Warning: mysql_result() expects parameter 1 to be resource, boolean given in D:\

[crave_kevin]

don't know what to do next with this error popping up,,,any help?

 

 

<?php

//set up the variables

$loguser = $_POST['loguser'];

$logpass = md5($_POST['logpass']);

$login = $_get['login'];

 

//connect with the server and the database

mysql_connect("localhost","root","") or die("Can't connect with the Server!");

mysql_select_db("comp3project") or die("Can't connect with the database!");

 

if($login="yes")

{

$get = mysql_query("SELECT count(cid) FROM customer_data WHERE username = '$loguser'

and password = '$logpass'");

$result = mysql_result($get,0);

 

if($result!=1){print "Login Failed!";}

else {print "Login Successful!";}

}

?>

[cs.punk]

Your query fails as far as I can tell.

 

Try adding this after the query:

 

echo "Mysql Error: " . mysql_error();

[denno020]

I don't think you need to put those quotes around the variables inside the query. You've enclosed your query in double quotes already, which means you can put variable names straight in there and it'll still work..

Try removing the single quotes around $loguser and $logpass.

 

Denno

[Pikachu2000]

Any string value in a query string needs to be quoted regardless of whether it's stored in a variable.

[cs.punk]

Also noticed this:

 

$login = $_get['login'];

 

It should be $_GET['login']