RE: [PHP-DB] Not a valid MySQL result resource
don't you need it to be SELECT login, password... lowercase first letter? - if you're comparing the columns user.login and user.password. -Original Message- From: DL Neil [mailto:[EMAIL PROTECTED]] Sent: 09 November 2001 01:34 To: MPropre Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Not a valid MySQL result resource ?php //.../... first part of the code is to connect to the right DB on a MySQL server. It works fine //This code to show the query. It runs well under MySQL and gives 1 result : $query=SELECT Login, Password FROM `user` WHERE user.login=''$login'' and user.password=''$password''; //This var should contain a query result ressource: $result_query=mysql_query($query); //Here's my error message after executing: // Supplied argument is not a valid MySQL result resource in this line: $result_table=mysql_fetch_array($result_query); // I thought that this var ($result_table) should contain the row of 2 cells //And I expected to read $result_table['login'] or $result_table[0] as the first cell of the row... // I any of you can help me, poor php newby... :o) Great thanks !! ? =Please check the archives for advice about checking the result of every mysql_...() call. It's much better to know/be told by PHP/MySQL than to say I thought that...I expected to ... =the $query assignment statement contains multiple double-quotes. These cannot be 'nested'. Use a mixture of single and double-quotes or 'escape' the inner set(s). =also (and this may be a function of our email packages not a PHP thing) are those single quotes around user? =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] --- Incoming mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.295 / Virus Database: 159 - Release Date: 01/11/01 --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (http://www.grisoft.com). Version: 6.0.295 / Virus Database: 159 - Release Date: 01/11/01 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] Not a valid MySQL result resource
?php //.../... first part of the code is to connect to the right DB on a MySQL server. It works fine //This code to show the query. It runs well under MySQL and gives 1 result : $query=SELECT Login, Password FROM `user` WHERE user.login=''$login'' and user.password=''$password''; //This var should contain a query result ressource: $result_query=mysql_query($query); //Here's my error message after executing: // Supplied argument is not a valid MySQL result resource in this line: $result_table=mysql_fetch_array($result_query); // I thought that this var ($result_table) should contain the row of 2 cells //And I expected to read $result_table['login'] or $result_table[0] as the first cell of the row... // I any of you can help me, poor php newby... :o) Great thanks !! ? -- 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] Not a valid MySQL result resource
At 2:04 AM +0100 11/9/01, MPropre wrote:
?php
//.../... first part of the code is to connect to the right DB on a MySQL
server. It works fine
//This code to show the query. It runs well under MySQL and gives 1 result :
$query=SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
user.password=''$password'';
//This var should contain a query result ressource:
$result_query=mysql_query($query);
//Here's my error message after executing:
Not so fast. Where's your error checking to verify that the
query actually succeeded?
if (!$result_query)
{
die (Gee, I guess error checking is a good thing after all!
. mysql_error ());
}
// Supplied argument is not a valid MySQL result resource in this line:
$result_table=mysql_fetch_array($result_query);
// I thought that this var ($result_table) should contain the row of 2 cells
//And I expected to read $result_table['login'] or $result_table[0] as the
first cell of the row...
// I any of you can help me, poor php newby... :o) Great thanks !!
?
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Re: [PHP-DB] Not a valid MySQL result resource
?php //.../... first part of the code is to connect to the right DB on a MySQL server. It works fine //This code to show the query. It runs well under MySQL and gives 1 result : $query=SELECT Login, Password FROM `user` WHERE user.login=''$login'' and user.password=''$password''; //This var should contain a query result ressource: $result_query=mysql_query($query); //Here's my error message after executing: // Supplied argument is not a valid MySQL result resource in this line: $result_table=mysql_fetch_array($result_query); // I thought that this var ($result_table) should contain the row of 2 cells //And I expected to read $result_table['login'] or $result_table[0] as the first cell of the row... // I any of you can help me, poor php newby... :o) Great thanks !! ? =Please check the archives for advice about checking the result of every mysql_...() call. It's much better to know/be told by PHP/MySQL than to say I thought that...I expected to ... =the $query assignment statement contains multiple double-quotes. These cannot be 'nested'. Use a mixture of single and double-quotes or 'escape' the inner set(s). =also (and this may be a function of our email packages not a PHP thing) are those single quotes around user? =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] Not a valid MySQL result resource
I'd check for mysql_num_rows( $result_query) 0 as well.
It's entirely possible the user mis-typed the login or password values.
Miles
At 07:26 PM 11/8/01 -0600, Paul DuBois wrote:
At 2:04 AM +0100 11/9/01, MPropre wrote:
?php
//.../... first part of the code is to connect to the right DB on a MySQL
server. It works fine
//This code to show the query. It runs well under MySQL and gives 1 result :
$query=SELECT Login, Password FROM `user` WHERE user.login=''$login'' and
user.password=''$password'';
//This var should contain a query result ressource:
$result_query=mysql_query($query);
//Here's my error message after executing:
Not so fast. Where's your error checking to verify that the
query actually succeeded?
if (!$result_query)
{
die (Gee, I guess error checking is a good thing after all!
. mysql_error ());
}
// Supplied argument is not a valid MySQL result resource in this line:
$result_table=mysql_fetch_array($result_query);
// I thought that this var ($result_table) should contain the row of 2 cells
//And I expected to read $result_table['login'] or $result_table[0] as the
first cell of the row...
// I any of you can help me, poor php newby... :o) Great thanks !!
?
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