[PHP-DB] query error
I've got a database table with a whole list of windows file paths. eg Advent Tower PC\C\Program Files\Microsoft Office\Microsoft Binder.lnk Advent Tower PC\C\Program Files\Microsoft Office\Microsoft Excel.lnk Advent Tower PC\C\Program Files\Microsoft Office\Microsoft Office Setup.lnk Advent Tower PC\C\Program Files\Microsoft Office\Microsoft Outlook.lnk Advent Tower PC\C\Program Files\Microsoft Office\Microsoft PowerPoint.lnk Advent Tower PC\C\Program Files\Microsoft Office\Microsoft Word.lnk Advent Tower PC\C\Program Files\Microsoft Office\MS Access Workgroup Administrator.lnk Advent Tower PC\C\Program Files\Microsoft Office\MSCREATE.DIR Advent Tower PC\C\Program Files\Microsoft Office\OF97SPEC.INI Advent Tower PC\C\Program Files\Microsoft Office\Office Advent Tower PC\C\Program Files\Microsoft Office\Office\1033 Now I want my DB query to search for directories .. here's a sample query select * from ff_files where full_path like 'Advent Tower PC\C\freeserve\help\images\%' doesnt work as the % is taken as a literal '%' I've aslso tried like 'Advent Tower PC\\C\\freeserve\\help\\images\%' I'm completely lost .. all I want is to return all rows that start with 'Advent Tower PC\C\freeserve\help\images\' tia Pete -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Query Error
I've been racking my brain and the web for finding a solution and no
luck so far.
Here's the form:
$display_block .= HRInput item: form method=post
action=\additem.php\nbsp;nbsp;nbsp;nbsp;Item ID:input
type=\text\ name=\sel_item_id\BRQuantity: select
name=\sel_item_qty\;
$display_block .= option value=\0\0/optionoption value=\1\
selected1/option;
for ($i=2; $i 301; $i++){
$display_block .= option value=\$i\$i/option;
}
$display_block .= /selectinput type=\hidden\ name=\SESSID\
value=\$PHPSESSID\/tdtdinput type=\hidden\ name=\url\
value=\$_SERVER[PHP_SELF]\BRPrice:input type=\text\
name=\sel_item_price\ size=\5\input type=\submit\
name=\submit\ value=\Add to Cart\/form;
And here is the additem.php file where it check for the info in the
field and it is erroring out.
//validate item and get item and price
// $get_iteminfo = select item_num, description from GCN_items where
'item_num' LIKE '%$_POST[sel_item_id]';
$get_iteminfo = select * from GCN_items where 'item_num' LIKE
'%$_POST[sel_item_id]%';
$get_iteminfo_res = mysql_query($get_iteminfo) or die(mysql_error());
if (mysql_num_rows($get_iteminfo_res) 1){
//invalid id, send away and show query codes
//header(Location: ohcrud.php);
print the query I just ran was: $get_iteminfoBR\n;
print the query I just ran was: $get_iteminfo_resBR\n;
echo mysql_num_rows($get_iteminfo_res);
exit;
I did have = instead of LIKE before, but the results are the same.
GCN_items has the code I am looking for in it but the query is showing
no results.
HELP!
Robert
~~~
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Re: [PHP-DB] Query Error
From: Robert Sossomon [EMAIL PROTECTED] And here is the additem.php file where it check for the info in the field and it is erroring out. It would be rather useful to know the error... $get_iteminfo = select * from GCN_items where 'item_num' LIKE '%$_POST[sel_item_id]%'; Take away the single quotes around your column name. You're asking for a 'string' LIKE 'another string' and not comparing anything in your column at all. You can use backticks around table and column names, though. Maybe that's what you're trying to achieve. $get_iteminfo = select * from GCN_items where `item_num` LIKE '%$_POST[sel_item_id]%'; ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Query Error
The errors as it prints are: The query I just ran was: select * from GCN_items where `item_num` = '%fm-a294%' The query I just ran was: Resource id #2 0 -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Tuesday, November 04, 2003 11:04 AM To: Robert Sossomon; [EMAIL PROTECTED] Subject: Re: [PHP-DB] Query Error From: Robert Sossomon [EMAIL PROTECTED] And here is the additem.php file where it check for the info in the field and it is erroring out. It would be rather useful to know the error... $get_iteminfo = select * from GCN_items where 'item_num' LIKE '%$_POST[sel_item_id]%'; Take away the single quotes around your column name. You're asking for a 'string' LIKE 'another string' and not comparing anything in your column at all. You can use backticks around table and column names, though. Maybe that's what you're trying to achieve. $get_iteminfo = select * from GCN_items where `item_num` LIKE '%$_POST[sel_item_id]%'; ---John Holmes... -- 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] Query Error
From: Robert Sossomon [EMAIL PROTECTED] The errors as it prints are: The query I just ran was: select * from GCN_items where `item_num` = '%fm-a294%' The query I just ran was: Resource id #2 0 Those aren't errors; it's just what you asked the script to display. Your query simply isn't returning any rows. You should be using LIKE instead of the equal sign, since I assume you're looking for a pattern. select * from GCN_items where `item_num` LIKE '%fm-a294%' otherwise you're looking for the literal string %fm-a294% ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Query Error
