[RESOLVED] What's wrong with this command?

[Tim_Silva]

I don't understand, when I run this command in mySql directly it works but when I try to do it in a form it does not. I think the mistake is in the UPDATE somewhere:

This is how I do it in mySql, and it works fine:

UPDATE `myTable` SET `sourceId` = `1021` WHERE `source` = `University Of Toronto` ;

This is how I am trying to do in a form but nothing happens:

<?php 

$hostname="someserver.net";
$username="someusername";
$password="somepassword";
$dbname="someusername";
$usertable="sometable";
$sourceId=$_POST['sourceId'];
$source=$_POST['source'];

mysql_connect($hostname,$username,$password) or die(mysql_error());
mysql_select_db("$dbname") or die(mysql_error());

mysql_query("UPDATE `$usertable` SET `sourceId` = `$sourceId` WHERE `source` = `$source`");

Print "Your information has been successfully added to the database.";

?>

[bradgrafelman]

Backticks (``) denote database/table/column names - NOT string values.

For string values, use single quotes.

[Tim_Silva]

Thanks. That was it. I didn't know the difference.

For future reference, this was the right version:

mysql_query("UPDATE `$usertable` SET `sourceId` = '$sourceId' WHERE `source` = '$source'");

Tim

[bradgrafelman]

Don't forget to protect yourself against SQL injection attacks using [man]mysql_real_escape_string/man as well.