Re: [PHP-DB] Parse error
There are further problems with the mySQL query: 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 ') AND CURDATE() <= DATE(FROM_UNIXTIME(1239508800)), 1, 0), IF(CURDATE() >= DATE' This is the whole query, what am I missing / doing wrong? SELECT GREATEST( IF(CURDATE() >= DATE_SUB(CONCAT(DATE(FROM_UNIXTIME(1239508800)), INTERVAL LEAST(14, (SELECT COUNT(*) FROM `verse_of_the_day_Bible_verses` WHERE seasonal_use = 1)) DAY) AND CURDATE() <= DATE(FROM_UNIXTIME(1239508800)), 1, 0), IF(CURDATE() >= DATE_SUB(CONCAT(YEAR(CURDATE()), '-12-25'), INTERVAL LEAST(14, (SELECT COUNT(*) FROM `verse_of_the_day_Bible_verses` WHERE seasonal_use = 2)) DAY) AND CURDATE() <= YEAR(CURDATE()), '-12-25'), 2, 0) ) AS verse_application On Thu, 2009-09-24 at 15:39 +1000, Chris wrote: > Ron Piggott wrote: > > Let me try this is a different way. > > > > What is the variable which I am able to retrieve? I am expecting the > > result to be a 0, 1 or 2. > > Your original problem was getting a parse error. > > Your query is not php code - step 1 is to fix that. > > Once you've done that, post the new code and new problem you're having. > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Parse error
Ron Piggott wrote: Let me try this is a different way. What is the variable which I am able to retrieve? I am expecting the result to be a 0, 1 or 2. Your original problem was getting a parse error. Your query is not php code - step 1 is to fix that. Once you've done that, post the new code and new problem you're having. -- Postgresql & php tutorials http://www.designmagick.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Parse error
Ron Piggott wrote:
This is the actual code. What should I have?
Same as when you do a query normally.
$query = "SELECT ... WHERE ";
$result = mysql_query($query);
.
On Thu, 2009-09-24 at 13:20 +1000, Chris wrote:
Ron Piggott wrote:
The following code gives me this error message:
Parse error: syntax error, unexpected ':'
in /home/thev4173/public_html/test.php on line 8
It is referencing the : that follows EasterDate (and will eventually get
mad at ChristmasDate)
mysql_connect('localhost',$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
IF(CURDATE() >= DATE_SUB(CONCAT(@EasterDate :=
DATE(FROM_UNIXTIME(easter_date(date('Y'))), INTERVAL LEAST(14, (SELECT
COUNT(*) FROM `verse_of_the_day_Bible_verses` WHERE seasonal_use = 1))
DAYS) AND CURDATE() <= @EasterDate, 1, 0)
IF(CURDATE() >= DATE_SUB(CONCAT(@ChristmasDate := YEAR(CURDATE()),
'-12-25'), INTERVAL LEAST(14, (SELECT COUNT(*) FROM
`verse_of_the_day_Bible_verses` WHERE seasonal_use = 2)) DAYS) AND
CURDATE() <= @ChristmasDate, 2, 0)
mysql_close();
What is the problem?
If this is your actual code, the queries aren't in php variables.
$query = "IF(curdate() ";
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Re: [PHP-DB] Parse error
Ron Piggott wrote:
The following code gives me this error message:
Parse error: syntax error, unexpected ':'
in /home/thev4173/public_html/test.php on line 8
It is referencing the : that follows EasterDate (and will eventually get
mad at ChristmasDate)
mysql_connect('localhost',$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
IF(CURDATE() >= DATE_SUB(CONCAT(@EasterDate :=
DATE(FROM_UNIXTIME(easter_date(date('Y'))), INTERVAL LEAST(14, (SELECT
COUNT(*) FROM `verse_of_the_day_Bible_verses` WHERE seasonal_use = 1))
DAYS) AND CURDATE() <= @EasterDate, 1, 0)
IF(CURDATE() >= DATE_SUB(CONCAT(@ChristmasDate := YEAR(CURDATE()),
'-12-25'), INTERVAL LEAST(14, (SELECT COUNT(*) FROM
`verse_of_the_day_Bible_verses` WHERE seasonal_use = 2)) DAYS) AND
CURDATE() <= @ChristmasDate, 2, 0)
mysql_close();
What is the problem?
If this is your actual code, the queries aren't in php variables.
$query = "IF(curdate() ";
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Re: [PHP-DB] Parse error: syntax error, unexpected '>' in C:\wamp\www\new.php on line 38
WOW,
Both of your suggestions worked like magic. Thanks a lot Patrick
-Sashi
Patrick Price wrote:
I believe your problem is that on line 33 you forgot your closing double
quote in the echo statement.
Thanks
patrick
Sent from: Decatur Ga United States.
On Fri, Feb 13, 2009 at 10:02 PM, Sashikanth Gurram wrote:
Hi all,
I have pasted the PHP code I wrote below. I do not know why, but I am
getting the parse error: unexpected '>'. It says the error is on line 38.
The exact line on line 38 is
* echo"\n*
Basically, I am trying to execute an sql query through PHP which tries to
extract data from the mysql database and display the data in the browser.
The query is running fine but I am getting the problems when I try to
display the data in the browser.
Can anyone help me with it?
Thanks,
Sashi
";
echo "OCCUPANCY";
echo "";
echo ";
while ($row=mysqli_fetch_array($data))
{
extract($row);
echo"\n
$building\n
$parking_lot\n
$month\n
$day\n
$occupancy\n
$empty_spaces\n
$special_days\n
\n";
echo "\n";
}
echo"\n";
?>
--
~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
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~
~
Sashikanth Gurram
Graduate Research Assistant
Department of Civil and Environmental Engineering
Virginia Tech
Blacksburg, VA 24060, USA
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Re: [PHP-DB] Parse error: syntax error, unexpected '>' in C:\wamp\www\new.php on line 38
I believe your problem is that on line 33 you forgot your closing double
quote in the echo statement.
Thanks
patrick
Sent from: Decatur Ga United States.
On Fri, Feb 13, 2009 at 10:02 PM, Sashikanth Gurram wrote:
> Hi all,
>
> I have pasted the PHP code I wrote below. I do not know why, but I am
> getting the parse error: unexpected '>'. It says the error is on line 38.
> The exact line on line 38 is
> * echo"\n*
>
> Basically, I am trying to execute an sql query through PHP which tries to
> extract data from the mysql database and display the data in the browser.
> The query is running fine but I am getting the problems when I try to
> display the data in the browser.
>
> Can anyone help me with it?
>
> Thanks,
> Sashi
>
> /* CONNECTS TO THE DATABASE*/
> $host="**";
> $user="**";
> $password="***";
> $dbname="*";
> $cxn=mysqli_connect($host, $user, $password, $dbname);
> if (!$cxn=mysqli_connect($host, $user, $password, $dbname))
> {
> $error=mysqli_error($cxn);
> echo "$error";
> die();
> }
> else
> {
> echo "Connection established successfully";
> }
> $sql="select * FROM OCCUPANCY";
> $data=mysqli_query($cxn,$sql);
> if (!$data=mysqli_query($cxn,$sql))
> {
> $error=mysqli_error($cxn);
> echo "$error";
> die();
> }
> else
> {
> echo "Query sent successfully";
> }
> echo "";
> echo "OCCUPANCY";
> echo "";
> echo ";
> while ($row=mysqli_fetch_array($data))
> {
> extract($row);
> echo"\n
>$building\n
>$parking_lot\n
>$month\n
>$day\n
>$occupancy\n
>$empty_spaces\n
>$special_days\n
>\n";
> echo "\n";
> }
> echo"\n";
> ?>
>
> --
> ~
> ~
> Sashikanth Gurram
> Graduate Research Assistant
> Department of Civil and Environmental Engineering
> Virginia Tech
> Blacksburg, VA 24060, USA
>
>
> --
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>
>
Re: [PHP-DB] Parse error: syntax error, unexpected T_ELSE
On Tue, Apr 15, 2008 at 5:56 PM, Javier Viegas <[EMAIL PROTECTED]> wrote: > Thanks Daniel i ´ve removed the unexpected else secction. Now it works but > it tells me that there might be a syntax error on the query, this is good > news because now i know i´m having connection with the database but also > there must be something wrong with the query. > > > $del_str = "DELETE FROM libros WHERE bnumero=".$_GET['Id']; [snip!] You didn't enclose the value in quotes, that's all (and you should sanitize the input --- NEVER accept data without sanitizing it first!). -- Ask me about: Dedicated servers starting @ $59.99/mo., VPS starting @ $19.99/mo., and shared hosting starting @ $2.50/mo. Unmanaged, managed, and fully-managed! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Parse error: syntax error, unexpected T_ELSE
Hi,
I suggest you echo $del_str to see exactly what your query is.
There may be something unexpected in $_GET['Id'].
Echo the whole query so you can see everything in context.
Gary
> -Original Message-
> From: Javier Viegas [mailto:[EMAIL PROTECTED]
> Sent: Tue, April 15, 2008 5:56 PM
> To: php-db@lists.php.net
> Subject: Re: [PHP-DB] Parse error: syntax error, unexpected T_ELSE
>
>
> Thanks Daniel i ´ve removed the unexpected else secction. Now
> it works but
> it tells me that there might be a syntax error on the query,
> this is good
> news because now i know i´m having connection with the
> database but also
> there must be something wrong with the query.
>
> $del_str = "DELETE FROM libros WHERE bnumero=".$_GET['Id'];
>
> Is this correct, i´ve checked and the correct syntax to use
> would be delete
> from libros where bnumero='something';
> Given 'something' is taken from a variable called Id it seems
> reasonable to
> me the way it is written. Do you see anything wrong?
>
> Thanks again.
>
>
> On Tue, Apr 15, 2008 at 6:03 PM, Daniel Brown
> <[EMAIL PROTECTED]> wrote:
>
> > On Tue, Apr 15, 2008 at 4:49 PM, Javier Viegas
> <[EMAIL PROTECTED]>
> > wrote:
> > > Hi, i have this script wich basically connects to a
> database and delete
> > a
> > > record according to the Id parameter given. The problem
> is that when i
> > test
> > > it i get this error:
> > >
> > > *Parse error*: syntax error, unexpected T_ELSE in *
> > > /var/www/biblio/scripts/delete.php* on line *31
> >
> > Javier,
> >
> >This block:
> >
> > } else {
> > $r_string = '&errorcode=4&';
> >
> > }
> >
> > is incorrect. You call an else condition on line
> 21, so PHP
> > expects that block to be the last for that if() condition. Either
> > remove the } else { and $r_string = '&errorcode=4&'; lines
> or rewrite
> > the condition.
