/doc/en/SEC448.html ... However, you do not need to enforce constraints to have a relationship. I just assume that's what you're getting at. 2. Yes, if you enforce constraints with the CASCADE option it will. ..see same document: http://www.mysql.com/doc/en/SEC448.html ... Now the catch, my version of phpMyAdmin 2.3.0 does not allow me to choose the table type of InnoDB, even though my database is version 3.23.54. So I assume phpMyAdmin does not allow one to manage these attributes. Someone please correct me if I'm wrong. Ryan -Original Message- From: Phanivas Vemuri [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 02, 2003 12:50 PM To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: [PHP-DB] Please help - Managing relational tables using PHP hi all, 1. How does phpMyAdmin manage tables with one to many or many to one relation ships. Is there any documentation about this. --or- 2. Does phpMyAdmin support editing data from more than one table that have a one to many or many to one relationship. Thanks and Wish you a happy new year, Phani. _ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php _ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Niles Audio Corporation This mail is confidential -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] please help!
You're not evaluating the variable correctly...
Instead of...
$site = .row['site'].;
It should be...
$site = $row[site];
Or
$site = $row['site']
I'm not sure why you have leading and trailing
periods, either. Looks like you were trying to
do string appends, but they're not being used
correctly
'Luck
-Szii
- Original Message -
From: Evans, Josh [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, April 23, 2002 11:26 AM
Subject: [PHP-DB] please help!
Database: sites
I have to columns: id and site. I will have approx 800 entries in the db
when I am finished and what I want it to do and start with 1st entry and
ping the site and display in a table on the screen if it was able to ping
site or not.
This is what I have so far. My problem is I am not sure how to make it
select a entry, perform the ping, record the results, and then continue to
the next entry.
?
$db = mysql_connect(localhost,,) or die(Problem connecting);
mysql_select_db(josh) or die(Problem selecting database);
$query = SELECT * FROM josh.sites;
$result = mysql_query($query) or die (Query failed);
$numofrows = mysql_num_rows($result);
for($i = 0; $i $numofrows; $i++) {
$row = mysql_fetch_array($result);
if($i % 2) { //this means if there is a remainder
$site = .row['site'].; its not pulling a site from the
database?
$host1 = `ping -n 4 $site`;
if (eregi(reply, $host1)) {
echo $site is up!;
} else {
echo ($site is down!);
}
echo (p);
?
Please Help!
Josh Evans
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Re: [PHP-DB] Please help count ?
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
you can do this...
$myvar = ecplode(' ', $postcode);
$firsthalf = $myvar[0];
But not everyone puts the space in there
S
On Thursday 04 April 2002 5:29 pm, Dave Carrera wrote:
Hi All
I have a variable returned by my application.
For clarity it is a post code so result could be
EX3 T56 or BG56 G67 or CA2 123
But I only need the first bit of the postcode to continue my search.
How do I set my var to only include the first bit of the postcode.
It is returned in the format shown, so my question might be how I read
only the first bit before the space.
I fully appreciate any kind of help with this and as always thank you in
advance for any help.
Dave Carrera
Php Developer
http://davecarrera.freelancers.net
http://www.davecarrera.com
- --
Shane
-BEGIN PGP SIGNATURE-
Version: GnuPG v1.0.6 (GNU/Linux)
Comment: For info see http://www.gnupg.org
iD8DBQE8rH/75DXg6dCMBrQRAlEUAJ9R/jfY0Ufdj5yCSiTRw7qB81B3TwCeJELP
GW03u2HKvFXy9fDEEdV1lfg=
=xALC
-END PGP SIGNATURE-
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RE: [PHP-DB] Please help count ?
$post = ereg_replace((.*) ,\\1, $old_post); If you want to search mysql for this that is something else -Original Message- From: Dave Carrera [mailto:[EMAIL PROTECTED]] Sent: Thursday, April 04, 2002 10:29 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Please help count ? Hi All I have a variable returned by my application. For clarity it is a post code so result could be EX3 T56 or BG56 G67 or CA2 123 But I only need the first bit of the postcode to continue my search. How do I set my var to only include the first bit of the postcode. It is returned in the format shown, so my question might be how I read only the first bit before the space. I fully appreciate any kind of help with this and as always thank you in advance for any help. Dave Carrera Php Developer http://davecarrera.freelancers.net http://www.davecarrera.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help count ?
