Re: [PHP] Still error messages!!

[David OBrien]

At 12:08 PM 1/28/2004, Radwan Aladdin wrote:
Hello all..

This is the PHP code :

<?php

include("Config.php");

$link = mysql_connect("$user_hostname", "$user_username", "$user_password");
mysql_select_db("$user_database", $link);
$UserName = $_GET['UserName'];
$Password = $_GET['Password'];
$LogoutTime = date("U");
$query1 = "UPDATE accounts SET LogoutTime=$LogoutTime";
$query2 = "SELECT LoginTime,Distance,LessonNumber FROM accounts WHERE 
UserName='$UserName' AND Password='$Password'";
$query3 = "UPDATE accounts SET LessonNumber=LessonNumber + 1";


right here do a

?> <pre> <?
echo $query1;
echo $query2;
echo $query3;
?> </pre> <?
and make sure the queries are what you want them to be...

$result = mysql_query($query1) or die("Query error: " . mysql_error());

$RightLoginTime = 'LoginTime';
$Distance = 'Distance';
$LessonNumber = 'LessonNumber';
$LessonsTimeLimit = "30";
$query4 = "UPDATE accounts SET (Distance='" . $LogoutTime - 
$RightLoginTime . "') WHERE UserName=" . $UserName . " AND Password=" . 
$Password . ")";
Do the same here for query4

$result2 = mysql_query($query4) or die("Query error: " . mysql_error());

if($Distance == $LessonsTimeLimit){
    $result3 = mysql_query($query3) or die("Query error: " . mysql_error());
}else{
    echo "Not yet!";
}
?>
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