> >
> >And if that's your real database login information, change it and
> > update all of your scripts and systems ASAP.
> >
> >
> > >
> > > This is the script:
> > >
> > > * > > /*
> > >deletescore.php: deletes record for passed id from
> highscores table
> > and
> > > returns status to Flash
> > > */
> > > // fill with correct data for your server configuration
> > > $server = "localhost";
> > > $username = "root";
> > > $password = "itsveryeasy";
> > > $database = "biblioteca";
> > >
> > > //connect to database added by calm
> > > mysql_connect($server, $username, $password);
> > >
> > > if (!mysql_connect($server, $username, $password)) {
> > >$r_string = '&errorcode=1&';
> > >
> > > } elseif (!mysql_select_db($database)) {
> > >$r_string = '&errorcode=2&';
> > >
> > > } else {
> > >
> > > $del_str = "DELETE FROM libros WHERE bnumero=".$_GET['Id'];
> > >
> > > if (!mysql_query ($del_str)) {
> > > $msg = mysql_error();
> > > $r_string = '&errorcode=3&msg='.$msg;
> > > } else {
> > > $r_string = '&errorcode=0&';
> > > }
> > >} else {
> > > $r_string = '&errorcode=4&';
> > >
> > > }
> > >
> > > echo $r_string;
> > > ?>
> > >
> > > Wha am i doing wrong??
> > >
> > > Thanks.
> > >
> > > Javier
> > >
> >
> >
> >
> > --
> >
> > Ask me about:
> > Dedicated servers starting @ $59.99/mo., VPS starting @ $19.99/mo.,
> > and shared hosting starting @ $2.50/mo.
> > Unmanaged, managed, and fully-managed!
> >
>
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Re: [PHP-DB] Parse error: syntax error, unexpected T_ELSE
Thanks Daniel i ´ve removed the unexpected else secction. Now it works but
it tells me that there might be a syntax error on the query, this is good
news because now i know i´m having connection with the database but also
there must be something wrong with the query.
$del_str = "DELETE FROM libros WHERE bnumero=".$_GET['Id'];
Is this correct, i´ve checked and the correct syntax to use would be delete
from libros where bnumero='something';
Given 'something' is taken from a variable called Id it seems reasonable to
me the way it is written. Do you see anything wrong?
Thanks again.
On Tue, Apr 15, 2008 at 6:03 PM, Daniel Brown <[EMAIL PROTECTED]> wrote:
> On Tue, Apr 15, 2008 at 4:49 PM, Javier Viegas <[EMAIL PROTECTED]>
> wrote:
> > Hi, i have this script wich basically connects to a database and delete
> a
> > record according to the Id parameter given. The problem is that when i
> test
> > it i get this error:
> >
> > *Parse error*: syntax error, unexpected T_ELSE in *
> > /var/www/biblio/scripts/delete.php* on line *31
>
> Javier,
>
>This block:
>
> } else {
> $r_string = '&errorcode=4&';
>
> }
>
> is incorrect. You call an else condition on line 21, so PHP
> expects that block to be the last for that if() condition. Either
> remove the } else { and $r_string = '&errorcode=4&'; lines or rewrite
> the condition.
>
>And if that's your real database login information, change it and
> update all of your scripts and systems ASAP.
>
>
> >
> > This is the script:
> >
> > * > /*
> >deletescore.php: deletes record for passed id from highscores table
> and
> > returns status to Flash
> > */
> > // fill with correct data for your server configuration
> > $server = "localhost";
> > $username = "root";
> > $password = "itsveryeasy";
> > $database = "biblioteca";
> >
> > //connect to database added by calm
> > mysql_connect($server, $username, $password);
> >
> > if (!mysql_connect($server, $username, $password)) {
> >$r_string = '&errorcode=1&';
> >
> > } elseif (!mysql_select_db($database)) {
> >$r_string = '&errorcode=2&';
> >
> > } else {
> >
> > $del_str = "DELETE FROM libros WHERE bnumero=".$_GET['Id'];
> >
> > if (!mysql_query ($del_str)) {
> > $msg = mysql_error();
> > $r_string = '&errorcode=3&msg='.$msg;
> > } else {
> > $r_string = '&errorcode=0&';
> > }
> >} else {
> > $r_string = '&errorcode=4&';
> >
> > }
> >
> > echo $r_string;
> > ?>
> >
> > Wha am i doing wrong??
> >
> > Thanks.
> >
> > Javier
> >
>
>
>
> --
>
> Ask me about:
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>
Re: [PHP-DB] Parse error: syntax error, unexpected T_ELSE
On Tue, Apr 15, 2008 at 4:49 PM, Javier Viegas <[EMAIL PROTECTED]> wrote:
> Hi, i have this script wich basically connects to a database and delete a
> record according to the Id parameter given. The problem is that when i test
> it i get this error:
>
> *Parse error*: syntax error, unexpected T_ELSE in *
> /var/www/biblio/scripts/delete.php* on line *31
Javier,
This block:
} else {
$r_string = '&errorcode=4&';
}
is incorrect. You call an else condition on line 21, so PHP
expects that block to be the last for that if() condition. Either
remove the } else { and $r_string = '&errorcode=4&'; lines or rewrite
the condition.
And if that's your real database login information, change it and
update all of your scripts and systems ASAP.
>
> This is the script:
>
> * /*
>deletescore.php: deletes record for passed id from highscores table and
> returns status to Flash
> */
> // fill with correct data for your server configuration
> $server = "localhost";
> $username = "root";
> $password = "itsveryeasy";
> $database = "biblioteca";
>
> //connect to database added by calm
> mysql_connect($server, $username, $password);
>
> if (!mysql_connect($server, $username, $password)) {
>$r_string = '&errorcode=1&';
>
> } elseif (!mysql_select_db($database)) {
>$r_string = '&errorcode=2&';
>
> } else {
>
> $del_str = "DELETE FROM libros WHERE bnumero=".$_GET['Id'];
>
> if (!mysql_query ($del_str)) {
> $msg = mysql_error();
> $r_string = '&errorcode=3&msg='.$msg;
> } else {
> $r_string = '&errorcode=0&';
> }
>} else {
> $r_string = '&errorcode=4&';
>
> }
>
> echo $r_string;
> ?>
>
> Wha am i doing wrong??
>
> Thanks.
>
> Javier
>
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Re: [PHP-DB] parse error in create statement.
Missing a . in the line before the last quote GRANT CREATE TRIGGER TO ".$adduser" "$sqlstmt= "CREATE USER ".$adduser." PROFILE DEFAULT IDENTIFIED BY ".$addpass." DEFAULT TABLESPACE USERS TEMPORARY TABLESPACE TEMP QUOTA UNLIMITED ON USERS ACCOUNT UNLOCK; GRANT CREATE TABLE TO ".$adduser." GRANT CREATE TRIGGER TO ".$adduser." GRANT CONNECT TO ".$adduser;". bastien From: Chris Ramsay <[EMAIL PROTECTED]> Reply-To: Chris Ramsay <[EMAIL PROTECTED]> To: babu <[EMAIL PROTECTED]> CC: php-db@lists.php.net Subject: Re: [PHP-DB] parse error in create statement. Date: Tue, 14 Jun 2005 10:56:24 +0100 Check your placement of all the '.' for a start. raz On 6/14/05, babu <[EMAIL PROTECTED]> wrote: > HI, > > Whats the error in this code. > > "$sqlstmt= "CREATE USER ".$adduser." PROFILE DEFAULT > IDENTIFIED BY ".$addpass." DEFAULT TABLESPACE USERS > TEMPORARY TABLESPACE TEMP > QUOTA UNLIMITED > ON USERS > ACCOUNT UNLOCK; > GRANT CREATE TABLE TO ".$adduser." > GRANT CREATE TRIGGER TO ".$adduser" > GRANT CONNECT TO ".$adduser;". > > > any suggestions. > > > > - > Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail > -- 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] parse error in create statement.
"$sqlstmt= "CREATE USER ".$adduser." PROFILE DEFAULT IDENTIFIED BY ".$addpass." DEFAULT TABLESPACE USERS TEMPORARY TABLESPACE TEMP QUOTA UNLIMITED ON USERS ACCOUNT UNLOCK; GRANT CREATE TABLE TO ".$adduser." GRANT CREATE TRIGGER TO ".$adduser." GRANT CONNECT TO ".$adduser."; for the .'s
Re: [PHP-DB] parse error in create statement.
Also, correct me if I'm wrong, but this needs to be split into two seperate query strings and executed seperately... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] parse error in create statement.
Check your placement of all the '.' for a start. raz On 6/14/05, babu <[EMAIL PROTECTED]> wrote: > HI, > > Whats the error in this code. > > "$sqlstmt= "CREATE USER ".$adduser." PROFILE DEFAULT > IDENTIFIED BY ".$addpass." DEFAULT TABLESPACE USERS > TEMPORARY TABLESPACE TEMP > QUOTA UNLIMITED > ON USERS > ACCOUNT UNLOCK; > GRANT CREATE TABLE TO ".$adduser." > GRANT CREATE TRIGGER TO ".$adduser" > GRANT CONNECT TO ".$adduser;". > > > any suggestions. > > > > - > Yahoo! Messenger NEW - crystal clear PC to PCcalling worldwide with voicemail > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] parse error in php backend sqlite
Aravalli Sai wrote:
hi
this is php code for inventory management system
which is done using SQLite.this is an user
interface where an user wil giv values to the text
boxes and it should be saved in the backend database
which is sqlite..
but when i run this code on browser it is giving an
error ... Parse error: parse error in
/home/saravall/.HTML/inv.php on line 63
$db ="/home/saravall/office.db";
$handle = sqlite_open($db) or die("could not open
database");
if(isset($_POST['submit'])) {
if (!empty($_post['tagno']) && !empty
($_post['desc']) &&
!empty ($_post['acqdt']) && !empty
($_post['manufacturer']) && !empty
($_post['model']) &&!empty($_post['serialid']) &&
!empty( $_post
['custloc']) && !empty($_post['totcost']))
{
$insquery = "insert into inventory
(tagno,desc,acqdt,manufacturer,model, serial
id,custloc,totcost) values (\"".sqlite_escape_string(
$_post['tagno'])."\",\"".sqlite_escape_string($_post['desc'])."\",\""
.sqlite_escape_string($_post['serialid'])."\",\"".sqlite_escape_string($_post
['custloc'])."\",\"".sqlite_escape_string($_post['totcost'])."\")";
$insresult = sqlite_query($handle,$insquery) or
die("error:".sqlite_error_string(sqlite_last_error($handle)));
echo "successfully inserted!";
}
else {
echo "incomplete from input!";
$query = "SELECT * FROM inventory ";
$result = sqlite_query($handle,$query) or
die("err:".sqlite_error_string(sqlite_last_error($handle)));
if (sqlite_num_rows($result) > 0)
{
echo "";
while ($row = sqlite_fetch_array($result))
{
echo "";
echo "".$row[0]."";
echo "".$row[1]."";
echo "".$row[2]."";
echo "".$row[3]."";
echo "".$row[4]."";
echo "".$row[5]."";
echo "".$row[6]."";
echo "".$row[7]."";
echo "".$row[8]."";
echo "".$row[9]."";
echo "";
}
echo "";
}
sqlite_close($handle);
?>
i would appreciate if you can help me in correcting
this error..
thanks in advance..
sai
Your braces don't match up. The problem starts at the ELSE statement in line 33.
