I Am having problems with a simple script i wrote to login a username and password from a mysql database.
i get the error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in H:\steadyology\steadyology\account.php on line **21

I ran the script on my localhost where i have php and mysql running. the following is the query i perfomrmed on the datbase table user

header("Expires: Mon, 26 Jul 1997 05:00:00 GMT");    // Date in the past
header("Last-Modified: " . gmdate("D, d M Y H:i:s") . " GMT"); 
                                                     // always modified
header("Cache-Control: no-store, no-cache, must-revalidate");  // HTTP/1.1
header("Cache-Control: post-check=0, pre-check=0", false);
header("Pragma: no-cache");                          // HTTP/1.0


 $user=$_POST["user"]; $user=strtoupper($user);
$query="select password from user where user=$user";

$result=mysql_query("select * from user where user=$user",$db);

if(!($arr=mysql_fetch_array($result))){//rest that username exists


if(!($done==0)){  // test that password is correct


The line number with error is ****if(!($arr=mysql_fetch_array($result)))

the user table has the following fields
user_id (primary key)
user (contains the username)
pass (the password)

Any help will be appreciated.

Thank you

Recommended Answers

All 2 Replies

1) Echo out your query to the screen. There is an error in it somewhere and that's the best way to find it.

2) Use [ code][/code ] tags when displaying your code. It makes it easier for us to read. :)

The line number with error is if(!($arr=mysql_fetch_array($result)))

Without seeing the actual error output, I would have to assume your error is being caused where you are placing your !($arr ..

Try this usage instead:

$result = mysql_query(.............);
if (mysql_num_rows($result) > 0)
    // we do indeed have results.. collect the array
    $arr = mysql_fetch_array($result);
    // do something here if we don't have any rows returned


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