Dave, Try: $postcode_parts = explode( , $postcode); The first part of the postcode will be available in $postcode_parts[0]. - Chris -Original Message- From: Dave Carrera [SMTP:[EMAIL PROTECTED]] Sent: Thursday, April 04, 2002 5:29 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Please help count ? Hi All I have a variable returned by my application. For clarity it is a post code so result could be EX3 T56 or BG56 G67 or CA2 123 But I only need the first bit of the postcode to continue my search. How do I set my var to only include the first bit of the postcode. It is returned in the format shown, so my question might be how I read only the first bit before the space. I fully appreciate any kind of help with this and as always thank you in advance for any help. Dave Carrera Php Developer http://davecarrera.freelancers.net http://www.davecarrera.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Please help me! :(
Can I ask why you're not using autoincrement for your
id field? If you used this you would never have the problem
your having.
Everytime a record is added it will increment the id field
by one, so you don't have to use all the code your using.
Howard
-Original Message-
From: Leif K-Brooks [mailto:[EMAIL PROTECTED]]
Sent: Sunday, 24 March 2002 3:42 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Please help me! :(
I have a site where things in the database have ids. When something new is
added, it gets an id one higher than the highest existing id. I use code
something like this:
$gethighestid = mysql_fetch_array(mysql_query(select id from table order by
id desc limit 1));
$tobeid = $gethighestid[id]+1;
mysql_query(insert into table(id,othercolumn,othercolumn2)
values('$tobeid','something','something'));
The thing is, I just got two rows with duplicate ids. Aparantley, two
people must of added two things at just the right times to make the same id.
Is there any way that will reduce, or eliminate, the time gap between
getting id and inserting?
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Re: [PHP-DB] PLEASE HELP !!!!!!!
Manual reference: http://www.php.net/manual/en/function.mysql-connect.php resource mysql_connect ( [string server [, string username [, string password [, bool new_link) Seeing it is not a web page you're connecting to, try removing the http://;. localhost has particular meaning within the IP conventions - if cgi.xyz.com is remote then only use that server's name and remove the 'local' designation, ie use the name from your computer's perspective, not the remote computer's perspective. Use PING to ascertain that you have the server's name expressed correctly. The default values are server = 'localhost:3306' which may be causing some confusion here. I recommend simplifying by leaving out the port number (3306) for now - MySQL knows that (hopefully) and thus the remote computer's RDBMS should be 'listening' on that port - if it's not, we'll have to get downer-and-dirtier... Let us know how you get on! =dn - Original Message - From: Beau Lebens [EMAIL PROTECTED] To: 'Inter-Media Webmaster' [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: 14 March 2002 00:21 Subject: RE: [PHP-DB] PLEASE HELP !!! try an IP // -Original Message- // From: Inter-Media Webmaster [mailto:[EMAIL PROTECTED]] // Sent: Thursday, 14 March 2002 6:00 AM // To: [EMAIL PROTECTED] // Subject: [PHP-DB] PLEASE HELP !!! // // // PLEASE HELP !!! // HOW I CAN GET CONNETCTION TO REMOTE HOST IN OTHER SERVER WITH // mysql_connect(); // e.g. // my php file is in www.abc.com/index.php // mysql server is localhost on cgi.xyz.com // // can i do something like this // // mysql_connect(http://cgi.xyz.com:localhost:3306;, username, // password); // // PLEASE ANSWER to [EMAIL PROTECTED] // // // // // // -- // PHP Database Mailing List (http://www.php.net/) // To unsubscribe, visit: http://www.php.net/unsub.php // -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Please help, On phpnuke and phpgroupware.
On Thursday 10 January 2002 21:28, HiM wrote: I am sorry that last message haven't give the URL...:( http://nori.dns2go.com:8080/phpnuke/html/admin.php http://nori.dns2go.com:8080/phpgroupware/setup/config.php I was trying to use to phpnuke system and also the phpgroupware. I have followed the instruction listed in INSTALL step by step. But I don't know why there's so many error even in phpnuke / phpgroupware I am using Windows2000 server and apache, php 4.1.1 and mysql 3.23 The recommended php.ini settings for php 4.1.1 are more secure than previous versions BUT will break a lot of the existing PHP applications. Either change the php.ini settings to match those of an earlier version of PHP (eg 4.0.6) or wait until the author of the apps in question upgrade them to make it compatible with 4.1.1(!) -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* I don't want people to love me. It makes for obligations. -- Jean Anouilh */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Please help, On phpnuke and phpgroupware.