Doug
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Re: [PHP-DB] parse error in php backend sqlite
- Original message from Aravalli Sai : --
> but when i run this code on browser it is giving an
> error ... Parse error: parse error in
> /home/saravall/.HTML/inv.php on line 63
>
It would be helpful if you said what the parse error is and which line is 63.
> if(isset($_POST['submit'])) {
>
> if (!empty($_post['tagno']) && !empty
Why are you switching between $_POST and $_post? The superglobal array is $_POST.
> i would appreciate if you can help me in correcting
> this error..
> thanks in advance..
> sai
Your curly braces, {}, don't match up. You aren't closing them all. Indenting your
code will help you see this and fix the problem.
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RE: [PHP-DB] Parse error on array and SQL query
If you're not getting the parse error anymore, I'd suggest you echo $query
to the browser and check to see that the SQL statement is what you expect it
to be. Can't count the number of times doing that has pointed out some stray
apostrophe or comma or something that killed the query.
> -Original Message-
> From: John W. Holmes [mailto:[EMAIL PROTECTED]
> Sent: Thursday, November 20, 2003 1:09 PM
> To: Dillon, John
> Cc: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Parse error on array and SQL query
>
>
> Dillon, John wrote:
>
> > Tried that, doesn't work with:
> >
> > $query="INSERT INTO ReplyTbl VALUES ('" . $monster1 . "', '" .
> > $monster2 .
> > "', '" . $monster3 . "', NULL, NULL)";
> >
> > Don't suppose they use 'monster' as a reserved word...
>
> How does it "not work" again??
>
> --
> ---John Holmes...
>
> Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/
>
> php|architect: The Magazine for PHP Professionals www.phparch.com
>
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>
>
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Re: [PHP-DB] Parse error on array and SQL query
Dillon, John wrote:
> Tried that, doesn't work with:
>
> $query="INSERT INTO ReplyTbl VALUES ('" . $monster1 . "', '" .
> $monster2 .
> "', '" . $monster3 . "', NULL, NULL)";
>
> Don't suppose they use 'monster' as a reserved word...
How does it "not work" again??
--
---John Holmes...
Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/
php|architect: The Magazine for PHP Professionals – www.phparch.com
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RE: [PHP-DB] Parse error on array and SQL query
Dillon, John wrote:
> On line 188 and similar:
> line 187$this=$altIDs[$i];
> line 188$query .= "ID='$this'";
>
> I get the error: Parse error: parse error in
> /home/jdillon/public_html/provreport.php on line 188
>
> Could this be because some setting has been changed on my shared host?
$this is normally used in object oriented programming to signify the
current object. That may be causing the trouble. I'd use another
variable than $this.
--
---John Holmes...
Tried that, doesn't work with:
$query="INSERT INTO ReplyTbl VALUES ('" . $monster1 . "', '" . $monster2 .
"', '" . $monster3 . "', NULL, NULL)";
Don't suppose they use 'monster' as a reserved word...
but the similar code works in another file.
John
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Re: [PHP-DB] Parse error on array and SQL query
Dillon, John wrote: On line 188 and similar: line 187 $this=$altIDs[$i]; line 188 $query .= "ID='$this'"; I get the error: Parse error: parse error in /home/jdillon/public_html/provreport.php on line 188 Could this be because some setting has been changed on my shared host? $this is normally used in object oriented programming to signify the current object. That may be causing the trouble. I'd use another variable than $this. -- ---John Holmes... Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/ php|architect: The Magazine for PHP Professionals – www.phparch.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Parse error on array and SQL query
My host provider suggested escaping the SQL: $query .= "ID=\'$this\'"; which gets rid of the php error but gives invalid SQL. After a few exchanges with them it suddenly worked again as: $query .= "ID='$this'";. -Original Message- From: Hutchins, Richard [mailto:[EMAIL PROTECTED] Sent: 20 November 2003 14:43 To: 'Dillon, John'; [EMAIL PROTECTED] Subject: RE: [PHP-DB] Parse error on array and SQL query Speaking for myself only, I'm probably going to need to see more code than this to be able to help. These two lines appear to be OK. The error might be in what feeds line 187. > -Original Message- > From: Dillon, John [mailto:[EMAIL PROTECTED] > Sent: Thursday, November 20, 2003 9:16 AM > To: [EMAIL PROTECTED] > Subject: [PHP-DB] Parse error on array and SQL query > > > > On line 188 and similar: > line 187$this=$altIDs[$i]; > line 188$query .= "ID='$this'"; > > I get the error: Parse error: parse error in > /home/jdillon/public_html/provreport.php on line 188 > > Could this be because some setting has been changed on my shared host? > > John > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > http://www.cantor.com > CONFIDENTIAL: This e-mail, including its contents and > attachments, if any, are confidential. If you are not the > named recipient please notify the sender and immediately > delete it. You may not disseminate, distribute, or forward > this e-mail message or disclose its contents to anybody else. > Copyright and any other intellectual property rights in its > contents are the sole property of Cantor Fitzgerald. > E-mail transmission cannot be guaranteed to be secure or > error-free. The sender therefore does not accept liability > for any errors or omissions in the contents of this message > which arise as a result of e-mail transmission. If > verification is required please request a hard-copy version. > Although we routinely screen for viruses, addressees > should check this e-mail and any attachments for viruses. We > make no representation or warranty as to the absence of > viruses in this e-mail or any attachments. Please note that > to ensure regulatory compliance and for the protection of our > customers and business, we may monitor and read e-mails sent > to and from our server(s). > > For further important information, please read the Important > Legal Information and Legal Statement at > http://www.cantor.com/legal_information.html > > -- > 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] Parse error on array and SQL query
Speaking for myself only, I'm probably going to need to see more code than this to be able to help. These two lines appear to be OK. The error might be in what feeds line 187. > -Original Message- > From: Dillon, John [mailto:[EMAIL PROTECTED] > Sent: Thursday, November 20, 2003 9:16 AM > To: [EMAIL PROTECTED] > Subject: [PHP-DB] Parse error on array and SQL query > > > > On line 188 and similar: > line 187$this=$altIDs[$i]; > line 188$query .= "ID='$this'"; > > I get the error: Parse error: parse error in > /home/jdillon/public_html/provreport.php on line 188 > > Could this be because some setting has been changed on my shared host? > > John > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > http://www.cantor.com > CONFIDENTIAL: This e-mail, including its contents and > attachments, if any, are confidential. If you are not the > named recipient please notify the sender and immediately > delete it. You may not disseminate, distribute, or forward > this e-mail message or disclose its contents to anybody else. > Copyright and any other intellectual property rights in its > contents are the sole property of Cantor Fitzgerald. > E-mail transmission cannot be guaranteed to be secure or > error-free. The sender therefore does not accept liability > for any errors or omissions in the contents of this message > which arise as a result of e-mail transmission. If > verification is required please request a hard-copy version. > Although we routinely screen for viruses, addressees > should check this e-mail and any attachments for viruses. We > make no representation or warranty as to the absence of > viruses in this e-mail or any attachments. Please note that > to ensure regulatory compliance and for the protection of our > customers and business, we may monitor and read e-mails sent > to and from our server(s). > > For further important information, please read the Important > Legal Information and Legal Statement at > http://www.cantor.com/legal_information.html > > -- > 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] parse error.
Hi Micella
You need to either slash-escape your insert query or use single quotes:
$query="insert into dene (day,start_time) values (\"2003.09.25\",\"22:30\") " ;
or
$query="insert into dene (day,start_time) values ('2003.09.25','22:30') " ;
steve
M|cella Erdem Efe wrote:
I am a new programmer of PHP4.3.3 an MySql 3.23 .
When below script is executed
"Parse error: parse error, unexpected T_DNUMBER in c:\program
files\apache group\apache\htdocs\ilkdeneme21.php on line 17"
is being seen.
This may be very simple but I could not solve it.Could you help me for this
error?
