SELECT LIKE returns no result

[Germaris]

Hi there!

In my script I use the following code:

$table = "_lisausers"; 
$mypassword; //this is a 'POST' variable
$query =mysql_query("SELECT pwd FROM `$table` WHERE pwd LIKE '$mypassword'" ) or die();
if (mysql_num_rows( $query ) > 0) {
	print "&retour=true";
}
else {
	print "&retour=false";
}

Despite the fact that $mypassword already exists in the $table, this query always gives me a wrong result if I ask an existing or not pwd in this table !!!

Would you tell me what's wrong in this query and the if/else block?

Many thanks in advance for your help!

Regards,
Gerry

[Markus]

I think when you use LIKE you also have to use the % operator.

[php]
SELECT * FROM `table` WHERE `column` LIKE '%something%';
[/php]

Also, in your: 'or die()' add this: or die(mysql_error ());
for error checking.

[nomad]

I think when you use LIKE you also have to use the % operator.

[php]
SELECT * FROM `table` WHERE `column` LIKE '%something%';
[/php]

Also, in your: 'or die()' add this: or die(mysql_error ());
for error checking.

or you could do this as well
SELECT * FROM `table` WHERE item_name LIKE 'A%';

This would find any item that starts with the letter A

nomad

[nomad]

I think when you use LIKE you also have to use the % operator.

[php]
SELECT * FROM `table` WHERE `column` LIKE '%something%';
[/php]

Also, in your: 'or die()' add this: or die(mysql_error ());
for error checking.

sorry I thought the first one did not go thro...

nomad

[Germaris]

Thanks very much to all of you folks for your replies.

The problem is solved.
The PHP is okay, it's the way I manipulate the results in my .swf file which was wrong.
Silly of me !

Best regards,
Gerry