Jason, But after downgrade to 4.0.6, is there any new feature I can't use? Nori Chan - Original Message - From: Jason Wong [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, January 10, 2002 9:41 PM Subject: Re: [PHP-DB] Please help, On phpnuke and phpgroupware. On Thursday 10 January 2002 21:28, HiM wrote: I am sorry that last message haven't give the URL...:( http://nori.dns2go.com:8080/phpnuke/html/admin.php http://nori.dns2go.com:8080/phpgroupware/setup/config.php I was trying to use to phpnuke system and also the phpgroupware. I have followed the instruction listed in INSTALL step by step. But I don't know why there's so many error even in phpnuke / phpgroupware I am using Windows2000 server and apache, php 4.1.1 and mysql 3.23 The recommended php.ini settings for php 4.1.1 are more secure than previous versions BUT will break a lot of the existing PHP applications. Either change the php.ini settings to match those of an earlier version of PHP (eg 4.0.6) or wait until the author of the apps in question upgrade them to make it compatible with 4.1.1(!) -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* I don't want people to love me. It makes for obligations. -- Jean Anouilh */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Please help, On phpnuke and phpgroupware.
On Thursday 10 January 2002 21:53, HiM wrote: Jason, But after downgrade to 4.0.6, is there any new feature I can't use? Nori Chan If you're going to be using only PHPNuke and/or PHPGroupware and they haven't been updated for 4.1.1 then they won't be using any of the new features of 4.1.1. Read the changelog for 4.1.1 to see what's added and fixed. -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* Let's not complicate our relationship by trying to communicate with each other. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Please help, On phpnuke and phpgroupware.
Jason, phpNuke problem fixed after downgrade to 4.0.6 Thanks ! And I am still solving to problem with phpgroupware. Nori - - Original Message - From: Jason Wong [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, January 10, 2002 10:07 PM Subject: Re: [PHP-DB] Please help, On phpnuke and phpgroupware. On Thursday 10 January 2002 21:53, HiM wrote: Jason, But after downgrade to 4.0.6, is there any new feature I can't use? Nori Chan If you're going to be using only PHPNuke and/or PHPGroupware and they haven't been updated for 4.1.1 then they won't be using any of the new features of 4.1.1. Read the changelog for 4.1.1 to see what's added and fixed. -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* Let's not complicate our relationship by trying to communicate with each other. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Please help, On phpnuke and phpgroupware.
On Thursday 10 January 2002 22:15, HiM wrote: Jason, phpNuke problem fixed after downgrade to 4.0.6 Glad to hear. Thanks ! You're welcome :) -- Jason Wong - Gremlins Associates - www.gremlins.com.hk /* Falling in love is a lot like dying. You never get to do it enough to become good at it. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Please help asap
In the printf() line after 5 6 7 you try to print %s 24 times but only provide 10
values. That may have something to do with it...
Bogdan
Jeff Moncrieff wrote:
Hello
I am trying make a script fatch my data form my mysql database. I use
this script but when execute it is says
Warning: printf(): too few arguments in
/home/httpd/html/larken/database.php on line 47
any hints
Thanks Jeff
html
body
?php
$db = mysql_connect(, root);
mysql_select_db(larken,$db);
$result = mysql_query(SELECT * FROM custgeninfo,$db);
echo table border=1\n;
echo trtdName of contact/tdtdName of company/tdtd
Address/tdtdPostal Code/tdtdFormat Mail/tdtdEmail
Addresstdmodel/tdtdComment/tdtdtelephone/tdtdNotes/tdtdCounty/tdtd
ID /td/tr\n;
while ($myrow = mysql_fetch_row($result)) {
// 1 2 34
5 6 7
printf(trtd%s%s/tdtd%s%s/tdtd%s%s/tdtd%s%s%s/tdtd%s%s/tdtd%s/tdtd%s%s%s/tdtd%s/tdtd%s%s/tdtd%s/tdtd%s%s%s%s%s/td/tr\n,$myrow[1],$myrow[2],$myrow[3],$myrow[4],$myrow[5],$myrow[6],$myrow[7],$myrow[8],$myrow[9],$myrow[10]);
}
echo /table
?