That fixed the output to actually find the information, however I'll be danged if it is not adding the items to the database. In fact I am seeing errors with it adding half the stuff to the database from stuff that was working. I rebooted the system. Removed the database, stopped the server. Started the server. Reloaded the database. Refreshed the data. Now I am not seeing stuff added to the system and I have no idea why. According to the scripts everything is working fine. The information is being passed to the pages correctly and even the output on the page showed that the data was correct for insertion, however when I look at the database where the information should be, it is not there. I'm losing my mind! Any thoughts? ~~~ A horse! A horse! My kingdom for a horse! -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Tuesday, November 04, 2003 1:31 PM To: Robert Sossomon; [EMAIL PROTECTED] Subject: Re: [PHP-DB] Query Error From: Robert Sossomon [EMAIL PROTECTED] The errors as it prints are: The query I just ran was: select * from GCN_items where `item_num` = '%fm-a294%' The query I just ran was: Resource id #2 0 Those aren't errors; it's just what you asked the script to display. Your query simply isn't returning any rows. You should be using LIKE instead of the equal sign, since I assume you're looking for a pattern. select * from GCN_items where `item_num` LIKE '%fm-a294%' otherwise you're looking for the literal string %fm-a294% ---John Holmes... -- 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-DB] Query error
Hello guys Does anybody have a clue on what wrong with the following sentence? select author_code from author where author_code not in (select author_code from authorxcat); Thanks *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Query error
if you are using mysql, it doesn't support the sub-select. Wilmar Perez [EMAIL PROTECTED] To: [EMAIL PROTECTED] ea.edu.cocc: Subject: [PHP-DB] Query error 10/02/2002 10:54 AM Hello guys Does anybody have a clue on what wrong with the following sentence? select author_code from author where author_code not in (select author_code from authorxcat); Thanks *** Wilmar Pérez Network Administrator Library System Tel: ++57(4)2105145 University of Antioquia Medellín - Colombia 2002 *** -- 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] Query error
If it's on MySQL, then the problem is that it doesn't support nested queries :) If it's on something else, i'm afraid i don't know :/ Back to MySQL, you can get around that by doing your select author_code from authorxcat component query, then put the results into 'x', 'y', 'z' format, then do something like select author_code from author where author_code not in ($previousResults) HTH Beau // -Original Message- // From: Wilmar Perez [mailto:[EMAIL PROTECTED]] // Sent: Wednesday, 2 October 2002 10:54 PM // To: [EMAIL PROTECTED] // Subject: [PHP-DB] Query error // // // Hello guys // // Does anybody have a clue on what wrong with the following sentence? // // select author_code from author where author_code not in // (select author_code from authorxcat); // // Thanks // // *** // Wilmar Pérez // Network Administrator //Library System // Tel: ++57(4)2105145 // University of Antioquia //Medellín - Colombia // 2002 // *** // // // // // -- // 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-DB] query error
when i perform query an mysql db, to insert a new record, the PHP says:
Warning: Supplied argument is not a valid MySQL result resource in
g:\wwwroot\phpusermanager-1.0\add_order.php on line 24
the snippet is as follows
?
// connect to the database
$conn = mysql_connect ($db_host, $db_user, $db_passwd);
// select database
mysql_select_db ($db_used, $conn);
// SQL statement
$query = INSERT into order (Ctr_no, OID, Person_in_charge, Source,
Destination, Status)
VALUES ('$container_num', '$order', '$referrer', '$source', '$dest',
'$status');
mysql_query ($query, $conn);
mysql_close($conn);
?
why the argument isn't valid? how to solve this problem?
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Re: [PHP-DB] query error
On Sunday 05 May 2002 14:44, erich wrote:
when i perform query an mysql db, to insert a new record, the PHP says:
Warning: Supplied argument is not a valid MySQL result resource in
g:\wwwroot\phpusermanager-1.0\add_order.php on line 24
the snippet is as follows
?
// connect to the database
$conn = mysql_connect ($db_host, $db_user, $db_passwd);
// select database
mysql_select_db ($db_used, $conn);
// SQL statement
$query = INSERT into order (Ctr_no, OID, Person_in_charge, Source,
Destination, Status)
VALUES ('$container_num', '$order', '$referrer', '$source', '$dest',
'$status');
mysql_query ($query, $conn);
mysql_close($conn);
?
why the argument isn't valid? how to solve this problem?
Find out why by putting in some error checking code. See examples in manual
and check out mysql_error().
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Re: [PHP-DB] query error
Echo your INSERT statement, so that you know what it looks like, if you
have the values you expect, field names are spelled correctly. that you're
not inserting char into int, etc.