Thanks for your attention
[EMAIL PROTECTED]
mysql_connect("localhost","username");
if (mysql_select_db($dbname))
{ echo "connected";}
else
{ echo "not connected";}
$query="insert into dene (day,start_time) values ("2003.09.25","22:30") " ;
$result=mysql_query($query);
if ($result))
{ echo "data girildi";}
else
{ echo "data girilemedi";
exit;}
?>
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Re: [PHP-DB] Parse error
[EMAIL PROTECTED] wrote: IBI> I am having a problem with a parse error. IBI> error: IBI> Parse error: parse error, unexpected ')' in IBI> /Users/timbest/Sites/cajunmikes/inc/menupage.php on line 75 IBI> Fatal error: Cannot instantiate non-existent class: menu in IBI> /Users/timbest/Sites/cajunmikes/TMPfj18cc2i20.php on line 7 IBI> The issue I am having is line 75 has no ')'. I am sure this error is IBI> causing the next fatal error. Anyone seen this before? I'm not sure what IBI> to do next since there is nothing on line 75... IBI> Thanks IBI> /T Post some code here... WBR, Max 'AMiGo' Gashkov [EMAIL PROTECTED] ]=[ http://diary.otaku.ru/amigo http://www.journals.ru/users/endymion Distributed.net participant [408228][RC5-72] __ NP: [DJ DADO/ FINE ARTS] next madness -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Parse Error
if ($packages == "1"){
if ($airporttransfer == "car"){
$airporttransfer2 = "12.00"};
if ($airporttransfer == "bus"){
$airporttransfer2 = "10.00"};
if ($airporttransfer == "none"){
$airporttransfer2 = "0.00"};
};
() round brackets round the condition {} braces around the 'do it' bit :)
HTH
Peter
-Original Message-
From: Chris Payne [mailto:[EMAIL PROTECTED]
Sent: 27 February 2003 00:43
To: php
Subject: [PHP-DB] Parse Error
Hi there everyone,
I'm trying to run an if statement and inside that have other if's, but it
keeps telling me I have a parse error on the first if in the if statement
(The second if down). Can anyone see what i'm doing wrong?
if ($packages == "1")(
if ($airporttransfer == "car")(
$airporttransfer2 = "12.00");
if ($airporttransfer == "bus")(
$airporttransfer2 = "10.00");
if ($airporttransfer == "none")(
$airporttransfer2 = "0.00");
);
Thanks for your help :-)
Chris
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Re: [PHP-DB] Parse Error
If statements use curly braces. Do this:
if ($packages == "1"){
if ($airporttransfer == "car")
$airporttransfer2 = "12.00");
if ($airporttransfer == "bus")
$airporttransfer2 = "10.00");
if ($airporttransfer == "none")
$airporttransfer2 = "0.00");
}
On Wed, 2003-02-26 at 16:42, Chris Payne wrote:
> Hi there everyone,
>
> I'm trying to run an if statement and inside that have other if's, but it keeps
> telling me I have a parse error on the first if in the if statement (The second if
> down). Can anyone see what i'm doing wrong?
>
> if ($packages == "1")(
>
> if ($airporttransfer == "car")(
> $airporttransfer2 = "12.00");
> if ($airporttransfer == "bus")(
> $airporttransfer2 = "10.00");
> if ($airporttransfer == "none")(
> $airporttransfer2 = "0.00");
>
> );
>
> Thanks for your help :-)
>
> Chris
Re: [PHP-DB] Parse Error...
Thank you!! This appears to have solved it... I didn't even think of using
the "$message" variable instead of the "print()"... Sheesh... I still have
LOTS to learn!!
Chase
"Micah Stevens" <[EMAIL PROTECTED]> wrote in message
1038878012.23696.129.camel@tyrell">news:1038878012.23696.129.camel@tyrell...
> Remember, the print() function sends stuff to the browser. If you wish
> to make the code output to the $message variable, you must use
> additional assignment statements.
>
> I made some quick changes to the code below, try that:
>
> $server = "";
> $user = "";
> $pass = "";
> $db = "";
> $table = "";
> ?>
>
> $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> $to .= "Chase <[EMAIL PROTECTED]>";
>
> $subject = "Price File Access History";
>
> $message = '
>
> Price File Access
>
>
>
>
>
> Date / Time
>
>
> Logged IP Address
>
>
> Logged Username
>
>
> ';
>
> $link = mysql_connect("$server", "$user", "$pass");
> mysql_select_db("$db", $link);
>
> $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> DESC",$link);
> $num_rows = mysql_num_rows($result);
>
> if($num_rows) {
> // print("");
>while($row = mysql_fetch_row($result))
>
>{
>
> $message .= "";
> $message .= "$row[0]";
> $message .= "$row[1]";
> $message .= "$row[2]";
> $message .= "";
>
>
>}
>$message .= "";
> } else {
>$message .= "No Files To Show!!";
>
> }
> $message .= "
>
> ";
>
> $headers = "MIME-Version: 1.0\r\n";
> $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
>
> $headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
>
> mail($to, $subject, $message, $headers);
>
> ?>
>
>
>
> On Mon, 2002-12-02 at 16:30, Chase wrote:
> > There is a single quote and semicolon on the line under that I
> > thought would be defining the end of the assignment.
> > Should I have used double quotes instead of single?
> >
> > Chase
> >
> >
> > "Micah Stevens" <[EMAIL PROTECTED]> wrote in message
> > 1038870535.23725.123.camel@tyrell">news:1038870535.23725.123.camel@tyrell...
> > > There aren't any closing quotes on your $message variable or semicolon
> > > to tell the parser that you're done with the assignment statement.
> > >
> > > -Micah
> > >
> > >
> > >
> > >
> > >
> > > On Mon, 2002-12-02 at 15:05, Chase wrote:
> > > > When trying to execute the following script to send the contents of
a
> > table
> > > > via "mail()" I am getting this error...
> > > >
> > > > Parse error: parse error, unexpected T_STRING in FILENAME.PHP on
line 36
> > > >
> > > > The portion of this code inside the HTML tags works great for
displaying
> > the
> > > > table on the screen, but when I include it in the larger script for
> > sending
> > > > a mail message, I get this error.
> > > >
> > > > I am sure that it is something simple that I have been too stupid to
> > catch,
> > > > but any help would be great!
> > > >
> > > > Chase
> > > >
> > > >
> > > > Code Follows --->
> > > >
> > > > > > > $server = "";
> > > > $user = "";
> > > > $pass = "";
> > > > $db = "";
> > > > $table = "";
> > > > ?>
> > > >
> > > > > > > $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> > > > $to .= "Chase <[EMAIL PROTECTED]>";
> > > >
> > > > $subject = "Price File Access History";
> > > >
> > > > $message = '
> > > >
> > > > Price File Access
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Date / Time
> > > >
> > > >
> > > > Logged IP Address
> > > >
> > > >
> > > > Logged Username
> > > >
> > > >
> > > >
> > > > $link = mysql_connect("$server", "$user", "$pass");
> > > > mysql_select_db("$db", $link);
> > > >
> > > > $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> > > > DESC",$link);
> > > > $num_rows = mysql_num_rows($result);
> > > >
> > > > if($num_rows) {
> > > > // print("");
> > > >while($row = mysql_fetch_row($result))
> > > >
> > > >{
> > > >
> > > > print("");
> > > > print("$row[0]");
> > > > print("$row[1]");
> > > > print("$row[2]");
> > > > print("");
> > > >
> > > >
> > > >}
> > > >print("");
> > > > } else {
> > > >print("No Files To Show!!");
> > > >
> > > > }
> > > >
> > > >
> > > >
> > > > ';
> > > >
> > > > $headers = "MIME-Version: 1.0\r\n";
> > > > $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
> > > >
> > > > $headers .= "From: MIC Price File Monitor
<[EMAIL PROTECTED]>\r\n";
> > > >
> > > > mail($to, $subject, $message, $headers);
> > > >
> > > > ?>
> > > >
> > > --
> > > Raincross Technologies
> > > Development and Consulting Services
> > > http://www.raincross-tech.com
> > >
> > >
> --
> Raincross Technologies
> Development and Consulting Services
> http://www.raincross-tech.com
>
>
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PHP Database Mailing List (http://www.php.net/)
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Re: [PHP-DB] Parse Error...
Remember, the print() function sends stuff to the browser. If you wish
to make the code output to the $message variable, you must use
additional assignment statements.
I made some quick changes to the code below, try that:
" . ", " ;
$to .= "Chase <[EMAIL PROTECTED]>";
$subject = "Price File Access History";
$message = '
Price File Access
Date / Time
Logged IP Address
Logged Username
';
$link = mysql_connect("$server", "$user", "$pass");
mysql_select_db("$db", $link);
$result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
DESC",$link);
$num_rows = mysql_num_rows($result);
if($num_rows) {
// print("");
while($row = mysql_fetch_row($result))
{
$message .= "";
$message .= "$row[0]";
$message .= "$row[1]";
$message .= "$row[2]";
$message .= "";
}
$message .= "";
} else {
$message .= "No Files To Show!!";
}
$message .= "
";
$headers = "MIME-Version: 1.0\r\n";
$headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
$headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
mail($to, $subject, $message, $headers);
?>
On Mon, 2002-12-02 at 16:30, Chase wrote:
> There is a single quote and semicolon on the line under that I
> thought would be defining the end of the assignment.
> Should I have used double quotes instead of single?
>
> Chase
>
>
> "Micah Stevens" <[EMAIL PROTECTED]> wrote in message
> 1038870535.23725.123.camel@tyrell">news:1038870535.23725.123.camel@tyrell...
> > There aren't any closing quotes on your $message variable or semicolon
> > to tell the parser that you're done with the assignment statement.
> >
> > -Micah
> >
> >
> >
> >
> >
> > On Mon, 2002-12-02 at 15:05, Chase wrote:
> > > When trying to execute the following script to send the contents of a
> table
> > > via "mail()" I am getting this error...
> > >
> > > Parse error: parse error, unexpected T_STRING in FILENAME.PHP on line 36
> > >
> > > The portion of this code inside the HTML tags works great for displaying
> the
> > > table on the screen, but when I include it in the larger script for
> sending
> > > a mail message, I get this error.
> > >
> > > I am sure that it is something simple that I have been too stupid to
> catch,
> > > but any help would be great!
> > >
> > > Chase
> > >
> > >
> > > Code Follows --->
> > >
> > > > > $server = "";
> > > $user = "";
> > > $pass = "";
> > > $db = "";
> > > $table = "";
> > > ?>
> > >
> > > > > $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> > > $to .= "Chase <[EMAIL PROTECTED]>";
> > >
> > > $subject = "Price File Access History";
> > >
> > > $message = '
> > >
> > > Price File Access
> > >
> > >
> > >
> > >
> > >
> > > Date / Time
> > >
> > >
> > > Logged IP Address
> > >
> > >
> > > Logged Username
> > >
> > >
> > >
> > > $link = mysql_connect("$server", "$user", "$pass");
> > > mysql_select_db("$db", $link);
> > >
> > > $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> > > DESC",$link);
> > > $num_rows = mysql_num_rows($result);
> > >
> > > if($num_rows) {
> > > // print("");
> > >while($row = mysql_fetch_row($result))
> > >
> > >{
> > >
> > > print("");
> > > print("$row[0]");
> > > print("$row[1]");
> > > print("$row[2]");
> > > print("");
> > >
> > >
> > >}
> > >print("");
> > > } else {
> > >print("No Files To Show!!");
> > >
> > > }
> > >
> > >
> > >
> > > ';
> > >
> > > $headers = "MIME-Version: 1.0\r\n";
> > > $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
> > >
> > > $headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
> > >
> > > mail($to, $subject, $message, $headers);
> > >
> > > ?>
> > >
> > --
> > Raincross Technologies
> > Development and Consulting Services
> > http://www.raincross-tech.com
> >
> >
--
Raincross Technologies
Development and Consulting Services
http://www.raincross-tech.com
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Re: [PHP-DB] Parse Error...