/body
/html
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Re: [PHP-DB] Please Help Me! thanks!!
Try using fsockopen() function : $file = fsockopen (www.yahoo.com, 80, $err_num, $err_msg, 30); You can find more informations about this fuction in http://www.php.net/manual/en/function.fsockopen.php May it help you. Andy www.mediahostnet.com --- G [EMAIL PROTECTED] wrote: Hello! i would like to useing fopen() fuction to open the other Url, but i am using a shared server now. When i useing fopen() or file() fuctions, always have a error message: Warning: php_hostconnect: connect failed in /usr/xxx/xxx/xxx/html/test.php on line 3 Warning: file(http://www.yahoo.com;) - Bad file descriptor in /usr/xxx/xxx/xxx/html/test.php on line 3 How can i do?have any other fuctions can do that? Thank You Very Much!! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] __ Do You Yahoo!? Everything you'll ever need on one web page from News and Sport to Email and Music Charts http://uk.my.yahoo.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP-DB] Please help!
use that counter variable. check if its equal to 2 before incrementing, if
it is then BR and reset to zero, otherwise just increment.
if ($counter==2) {
echo BR;
$counter=0;
} else {
$counter++;
}
marios
- Original Message - Original Message -
From: Matt C [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, August 22, 2001 9:37 PM
Subject: [PHP-DB] Please help!
Thats my code!
What I want to be able to do is have it so instead of one column there are
two columns of pictures:
pic1 | pic2
pic3 | pic4
etc..
How can I do this?
Thanks in advance!
?php
$mysql_access = mysql_connect(, **, *);
mysql_select_db(*, $mysql_access);
$query = SELECT * FROM Pictures;
$result = mysql_query($query, $mysql_access);
if(mysql_num_rows($result)) {
$counter = 0;
while($row = mysql_fetch_row($result)) {
printf(a href=\http://www.charmed-guide.com/pictures/$row[2].jpg\;
target=_blank\n);
printf(img src=\http://www.charmed-guide.com/pictures/$row[2]_th.jpg\;
border=0\n);
printf(/ap\n);
$counter++;
}
}
?
_
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Re: [PHP-DB] Please help!
I realize this is trivial... but if you're using things like the code
below alot it can clean things up a bit by using:
if (($counter % 2) == 0 ($counter != 0))
echo BR;
$counter++;
echo img src=whatever...\n;
Just reads easier... less branching, less code. In an ideal world it'd
optimize into machine code better than the below snippet too -- but that's
get anal retentive about the situtation :)
Justin Buist
Trident Technology, Inc.
4700 60th St. SW, Suite 102
Grand Rapids, MI 49512
Ph. 616.554.2700
Fx. 616.554.3331
Mo. 616.291.2612
On Wed, 22 Aug 2001, Marios Moutzouris wrote:
use that counter variable. check if its equal to 2 before incrementing, if
it is then BR and reset to zero, otherwise just increment.
if ($counter==2) {
echo BR;
$counter=0;
} else {
$counter++;
}
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Re: [PHP-DB] Please help!
you do not need a counter, just use a 2 column table
| Thats my code!
|
| What I want to be able to do is have it so instead of one column there are
| two columns of pictures:
|
|
| pic1 | pic2
| pic3 | pic4
|
| etc..
|
|
| How can I do this?
|
| Thanks in advance!
|
|
|
|
| ?php
|
| $mysql_access = mysql_connect(, **, *);
| mysql_select_db(*, $mysql_access);
|
| $query = SELECT * FROM Pictures;
| $result = mysql_query($query, $mysql_access);
|
| if(mysql_num_rows($result)) {
| $counter = 0;
| while($row = mysql_fetch_row($result)) {
|
| printf(a href=\http://www.charmed-guide.com/pictures/$row[2].jpg\;
| target=_blank\n);
| printf(img src=\http://www.charmed-guide.com/pictures/$row[2]_th.jpg\;
| border=0\n);
| printf(/ap\n);
|
| $counter++;
| }
|
| }
| ?
|
| _
| Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp
|
|
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|
|
[EMAIL PROTECTED]
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Plans from $17.95 www.yourname.com
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RE: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
It might help us to see exactly what the query is...
