parse error

[seanlyons]

anyone got any idea why this gives me a parse error on the $result=mysql_query line

 
$linkID = @mysql_connect("sql-01.portlandpremium", "myusername", "*****"); 
mysql_select_db("mydb", $linkID);


$result=mysql_query("SELECT * FROM login WHERE username="$logged_on"" , $linkID) or die ("Sorry, mistake somewhere.");
$row=mysql_fetch_array($result);
$school_id_session = $row["school_id"]; 
$logged_on = $HTTP_POST_VARS['username']; 

//Register as session variables:
session_register("school_id_session");
$school_id_session = $row["school_id"];
session_register("logged_on");

Assume all dtabase identifiers are correct...whats going on?

[PHP-Editors_com]

patience is a virtue, my friend.

check your double quote usage (remember to use escape chars), also you SQL structure.

I KNOW YOU LIKE YOUR DOUBLE QUOTES BUT - PLEASE DO NOT DOUBLE POST IT MAKES A MESS OF THE FORUM.

[seanlyons]

there's nothing like a supportive message.

THANKS