Okay, you just jumped WAY outside my basic knowledge... However, it does
make perfect sense, so I am off to the library...
Chase
"Jim Hunter" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
In that case, your logic is not going to work. You have live PHP code inside
a string. You are going to need to build the $message in pieces. Start with
the HTML, then add the result from each itteration of your PHP loop to
$message then add the trailing HTML to form the complete $message.
Your close, but it's just not going to work like you think it will.
Jim
---Original Message---
From: Chase
Date: Monday, December 02, 2002 04:34:18 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error...
There is a single quote and semicolon on the line under that I
thought would be defining the end of the assignment.
Should I have used double quotes instead of single?
Chase
"Micah Stevens" <[EMAIL PROTECTED]> wrote in message
1038870535.23725.123.camel@tyrell">news:1038870535.23725.123.camel@tyrell...
> There aren't any closing quotes on your $message variable or semicolon
> to tell the parser that you're done with the assignment statement.
>
> -Micah
>
>
>
>
>
> On Mon, 2002-12-02 at 15:05, Chase wrote:
> > When trying to execute the following script to send the contents of a
table
> > via "mail()" I am getting this error...
> >
> > Parse error: parse error, unexpected T_STRING in FILENAME.PHP on line 36
> >
> > The portion of this code inside the HTML tags works great for displaying
the
> > table on the screen, but when I include it in the larger script for
sending
> > a mail message, I get this error.
> >
> > I am sure that it is something simple that I have been too stupid to
catch,
> > but any help would be great!
> >
> > Chase
> >
> >
> > Code Follows --->
> >
> > > $server = "";
> > $user = "";
> > $pass = "";
> > $db = "";
> > $table = "";
> > ?>
> >
> > > $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> > $to .= "Chase <[EMAIL PROTECTED]>";
> >
> > $subject = "Price File Access History";
> >
> > $message = '
> >
> > Price File Access
> >
> >
> >
> >
> >
> > Date / Time
> >
> >
> > Logged IP Address
> >
> >
> > Logged Username
> >
> >
> >
> > $link = mysql_connect("$server", "$user", "$pass");
> > mysql_select_db("$db", $link);
> >
> > $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> > DESC",$link);
> > $num_rows = mysql_num_rows($result);
> >
> > if($num_rows) {
> > // print("");
> > while($row = mysql_fetch_row($result))
> >
> > {
> >
> > print("");
> > print("$row[0]");
> > print("$row[1]");
> > print("$row[2]");
> > print("");
> >
> >
> > }
> > print("");
> > } else {
> > print("No Files To Show!!");
> >
> > }
> >
> >
> >
> > ';
> >
> > $headers = "MIME-Version: 1.0\r\n";
> > $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
> >
> > $headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
> >
> > mail($to, $subject, $message, $headers);
> >
> > ?>
> >
> --
> Raincross Technologies
> Development and Consulting Services
> http://www.raincross-tech.com
>
>
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Re: [PHP-DB] Parse Error...
In that case, your logic is not going to work. You have live PHP code inside
a string. You are going to need to build the $message in pieces. Start with
the HTML, then add the result from each itteration of your PHP loop to
$message then add the trailing HTML to form the complete $message.
Your close, but it's just not going to work like you think it will.
Jim
---Original Message---
From: Chase
Date: Monday, December 02, 2002 04:34:18 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error...
There is a single quote and semicolon on the line under that I
thought would be defining the end of the assignment.
Should I have used double quotes instead of single?
Chase
"Micah Stevens" <[EMAIL PROTECTED]> wrote in message
1038870535.23725.123.camel@tyrell">news:1038870535.23725.123.camel@tyrell...
> There aren't any closing quotes on your $message variable or semicolon
> to tell the parser that you're done with the assignment statement.
>
> -Micah
>
>
>
>
>
> On Mon, 2002-12-02 at 15:05, Chase wrote:
> > When trying to execute the following script to send the contents of a
table
> > via "mail()" I am getting this error...
> >
> > Parse error: parse error, unexpected T_STRING in FILENAME.PHP on line 36
> >
> > The portion of this code inside the HTML tags works great for displaying
the
> > table on the screen, but when I include it in the larger script for
sending
> > a mail message, I get this error.
> >
> > I am sure that it is something simple that I have been too stupid to
catch,
> > but any help would be great!
> >
> > Chase
> >
> >
> > Code Follows --->
> >
> > > $server = "";
> > $user = "";
> > $pass = "";
> > $db = "";
> > $table = "";
> > ?>
> >
> > > $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> > $to .= "Chase <[EMAIL PROTECTED]>";
> >
> > $subject = "Price File Access History";
> >
> > $message = '
> >
> > Price File Access
> >
> >
> >
> >
> >
> > Date / Time
> >
> >
> > Logged IP Address
> >
> >
> > Logged Username
> >
> >
> >
> > $link = mysql_connect("$server", "$user", "$pass");
> > mysql_select_db("$db", $link);
> >
> > $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> > DESC",$link);
> > $num_rows = mysql_num_rows($result);
> >
> > if($num_rows) {
> > // print("");
> > while($row = mysql_fetch_row($result))
> >
> > {
> >
> > print("");
> > print("$row[0]");
> > print("$row[1]");
> > print("$row[2]");
> > print("");
> >
> >
> > }
> > print("");
> > } else {
> > print("No Files To Show!!");
> >
> > }
> >
> >
> >
> > ';
> >
> > $headers = "MIME-Version: 1.0\r\n";
> > $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
> >
> > $headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
> >
> > mail($to, $subject, $message, $headers);
> >
> > ?>
> >
> --
> Raincross Technologies
> Development and Consulting Services
> http://www.raincross-tech.com
>
>
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.
Re: [PHP-DB] Parse Error...
There is a single quote and semicolon on the line under that I
thought would be defining the end of the assignment.
Should I have used double quotes instead of single?
Chase
"Micah Stevens" <[EMAIL PROTECTED]> wrote in message
1038870535.23725.123.camel@tyrell">news:1038870535.23725.123.camel@tyrell...
> There aren't any closing quotes on your $message variable or semicolon
> to tell the parser that you're done with the assignment statement.
>
> -Micah
>
>
>
>
>
> On Mon, 2002-12-02 at 15:05, Chase wrote:
> > When trying to execute the following script to send the contents of a
table
> > via "mail()" I am getting this error...
> >
> > Parse error: parse error, unexpected T_STRING in FILENAME.PHP on line 36
> >
> > The portion of this code inside the HTML tags works great for displaying
the
> > table on the screen, but when I include it in the larger script for
sending
> > a mail message, I get this error.
> >
> > I am sure that it is something simple that I have been too stupid to
catch,
> > but any help would be great!
> >
> > Chase
> >
> >
> > Code Follows --->
> >
> > > $server = "";
> > $user = "";
> > $pass = "";
> > $db = "";
> > $table = "";
> > ?>
> >
> > > $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> > $to .= "Chase <[EMAIL PROTECTED]>";
> >
> > $subject = "Price File Access History";
> >
> > $message = '
> >
> > Price File Access
> >
> >
> >
> >
> >
> > Date / Time
> >
> >
> > Logged IP Address
> >
> >
> > Logged Username
> >
> >
> >
> > $link = mysql_connect("$server", "$user", "$pass");
> > mysql_select_db("$db", $link);
> >
> > $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> > DESC",$link);
> > $num_rows = mysql_num_rows($result);
> >
> > if($num_rows) {
> > // print("");
> >while($row = mysql_fetch_row($result))
> >
> >{
> >
> > print("");
> > print("$row[0]");
> > print("$row[1]");
> > print("$row[2]");
> > print("");
> >
> >
> >}
> >print("");
> > } else {
> >print("No Files To Show!!");
> >
> > }
> >
> >
> >
> > ';
> >
> > $headers = "MIME-Version: 1.0\r\n";
> > $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
> >
> > $headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
> >
> > mail($to, $subject, $message, $headers);
> >
> > ?>
> >
> --
> Raincross Technologies
> Development and Consulting Services
> http://www.raincross-tech.com
>
>
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Re: [PHP-DB] Parse Error...
There aren't any closing quotes on your $message variable or semicolon
to tell the parser that you're done with the assignment statement.
-Micah
On Mon, 2002-12-02 at 15:05, Chase wrote:
> When trying to execute the following script to send the contents of a table
> via "mail()" I am getting this error...
>
> Parse error: parse error, unexpected T_STRING in FILENAME.PHP on line 36
>
> The portion of this code inside the HTML tags works great for displaying the
> table on the screen, but when I include it in the larger script for sending
> a mail message, I get this error.
>
> I am sure that it is something simple that I have been too stupid to catch,
> but any help would be great!
>
> Chase
>
>
> Code Follows --->
>
> $server = "";
> $user = "";
> $pass = "";
> $db = "";
> $table = "";
> ?>
>
> $to = "CKnott <[EMAIL PROTECTED]>" . ", " ;
> $to .= "Chase <[EMAIL PROTECTED]>";
>
> $subject = "Price File Access History";
>
> $message = '
>
> Price File Access
>
>
>
>
>
> Date / Time
>
>
> Logged IP Address
>
>
> Logged Username
>
>
>
> $link = mysql_connect("$server", "$user", "$pass");
> mysql_select_db("$db", $link);
>
> $result = mysql_query("SELECT * FROM $table ORDER BY 'view_date'
> DESC",$link);
> $num_rows = mysql_num_rows($result);
>
> if($num_rows) {
> // print("");
>while($row = mysql_fetch_row($result))
>
>{
>
> print("");
> print("$row[0]");
> print("$row[1]");
> print("$row[2]");
> print("");
>
>
>}
>print("");
> } else {
>print("No Files To Show!!");
>
> }
>
>
>
> ';
>
> $headers = "MIME-Version: 1.0\r\n";
> $headers .= "Content-type: text/html; charset=iso-8859-1\r\n";
>
> $headers .= "From: MIC Price File Monitor <[EMAIL PROTECTED]>\r\n";
>
> mail($to, $subject, $message, $headers);
>
> ?>
>
--
Raincross Technologies
Development and Consulting Services
http://www.raincross-tech.com
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Re: [PHP-DB] Parse Error
According to your code, if the checkbox next to the word 'remove' is checked,
then the value of $cars passed to the form is "checkbox". Then in the
done2.php3 page, you are trying to delete where checkbox='array of values
listed'.