Could you insert an echo statement and let us know what SQL query it's
producing?
echo
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
;
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 1:03 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I am using an Access DBASE and trying to insert text from a form field.
When I use quotes in my text the code errors out, otherwise it works
fine. I have Magic Quotes turned on which is supposed to fix this
problem but IT DOES NOT!
I have posted in several forums but no one has been able to help me!
Here is my code:
--
$NowDate = date(m/d/y);
$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
);
Here is the error when using quotes in the NewsText form field:
Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
error (missing operator) in query expression ''can\'t use quotes!')'.,
SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
line 22
Thanks in advance...
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Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
Here is the echo output.
UPDATE News SET NewsTitle = 'Text',
NewsText = 'can\'t use quotes!' WHERE NewsID = 16
Matthew Loff [EMAIL PROTECTED] wrote in message
000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
It might help us to see exactly what the query is...
Could you insert an echo statement and let us know what SQL query it's
producing?
echo
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
;
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 1:03 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I am using an Access DBASE and trying to insert text from a form field.
When I use quotes in my text the code errors out, otherwise it works
fine. I have Magic Quotes turned on which is supposed to fix this
problem but IT DOES NOT!
I have posted in several forums but no one has been able to help me!
Here is my code:
--
$NowDate = date(m/d/y);
$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
);
Here is the error when using quotes in the NewsText form field:
Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
error (missing operator) in query expression ''can\'t use quotes!')'.,
SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
line 22
Thanks in advance...
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RE: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
That looks like a valid query to me... I know this would break SQL92
standard, but have you tried enclosing the values in double quotes? I
know that's valid with MySQL...
I don't have any other suggestions... Does anyone else?
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 3:56 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
Here is the echo output.
UPDATE News SET NewsTitle = 'Text',
NewsText = 'can\'t use quotes!' WHERE NewsID = 16
Matthew Loff [EMAIL PROTECTED] wrote in message
000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
It might help us to see exactly what the query is...
Could you insert an echo statement and let us know what SQL query it's
producing?
echo
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
;
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 1:03 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I am using an Access DBASE and trying to insert text from a form
field. When I use quotes in my text the code errors out, otherwise it
works fine. I have Magic Quotes turned on which is supposed to fix
this problem but IT DOES NOT!
I have posted in several forums but no one has been able to help me!
Here is my code:
--
--
--
$NowDate = date(m/d/y);
$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
);
Here is the error when using quotes in the NewsText form field:
--
--
Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
error (missing operator) in query expression ''can\'t use quotes!')'.,
SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
line 22
Thanks in advance...
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[EMAIL PROTECTED]
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Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I tried double quotes and many different combinations of single and
double but nothing has worked.
There HAS to be an answer out there! Do certain SQL statements just not work
in PHP / Access? Is this a bug in Access or PHP?
If I can't get this to work then I won't be able to use PHP for this site!
Someone please shed some light on this!
Matthew Loff [EMAIL PROTECTED] wrote in message
001701c10dae$53146280$0100a8c0@bang">news:001701c10dae$53146280$0100a8c0@bang...
That looks like a valid query to me... I know this would break SQL92
standard, but have you tried enclosing the values in double quotes? I
know that's valid with MySQL...
I don't have any other suggestions... Does anyone else?
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 3:56 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
Here is the echo output.
UPDATE News SET NewsTitle = 'Text',
NewsText = 'can\'t use quotes!' WHERE NewsID = 16
Matthew Loff [EMAIL PROTECTED] wrote in message
000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
It might help us to see exactly what the query is...
Could you insert an echo statement and let us know what SQL query it's
producing?
echo
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
;
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 1:03 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I am using an Access DBASE and trying to insert text from a form
field. When I use quotes in my text the code errors out, otherwise it
works fine. I have Magic Quotes turned on which is supposed to fix
this problem but IT DOES NOT!
I have posted in several forums but no one has been able to help me!