Use mysql_errno() and mysql_err() (Check those against manual to determine
what error you are getting. Use mysql_affected_rows() to determine the
number of rows which were inserted.
Finally, you aren't assigning the return value from mysql_query(), so if
you are testing a variable on line 24 it has nothing assigned to it.
HTH - Miles Thompson
At 02:44 PM 5/5/2002 +0800, erich wrote:
when i perform query an mysql db, to insert a new record, the PHP says:
Warning: Supplied argument is not a valid MySQL result resource in
g:\wwwroot\phpusermanager-1.0\add_order.php on line 24
the snippet is as follows
?
// connect to the database
$conn = mysql_connect ($db_host, $db_user, $db_passwd);
// select database
mysql_select_db ($db_used, $conn);
// SQL statement
$query = INSERT into order (Ctr_no, OID, Person_in_charge, Source,
Destination, Status)
VALUES ('$container_num', '$order', '$referrer', '$source', '$dest',
'$status');
mysql_query ($query, $conn);
mysql_close($conn);
?
why the argument isn't valid? how to solve this problem?
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[PHP-DB] query error...
Ok what is wrong with this sql statement? $sql = UPDATE $table_name SET c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\zip=\zi p\,phone=\$phone\; The error I recieve is as follows Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /path/to/wwwdemo_change.php3 on line 22 Parse error: parse error in /path/to/wwwdemo_change.php3 on line 22 I have looked on MySQL.com at the update function and it states I can use the SET under UPDATE for multiple fields separated by a , and it is not working... Any insight would be great. Thanks in advance, Jas -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] query error...
Try putting single quotes around your variables in your update statement instead of the escaped double quotes, they are easier to read. Also, were you wanting to update all entries in that table to the same, c_name, s_addy, city, state, zip and phone? If not, you will need to add 'where id=$id' (or something along those lines) to your update statement. Otherwise, the way you have it now, it will update every row in your database to the same values for the listed fields. jas [EMAIL PROTECTED] said: Ok what is wrong with this sql statement? $sql = UPDATE $table_name SET c_name=$c_name,s_addy=$s_addy,city=$city,state=statezip=zi p,phone=$phone; The error I recieve is as follows Warning: Unexpected character in input: '' (ASCII=92) state=1 in /path/to/wwwdemo_change.php3 on line 22 Parse error: parse error in /path/to/wwwdemo_change.php3 on line 22 I have looked on MySQL.com at the update function and it states I can use the SET under UPDATE for multiple fields separated by a , and it is not working... Any insight would be great. Thanks in advance, Jas -- 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] query error...
Ok jas Ok what is wrong with this sql statement? $sql = UPDATE $table_name SET c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\zip=\zi p\,phone=\$phone\; insufficient checking: comma missing between state and zip incorrectly escaped before phone ... I have looked on MySQL.com at the update function and it states I can use the SET under UPDATE for multiple fields separated by a , and it is not working... Any insight would be great. Insight: use ECHO to output $sql to the screen. If the results are less than satisfactory, copy-paste the sql code into native MySQL or a management package, the errmsgs are usually more informative/directive. Regards, =dn -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] query error...
Did you type this out or cut/paste? There was a comma missing between state\zip . Should work when you add the comma. Add a WHERE condition for the update, otherwise every row will be set to these values. Common practice is to use single quotes around the char variables, saves a lot of escaping, keeps things more readable. In connection with this, echoing $sql would probably have revealed the missing comma, and don't feel badly, we've all done it. Hope this sees you right - Miles At 02:43 AM 1/31/2002 -0700, jas wrote: Ok what is wrong with this sql statement? $sql = UPDATE $table_name SET c_name=\$c_name\,s_addy=\$s_addy\,city=\$city\,state=\state\zip=\zi p\,phone=\$phone\; The error I recieve is as follows Warning: Unexpected character in input: '\' (ASCII=92) state=1 in /path/to/wwwdemo_change.php3 on line 22 Parse error: parse error in /path/to/wwwdemo_change.php3 on line 22 I have looked on MySQL.com at the update function and it states I can use the SET under UPDATE for multiple fields separated by a , and it is not working... Any insight would be great. Thanks in advance, Jas -- 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-DB] query error
When I run any kind of query this is the value of result Resource id #2
Example
$result= mysql_query (" Select * from ACCT_TBL ") or die
("Error".mysql_error());
print result;
any ideas?
randy
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RE: [PHP-DB] query error
Looks fine, now retrieve the rows from the query.
-Original Message-
From: Randy Johnson [mailto:[EMAIL PROTECTED]]
Sent: Thursday, January 25, 2001 12:46 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] query error
When I run any kind of query this is the value of result
Resource id #2
Example
$result= mysql_query (" Select * from ACCT_TBL ") or die
("Error".mysql_error());
print result;
any ideas?
randy
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[EMAIL PROTECTED]
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