Is there a field in your database called checkbox? If so, what field type is
it set to? If not, there is your problem.
HTH
MB
jas <[EMAIL PROTECTED]> said:
> Here is the code from the file that queries the db and pulls the current
> contents of the db and provides a check box form for each record to delete
> the items in the db. I dont know if this will help but like I said before
> any insight would be great. Thanks in advance.
> Jas
>
> require '../scripts/db.php';
> $result = mysql_query("SELECT * FROM cur_inv",$dbh) or die("Could not
> execute query, please try again later");
> echo " name=\"rem_inv\" method=\"post\" action=\"done2.php3\">
> Current
> Inventory";
> $count = -1;
> while ($myrow = mysql_fetch_row($result)) {
> $count ++;
> echo "Type Of Car: ";
> printf(mysql_result($result,$count,"car_type"));
> echo " value=\"checkbox\">remove
> \n";
> echo "Model Of Car: ";
> printf(mysql_result($result,$count,"car_model"));
> echo "\n";
> echo "Year Of Car: ";
> printf(mysql_result($result,$count,"car_year"));
> echo "\n";
> echo "Price Of Car: $";
> printf(mysql_result($result,$count,"car_price"));
> echo "\n";
> echo "VIN Of Car: ";
> printf(mysql_result($result,$count,"car_vin"));
> echo "\n";
> }
> echo " value=\"delete\">";
> ?>
>
> After this file is pulled and the user selects which record to delete it
> jumps to the other snippit of code in done2.php3 which is
>
> > > $db_name = "test";
> > > $table_name = "inventory";
> > > $connection = @mysql_connect("localhost", "root", "password") or die
> > > ("Could
> > > not connect to database. Please try again later.");
> > > $db = @mysql_select_db("$db_name",$connection) or die ("Could not select
> > > database table. Please try again later.");
> > > $sql = "DELETE FROM $table_name WHERE $cars =
> > >
> \"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> > > r_num\"");
> > > $result = @mysql_query($sql, $connection) or die ("Could not execute
> > > query.
> > >
> > > Any insight would be great... thanks again.
> > > Jas
> > >
> > >
> > >
> > >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> >
> >
> >
> > --
> >
> >
> >
>
>
>
> --
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Re: [PHP-DB] Parse Error
Could you maybe give me an example of how to associate the checkbox with the
id? Thanks again,
Jas
"Gurhan Ozen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Ok then it sure will not work.. because you have :
>
> "DELETE FROM $table_name WHERE $cars
>
=\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$d
> lr_num\"");
>
> 1- $cars is set to the value "checkbox" which doesn't exist in your test
> table
> 2- The end of the statement $dlr_num\""); is not right , that
> parenthesis isn't supposed to be there
> 3- The delete statement is wrong...You are storing all that info in the
> separate fields but you are giving all combined as a condition in only one
> field ($cars).
> 3- Rewrite your delete statement to delete the rows identified by the id
> column as "DELETE FROM $table WHERE id ='id'"; Of course in the page where
> you list the cars to be deleted associate each checkbox with the car's id.
>
> Hope this helps..
>
> Gurhan
>
>
> -Original Message-
> From: jas [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, February 21, 2002 1:02 AM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Parse Error
>
>
> Here is the dump for the cars table...
> test (
>id varchar(30) NOT NULL auto_increment,
>car_type varchar(30),
>car_model varchar(30),
>car_year varchar(15),
>car_price varchar(15),
>car_vin varchar(25),
>dlr_num varchar(25),
>PRIMARY KEY (id),
>KEY id (id)
> test VALUES ('1', 'Ford', 'Bronco', '1969', '4500', 'vin897655', 'none');
>
> Sorry about that... still a bit of a newbie here. =)
> Jas
>
> "Gurhan Ozen" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > Jas..
> > First of all you don't have the variable $cars assigned..
> > Second of all, we still don't know what the "test" table looks like?
Does
> it
> > only have one column with all the infos about cars populated in it???
> >
> > Gurhan
> >
> >
> > -Original Message-
> > From: jas [mailto:[EMAIL PROTECTED]]
> > Sent: Thursday, February 21, 2002 12:48 AM
> > To: [EMAIL PROTECTED]
> > Subject: Re: [PHP-DB] Parse Error
> >
> >
> > Here is the code from the file that queries the db and pulls the current
> > contents of the db and provides a check box form for each record to
delete
> > the items in the db. I dont know if this will help but like I said
before
> > any insight would be great. Thanks in advance.
> > Jas
> >
> > > require '../scripts/db.php';
> > $result = mysql_query("SELECT * FROM cur_inv",$dbh) or die("Could not
> > execute query, please try again later");
> > echo " > name=\"rem_inv\" method=\"post\" action=\"done2.php3\">
> > Current
> > Inventory";
> > $count = -1;
> > while ($myrow = mysql_fetch_row($result)) {
> > $count ++;
> > echo "Type Of Car: ";
> > printf(mysql_result($result,$count,"car_type"));
> > echo " > value=\"checkbox\">remove
> > \n";
> > echo "Model Of Car: ";
> > printf(mysql_result($result,$count,"car_model"));
> > echo "\n";
> > echo "Year Of Car: ";
> > printf(mysql_result($result,$count,"car_year"));
> > echo "\n";
> > echo "Price Of Car: $";
> > printf(mysql_result($result,$count,"car_price"));
> > echo "\n";
> > echo "VIN Of Car: ";
> > printf(mysql_result($result,$count,"car_vin"));
> > echo "\n";
> > }
> > echo " > value=\"delete\">";
> > ?>
> >
> > After this file is pulled and the user selects which record to delete it
> > jumps to the other snippit of code in done2.php3 which is
> >
> > > > $db_name = "test";
> > > > $table_name = "inventory";
> > > > $connection = @mysql_connect("localhost", "root", "password") or die
> > > > ("Could
> > > > not connect to database. Please try again later.");
> > > > $db = @mysql_select_db("$db_name",$connection) or die ("Could not
> select
> > > > database table. Please try again later.");
> > > > $sql = "DELETE FROM $table_name WHERE $cars =
> > > >
> >
>
\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> > > > r_num\"");
> > > > $result = @mysql_query($sql, $connection) or die ("Could not execute
> > > > query.
> > > >
> > > > Any insight would be great... thanks again.
> > > > 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
> >
>
>
>
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RE: [PHP-DB] Parse Error
Ok then it sure will not work.. because you have :
"DELETE FROM $table_name WHERE $cars
=\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$d
lr_num\"");
1- $cars is set to the value "checkbox" which doesn't exist in your test
table
2- The end of the statement $dlr_num\""); is not right , that
parenthesis isn't supposed to be there
3- The delete statement is wrong...You are storing all that info in the
separate fields but you are giving all combined as a condition in only one
field ($cars).
3- Rewrite your delete statement to delete the rows identified by the id
column as "DELETE FROM $table WHERE id ='id'"; Of course in the page where
you list the cars to be deleted associate each checkbox with the car's id.
Hope this helps..
Gurhan
-Original Message-----
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 1:02 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the dump for the cars table...
test (
id varchar(30) NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id),
KEY id (id)
test VALUES ('1', 'Ford', 'Bronco', '1969', '4500', 'vin897655', 'none');
Sorry about that... still a bit of a newbie here. =)
Jas
"Gurhan Ozen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Jas..
> First of all you don't have the variable $cars assigned..
> Second of all, we still don't know what the "test" table looks like? Does
it
> only have one column with all the infos about cars populated in it???
>
> Gurhan
>
>
> -Original Message-
> From: jas [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, February 21, 2002 12:48 AM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Parse Error
>
>
> Here is the code from the file that queries the db and pulls the current
> contents of the db and provides a check box form for each record to delete
> the items in the db. I dont know if this will help but like I said before
> any insight would be great. Thanks in advance.
> Jas
>
> require '../scripts/db.php';
> $result = mysql_query("SELECT * FROM cur_inv",$dbh) or die("Could not
> execute query, please try again later");
> echo " name=\"rem_inv\" method=\"post\" action=\"done2.php3\">
> Current
> Inventory";
> $count = -1;
> while ($myrow = mysql_fetch_row($result)) {
> $count ++;
> echo "Type Of Car: ";
> printf(mysql_result($result,$count,"car_type"));
> echo " value=\"checkbox\">remove
> \n";
> echo "Model Of Car: ";
> printf(mysql_result($result,$count,"car_model"));
> echo "\n";
> echo "Year Of Car: ";
> printf(mysql_result($result,$count,"car_year"));
> echo "\n";
> echo "Price Of Car: $";
> printf(mysql_result($result,$count,"car_price"));
> echo "\n";
> echo "VIN Of Car: ";
> printf(mysql_result($result,$count,"car_vin"));
> echo "\n";
> }
> echo " value=\"delete\">";
> ?>
>
> After this file is pulled and the user selects which record to delete it
> jumps to the other snippit of code in done2.php3 which is
>
> > > $db_name = "test";
> > > $table_name = "inventory";
> > > $connection = @mysql_connect("localhost", "root", "password") or die
> > > ("Could
> > > not connect to database. Please try again later.");
> > > $db = @mysql_select_db("$db_name",$connection) or die ("Could not
select
> > > database table. Please try again later.");
> > > $sql = "DELETE FROM $table_name WHERE $cars =
> > >
>
\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> > > r_num\"");
> > > $result = @mysql_query($sql, $connection) or die ("Could not execute
> > > query.
> > >
> > > Any insight would be great... thanks again.
> > > 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
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Re: [PHP-DB] Parse Error
Here is the dump for the cars table...
test (
id varchar(30) NOT NULL auto_increment,
car_type varchar(30),
car_model varchar(30),
car_year varchar(15),
car_price varchar(15),
car_vin varchar(25),
dlr_num varchar(25),
PRIMARY KEY (id),
KEY id (id)
test VALUES ('1', 'Ford', 'Bronco', '1969', '4500', 'vin897655', 'none');
Sorry about that... still a bit of a newbie here. =)
Jas
"Gurhan Ozen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Jas..
