function getrow() not working

[simona]

need a little help with the following

function getRow()
{
global $link;
$result = mysql_query( "SELECT * FROM users, $link );
if ( ! $result )
die ( "getRow fatal error: ".mysql_error() );
return mysql_fetch_array( $result );
}

if ( getRow( "website", "login", $form[login] ) )
$message .= "Login \"$form[login]\" already exists. Try another<BR>\n";

this is so i can find if user already exists but the problem is, i can enter only one user into the database and the rest it comes up the the already exists message....can anyone figure out why?

will be glad to provide you with any code that is appropriate.
thanks in advance

simon

[gardner1]

$result = mysql_query( "SELECT * FROM users, $link );

first off you need to add a quote mark

$result = mysql_query( "SELECT * FROM users, $link ");

but even then, I plugged this into my test db

set $link to the name of my valid DB, and got the error

---

getRow fatal error No Database Selected

Warning: Supplied argument is not a valid MySQL result resource in
c:\program files\apache group\apache\htdocs\getrow.php on line 23

---

You didnt get an error message because the function failed, due to the missing quotation mark.

the $message is the next available line to execute and gives you your error... maybe????? dont know with out the full source

hope that helps in troubleshooting...

[gardner1]

sorry I meant...

first off you need to add a quote mark

$result = mysql_query( "SELECT * FROM users" , $link );