jaywhy13
12-24-2006, 08:44 PM
What am I doing wrong?
<?php
//header ("Context-Type: text/xml");
$usr = $_GET["usr"];
$psw = $_GET["psw"];
echo "Username is: $usr, and psw is: $psw";
$conn = mysql_connect ("localhost","jay","") or die ("No connection available");
$db = mysql_select_db ("rlaPool") or die ("Could not find database");
$query = "SELECT * FROM usrs WHERE usrs.usr=$usr";
$result = mysql_result ($result,$conn) or die ("<b>That didn't work!!!!</b>") ;
?>
I get error
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /var/www/jail/labtec/login.php on line 14
If I try
$result = mysql_result ($result) or die ("<b>That didn't work!!!!</b>") ;
I get wrong param count
Someone plz just quickly call me stupid and tell me what I did wrong... coz I can't belive I'm stuck at this point.
<?php
//header ("Context-Type: text/xml");
$usr = $_GET["usr"];
$psw = $_GET["psw"];
echo "Username is: $usr, and psw is: $psw";
$conn = mysql_connect ("localhost","jay","") or die ("No connection available");
$db = mysql_select_db ("rlaPool") or die ("Could not find database");
$query = "SELECT * FROM usrs WHERE usrs.usr=$usr";
$result = mysql_result ($result,$conn) or die ("<b>That didn't work!!!!</b>") ;
?>
I get error
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /var/www/jail/labtec/login.php on line 14
If I try
$result = mysql_result ($result) or die ("<b>That didn't work!!!!</b>") ;
I get wrong param count
Someone plz just quickly call me stupid and tell me what I did wrong... coz I can't belive I'm stuck at this point.