> First of all you don't have the variable $cars assigned..
> Second of all, we still don't know what the "test" table looks like? Does
it
> only have one column with all the infos about cars populated in it???
>
> Gurhan
>
>
> -Original Message-----
> From: jas [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, February 21, 2002 12:48 AM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] Parse Error
>
>
> Here is the code from the file that queries the db and pulls the current
> contents of the db and provides a check box form for each record to delete
> the items in the db. I dont know if this will help but like I said before
> any insight would be great. Thanks in advance.
> Jas
>
> require '../scripts/db.php';
> $result = mysql_query("SELECT * FROM cur_inv",$dbh) or die("Could not
> execute query, please try again later");
> echo " name=\"rem_inv\" method=\"post\" action=\"done2.php3\">
> Current
> Inventory";
> $count = -1;
> while ($myrow = mysql_fetch_row($result)) {
> $count ++;
> echo "Type Of Car: ";
> printf(mysql_result($result,$count,"car_type"));
> echo " value=\"checkbox\">remove
> \n";
> echo "Model Of Car: ";
> printf(mysql_result($result,$count,"car_model"));
> echo "\n";
> echo "Year Of Car: ";
> printf(mysql_result($result,$count,"car_year"));
> echo "\n";
> echo "Price Of Car: $";
> printf(mysql_result($result,$count,"car_price"));
> echo "\n";
> echo "VIN Of Car: ";
> printf(mysql_result($result,$count,"car_vin"));
> echo "\n";
> }
> echo " value=\"delete\">";
> ?>
>
> After this file is pulled and the user selects which record to delete it
> jumps to the other snippit of code in done2.php3 which is
>
> > > $db_name = "test";
> > > $table_name = "inventory";
> > > $connection = @mysql_connect("localhost", "root", "password") or die
> > > ("Could
> > > not connect to database. Please try again later.");
> > > $db = @mysql_select_db("$db_name",$connection) or die ("Could not
select
> > > database table. Please try again later.");
> > > $sql = "DELETE FROM $table_name WHERE $cars =
> > >
>
\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> > > r_num\"");
> > > $result = @mysql_query($sql, $connection) or die ("Could not execute
> > > query.
> > >
> > > Any insight would be great... thanks again.
> > > Jas
> > >
> > >
> > >
> > >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> >
> >
> >
> > --
> >
> >
> >
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RE: [PHP-DB] Parse Error
Jas..
First of all you don't have the variable $cars assigned..
Second of all, we still don't know what the "test" table looks like? Does it
only have one column with all the infos about cars populated in it???
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 12:48 AM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Parse Error
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I said before
any insight would be great. Thanks in advance.
Jas
Current
Inventory";
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo "Type Of Car: ";
printf(mysql_result($result,$count,"car_type"));
echo "remove
\n";
echo "Model Of Car: ";
printf(mysql_result($result,$count,"car_model"));
echo "\n";
echo "Year Of Car: ";
printf(mysql_result($result,$count,"car_year"));
echo "\n";
echo "Price Of Car: $";
printf(mysql_result($result,$count,"car_price"));
echo "\n";
echo "VIN Of Car: ";
printf(mysql_result($result,$count,"car_vin"));
echo "\n";
}
echo "";
?>
After this file is pulled and the user selects which record to delete it
jumps to the other snippit of code in done2.php3 which is
> > $db_name = "test";
> > $table_name = "inventory";
> > $connection = @mysql_connect("localhost", "root", "password") or die
> > ("Could
> > not connect to database. Please try again later.");
> > $db = @mysql_select_db("$db_name",$connection) or die ("Could not select
> > database table. Please try again later.");
> > $sql = "DELETE FROM $table_name WHERE $cars =
> >
\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> > r_num\"");
> > $result = @mysql_query($sql, $connection) or die ("Could not execute
> > query.
> >
> > Any insight would be great... thanks again.
> > Jas
> >
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>
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RE: [PHP-DB] Parse Error
Not enough information given to figure out what's going on, but looking at
your query i see:
"WHERE $cars =
"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dlr
_num\"");
do you keep 6 different values separated by commas in the field referenced
by $cars??? If this is correct what values you are passing for all those
variables? Is the $cars variable set to a valid column name in the "test"
table?? What is the data type for the column that $cars references?
Gurhan
-Original Message-
From: jas [mailto:[EMAIL PROTECTED]]
Sent: Thursday, February 21, 2002 12:26 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Parse Error
I hate to post this again but I have looked in a couple of php and mysql
books but cannot seem to figure this one out. I am getting a parse error
when trying to use php to delete records from a table. The error I am
recieving is as follows
Parse error: parse error in /path/to/php/done2.php3 on line 22
Here is the file that is giving me the error, any help would be great... I
think that my problem is that I left out a variable to hold the $cars data
but I am not sure. Here is the code...
$db_name = "test";
$table_name = "inventory";
$connection = @mysql_connect("localhost", "root", "password") or die
("Could
not connect to database. Please try again later.");
$db = @mysql_select_db("$db_name",$connection) or die ("Could not select
database table. Please try again later.");
$sql = "DELETE FROM $table_name WHERE $cars =
\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
r_num\"");
$result = @mysql_query($sql, $connection) or die ("Could not execute
query.
Any insight would be great... thanks again.
Jas
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Re: [PHP-DB] Parse Error
Here is the code from the file that queries the db and pulls the current
contents of the db and provides a check box form for each record to delete
the items in the db. I dont know if this will help but like I said before
any insight would be great. Thanks in advance.
Jas
Current
Inventory";
$count = -1;
while ($myrow = mysql_fetch_row($result)) {
$count ++;
echo "Type Of Car: ";
printf(mysql_result($result,$count,"car_type"));
echo "remove
\n";
echo "Model Of Car: ";
printf(mysql_result($result,$count,"car_model"));
echo "\n";
echo "Year Of Car: ";
printf(mysql_result($result,$count,"car_year"));
echo "\n";
echo "Price Of Car: $";
printf(mysql_result($result,$count,"car_price"));
echo "\n";
echo "VIN Of Car: ";
printf(mysql_result($result,$count,"car_vin"));
echo "\n";
}
echo "";
?>
After this file is pulled and the user selects which record to delete it
jumps to the other snippit of code in done2.php3 which is
> > $db_name = "test";
> > $table_name = "inventory";
> > $connection = @mysql_connect("localhost", "root", "password") or die
> > ("Could
> > not connect to database. Please try again later.");
> > $db = @mysql_select_db("$db_name",$connection) or die ("Could not select
> > database table. Please try again later.");
> > $sql = "DELETE FROM $table_name WHERE $cars =
> >
\"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> > r_num\"");
> > $result = @mysql_query($sql, $connection) or die ("Could not execute
> > query.
> >
> > Any insight would be great... thanks again.
> > Jas
> >
> >
> >
> >
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>
>
>
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>
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Re: [PHP-DB] Parse Error
Is $cars an array field? If not, you are trying to compare $cars to an
list/array of values (I am not sure this would work even if $cars was an
array field) 'WHERE $cars
= "$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dlr
_num\"");'.
Normally, you would have to compare each individual value in each requested
field to a variable.
Hard to say without more code.
MB
jas <[EMAIL PROTECTED]> said:
> I hate to post this again but I have looked in a couple of php and mysql
> books but cannot seem to figure this one out. I am getting a parse error
> when trying to use php to delete records from a table. The error I am
> recieving is as follows
>
> Parse error: parse error in /path/to/php/done2.php3 on line 22
>
> Here is the file that is giving me the error, any help would be great... I
> think that my problem is that I left out a variable to hold the $cars data
> but I am not sure. Here is the code...
>
> $db_name = "test";
> $table_name = "inventory";
> $connection = @mysql_connect("localhost", "root", "password") or die
> ("Could
> not connect to database. Please try again later.");
> $db = @mysql_select_db("$db_name",$connection) or die ("Could not select
> database table. Please try again later.");
> $sql = "DELETE FROM $table_name WHERE $cars =
> \"$car_type\",\"$car_model\",\"$car_year\",\"$car_price\",\"$car_vin\",\"$dl
> r_num\"");
> $result = @mysql_query($sql, $connection) or die ("Could not execute
> query.
>
> Any insight would be great... thanks again.
> Jas
>
>
>
>
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RE: [PHP-DB] Parse error
Hi Jenn, The code looked ok to me , so I copied and pasted the code you posted and it worked for me.. Where are you getting the parsing error? Is it the whole code anyway? Gurhan -Original Message- From: Jennifer Downey [mailto:[EMAIL PROTECTED]] Sent: Thursday, February 21, 2002 9:33 AM To: [EMAIL PROTECTED] Subject: [PHP-DB] Parse error Can someone tell me why I am getting a parse error in this little snip? Original value is $origVar"; $origVar += 25; echo "Added a value, now it's $origVar"; $origVar -= 12; echo "Subtracted a value, now it's $origVar"; $origVar .= " chickens"; echo "Final answer: $origVar"; ?> Thanks in advance Jen Downey -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Parse error
try: Original value is".$origVar.""; $origVar += 25; echo "Added a value, now it's ".$origVar.""; $origVar -= 12; echo "Subtracted a value, now it's ".$origVar.""; $origVar .= " chickens"; echo "Final answer: ".$origVar.""; ?> Cami -Original Message- From: Jennifer Downey [mailto:[EMAIL PROTECTED]] Sent: Thursday, February 21, 2002 2:33 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] Parse error Can someone tell me why I am getting a parse error in this little snip? Original value is $origVar"; $origVar += 25; echo "Added a value, now it's $origVar"; $origVar -= 12; echo "Subtracted a value, now it's $origVar"; $origVar .= " chickens"; echo "Final answer: $origVar"; ?> Thanks in advance Jen Downey -- 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] parse error
On Thu, 2001-12-13 at 23:49, Rob Day wrote:
As far as I can see, you're missing '; in line 360, unless it continues
further down and terminates the $message string there.
> Here is the offending code:
> 353 if ($moderation == "no"){
> 354 $message .= " checked";
> 355 }
> 356
> 357$message .= '
> 358>no
> 359Answer "yes" if you want all postings to the list to be
> sent to
> 360 a moderator for approval before distribution to the
> list.
>
> Any help would be greatly appreciated. Thanks!
--
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[EMAIL PROTECTED]
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RE: [PHP-DB] parse error
the way to start this is to just add a couple of letters to the message and
see if you're still getting an error -
try (on line 357)...