Here is my code:
--
--
--
$NowDate = date(m/d/y);
$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
);
Here is the error when using quotes in the NewsText form field:
--
--
Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
error (missing operator) in query expression ''can\'t use quotes!')'.,
SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
line 22
Thanks in advance...
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RE: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
It would be a hack, but if the data is only going to be used for the
site (e.g. nobody is going to use the data through access) you could
str_replace() the single quote to something else (perhaps an HTML ASCII
representation like #;) before the INSERT/UPDATE, and converting
back when SELECTing. It's ugly, but it would work...
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 7:21 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I tried double quotes and many different combinations of single and
double but nothing has worked.
There HAS to be an answer out there! Do certain SQL statements just not
work in PHP / Access? Is this a bug in Access or PHP?
If I can't get this to work then I won't be able to use PHP for this
site! Someone please shed some light on this!
Matthew Loff [EMAIL PROTECTED] wrote in message
001701c10dae$53146280$0100a8c0@bang">news:001701c10dae$53146280$0100a8c0@bang...
That looks like a valid query to me... I know this would break SQL92
standard, but have you tried enclosing the values in double quotes? I
know that's valid with MySQL...
I don't have any other suggestions... Does anyone else?
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 3:56 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
Here is the echo output.
UPDATE News SET NewsTitle = 'Text',
NewsText = 'can\'t use quotes!' WHERE NewsID = 16
Matthew Loff [EMAIL PROTECTED] wrote in message
000e01c10d89$6fa576c0$0100a8c0@bang">news:000e01c10d89$6fa576c0$0100a8c0@bang...
It might help us to see exactly what the query is...
Could you insert an echo statement and let us know what SQL query
it's
producing?
echo
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
;
-Original Message-
From: AKA Hook [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 15, 2001 1:03 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] PLEASE HELP! Problem with ODBC string insert
I am using an Access DBASE and trying to insert text from a form
field. When I use quotes in my text the code errors out, otherwise
it works fine. I have Magic Quotes turned on which is supposed to
fix this problem but IT DOES NOT!
I have posted in several forums but no one has been able to help me!
Here is my code:
--
--
--
$NowDate = date(m/d/y);
$DBASE = odbc_connect(mno,admin,admin);
$getnews = odbc_exec($DBASE,
INSERT INTO News
(NewsDate,NewsTitle,NewsText)
VALUES
('$NowDate','$NewsTitle','$NewsText')
);
Here is the error when using quotes in the NewsText form field:
--
--
Warning: SQL error: [Microsoft][ODBC Microsoft Access Driver] Syntax
error (missing operator) in query expression ''can\'t use
quotes!')'.,
SQL state 37000 in SQLExecDirect in W:\www\clanmno\news_post2.php on
line 22
Thanks in advance...
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Re: [PHP-DB] PLEASE HELP !!!
when you say an error - do you mean any error? or specific errors? ie., you're not getting it stuck in an infinite loop? (cpu time flips up to that when you stick in a loop) or is it all errors? PHPFAN [EMAIL PROTECTED] wrote in message 9eetk0$92d$[EMAIL PROTECTED]">news:9eetk0$92d$[EMAIL PROTECTED]... I am running PHP 4.0.5 on Windows 2000 with SQL server databases. If there is an error with my PHP code, my computer becomes very very slow and nothing works. The PHP process is using 99 % of the CPU. So I have to restart the computer. This happens everytime there is an error in my PHP code. For the work I am doing, I have to use PHP with backend SQL Server on Windows Platform. Any suggestions will be greatly appreciated. Thank you PHPFAN -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] Please help me with this script...
-Original Message- From: Slider [mailto:[EMAIL PROTECTED]] Sent: Monday, April 02, 2001 9:27 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Please help me with this script... Hello, I've been working on this script all day, but I'm getting the same failure report each time: Warning: Supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\motoerit\toon_alles.php on line 23 What could be wrong? This is the code I have: This usually means there was a problem with your query. For more info, including some useable sample code that should help you troubleshoot exactly what went wrong, see http://www.php.net/FAQ.php#7.12 --- Mark Roedel ([EMAIL PROTECTED]) || "There cannot be a crisis next week. Systems Programmer / WebMaster || My schedule is already full." LeTourneau University ||-- Henry Kissinger -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