$message .=">no\nHello world";
keep it all on one line, and then see if you're still getting an error - if
you are then i don't think this line is the one you have to worry about!
-Original Message-
From: Rob Day [mailto:[EMAIL PROTECTED]]
Sent: 13 December 2001 21:49
To: '[EMAIL PROTECTED]'
Subject: [PHP-DB] parse error
I've written a small script that processes a form from a webpage and sends
the submitted data as an HTML e-mail that has the form all filled out
already. I've gotten smaller versions of this script to work without any
problem just to test the idea.I can't figure out why I'm now getting the
following error:
Parse error: parse error in /home/httpd/cgi-bin/rday/lists/do_listapp.php3
on line 358
Here is the offending code:
353 if ($moderation == "no"){
354 $message .= " checked";
355 }
356
357$message .= '
358>no
359Answer "yes" if you want all postings to the list to be
sent to
360 a moderator for approval before distribution to the
list.
Any help would be greatly appreciated. Thanks!
Rob Day
Web Team Leader
Texas State Library and Archives Commission
phone: 512.936.4463 fax: 512.463.5436
[EMAIL PROTECTED]
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Re: [PHP-DB] parse error
Rob -
I think I see what you're doing: you're concatenating $message with something
else (the HTML output following line 357?). Problem is, you're just stating
the right hand side of what should be an assignment. The concatenation should
be:
$variable = $variable .= 'HTML output';
and it looks like you just have
$variable .= 'HTML output';
although you didn't post enough code for me to be sure of this.
If you didn't mean to concatenate something, trying taking the dot before the
equal sign out of the statement.
If I'm completely wrong as to what you're trying to do, apologies.
cheers,
db
Rob Day wrote:
> I've written a small script that processes a form from a webpage and sends
> the submitted data as an HTML e-mail that has the form all filled out
> already. I've gotten smaller versions of this script to work without any
> problem just to test the idea.I can't figure out why I'm now getting the
> following error:
> Parse error: parse error in /home/httpd/cgi-bin/rday/lists/do_listapp.php3
> on line 358
>
> Here is the offending code:
> 353 if ($moderation == "no"){
> 354 $message .= " checked";
> 355 }
> 356
> 357$message .= '
> 358>no
> 359Answer "yes" if you want all postings to the list to be
> sent to
> 360 a moderator for approval before distribution to the
> list.
>
> Any help would be greatly appreciated. Thanks!
>
> Rob Day
> Web Team Leader
> Texas State Library and Archives Commission
> phone: 512.936.4463 fax: 512.463.5436
> [EMAIL PROTECTED]
>
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RE: [PHP-DB] parse error
Is it possible that there's a string farther up with an unclosed
single-quote?
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> -Original Message-
> From: Rob Day [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, December 13, 2001 3:49 PM
> To: '[EMAIL PROTECTED]'
> Subject: [PHP-DB] parse error
>
>
> I've written a small script that processes a form from a
> webpage and sends
> the submitted data as an HTML e-mail that has the form all filled out
> already. I've gotten smaller versions of this script to work
> without any
> problem just to test the idea.I can't figure out why I'm now
> getting the
> following error:
> Parse error: parse error in
> /home/httpd/cgi-bin/rday/lists/do_listapp.php3
> on line 358
>
> Here is the offending code:
> 353 if ($moderation == "no"){
> 354 $message .= " checked";
> 355 }
> 356
> 357$message .= '
> 358>no
> 359Answer "yes" if you want all postings to
> the list to be
> sent to
> 360 a moderator for approval before distribution to the
> list.
>
> Any help would be greatly appreciated. Thanks!
>
> Rob Day
> Web Team Leader
> Texas State Library and Archives Commission
> phone: 512.936.4463 fax: 512.463.5436
> [EMAIL PROTECTED]
>
>
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Re: [PHP-DB] parse-error MySQL - PhP @ win2000
Thanks very much for the assistence: The curly-bracket thing did it! Now I
finally know the installation has been succesfull!
I'll try to start programming less sloppy, and spare you from these silly
errors!
I'll notify you when the site is up and running. Future site location:
www.hyperspectral.info
J.
**
Ir. Jelle G. Ferwerda
[EMAIL PROTECTED] - www.bio-vision.nl
**
Mankind:
Intelligent enough to make a strong impact on the environment,
But not strong enough to make an intelligent one!
**
- Original Message -
From: "Roel Mulder" <[EMAIL PROTECTED]>
To: "Jelle Ferwerda" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Saturday, November 10, 2001 10:39 PM
Subject: Re: [PHP-DB] parse-error MySQL - PhP @ win2000
> Dag Jelle,
> Als je printf(); gebruikt moet je dit wel correct doen, met print(); zou
> het goed moeten gaan.
>
> Dus regel 24 wordt: print ($resultaat[0]);
>
> (Excuse me muttering Dutch between the two of us :)
>
> M.vr.gr.
> Roel Mulder
>
> At 21:32 10-11-2001 +0100, you wrote:
> >Hi all,
> >
> >I have recently installed an apache server, PhP4 and MySql on my win2000
> >machine (I know, I should be running linux; That will have to come with
> >time). But now I am trying to learn PhP - MySQL. When running the script
> >below I get the error message:
> >
> >Parse error: parse error in c:\phpdev\www\public\test2.php on line 25
> >
> >
> >Is there anybody who knows what the problem might be? I think is in the
> >result display part, considering how things run properly when I leave it
> >out. Hopefully you ca get me going, since I do not know what I am doing
wrong!!
> >
> >Thanks for the help,
> >
> >Jelle
> >
> >
> > >$host="localhost";
> >$username="";
> >$password="";
> >$database="jelle_test";
> >$table="pet";
> >
> >
> >$db = mysql_connect($host)
> > or die ("verbinding maken met de server is mislukt");
> >Echo "gefeliciteerd, de verbinding is tot stand gebracht, de handle is
$db";
> >
> > @mysql_select_db($database, $db)
> > or die ("Helaas geen verbinding kunnen maken met de database");
> >
> >$sql_select = "Select * from pet";
> >$result = mysql_query($sql_select, $db);
> >if($result) {
> > echo "Selectie succesvol";
> > while ( $resultaat = mysql_fetch_row($result)) {
> > printf($resultaat[0]);
> >}
> >
> >mysql_close($db);
> >?>
> >
> >
> >--
> >
> >
> >**
> >Ir. Jelle G. Ferwerda
> >[EMAIL PROTECTED] - www.bio-vision.nl
>
> Mulder Technisch Advies
> Postbus 69
> NL-2740 AB WADDINXVEEN
> tel. 0182-640184 fax. 0182-640185
> http://www.mta.nl
>
>
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Re: [PHP-DB] parse-error MySQL - PhP @ win2000
Hi Jelle, - Original Message - I have recently installed an apache server, PhP4 and MySql on my win2000 machine (I know, I should be running linux; That will have to come with time). But now I am trying to learn PhP - MySQL. When running the script below I get the error message: Parse error: parse error in c:\phpdev\www\public\test2.php on line 25 Is there anybody who knows what the problem might be? I think is in the result display part, considering how things run properly when I leave it out. Hopefully you ca get me going, since I do not know what I am doing wrong!! Thanks for the help, Jelle =an error message that quotes a line number AFTER the end of the script indicates that some entity has been opened but not closed. In this case curly brackets. =Various source code formatting conventions exist to try to avoid the situation (that we have here) where the appearance of the source code works to deceive brain and eye. There has been some debate in PHP forums about the 'best' way (FWIW my opinion: whatever works=best), and this is an excellent illustration of why coding the opening bracket of a code block this way, may not be "best" way! The code-block indentation is not consistent which further disguises the omission. =Regards, =dn -- 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] parse-error MySQL - PhP @ win2000
Dag Jelle,
Als je printf(); gebruikt moet je dit wel correct doen, met print(); zou
het goed moeten gaan.
Dus regel 24 wordt: print ($resultaat[0]);
(Excuse me muttering Dutch between the two of us :)
M.vr.gr.
Roel Mulder
At 21:32 10-11-2001 +0100, you wrote:
>Hi all,
>
>I have recently installed an apache server, PhP4 and MySql on my win2000
>machine (I know, I should be running linux; That will have to come with
>time). But now I am trying to learn PhP - MySQL. When running the script
>below I get the error message:
>
>Parse error: parse error in c:\phpdev\www\public\test2.php on line 25
>
>
>Is there anybody who knows what the problem might be? I think is in the
>result display part, considering how things run properly when I leave it
>out. Hopefully you ca get me going, since I do not know what I am doing wrong!!
>
>Thanks for the help,
>
>Jelle
>
>
>$host="localhost";
>$username="";
>$password="";
>$database="jelle_test";
>$table="pet";
>
>
>$db = mysql_connect($host)
> or die ("verbinding maken met de server is mislukt");
>Echo "gefeliciteerd, de verbinding is tot stand gebracht, de handle is $db";
>
> @mysql_select_db($database, $db)
> or die ("Helaas geen verbinding kunnen maken met de database");
>
>$sql_select = "Select * from pet";
>$result = mysql_query($sql_select, $db);
>if($result) {
> echo "Selectie succesvol";
> while ( $resultaat = mysql_fetch_row($result)) {
> printf($resultaat[0]);
>}
>
>mysql_close($db);
>?>
>
>
>--
>
>
>**
>Ir. Jelle G. Ferwerda
>[EMAIL PROTECTED] - www.bio-vision.nl
Mulder Technisch Advies
Postbus 69
NL-2740 AB WADDINXVEEN
tel. 0182-640184 fax. 0182-640185
http://www.mta.nl
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Re: [PHP-DB] parse error
On Tue, Mar 13, 2001 at 07:56:31PM +0100, paauwe wrote: > what is a parse error? A simple search on dictionary.com came up with this definition for parse: Of fish, to have to remove the bones yourself. "I object to parsing fish", means "I don't want to get a whole fish, but a sliced one is okay". A 'parsed fish' has been deboned. There is some controversy over whether 'unparsed' should mean 'bony', or also mean `deboned'. With this knowledge, I feel it is safe to say that PHP is having a problem removing the bones from your fish. Perhaps you should sharpen your server's knife? A parse error is simply a syntax error. You probably forgot a closing ], ', ", `, ), ; or something similar. -- Jason Stechschulte [EMAIL PROTECTED] -- They can always run stderr through uniq. :-) -- Larry Wall in <[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]
