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guymrodgers

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Posts posted by guymrodgers

  1. Hey thanks so much Shocker!  Your code really helped.  However, now I am still having problems with the password access...  Here's my newly updated code that I tried:

    [code]
            mysql_select_db("argo", $con);

            $query=mysql_query("SELECT user FROM customers WHERE user = '$name'",$con);
    $correctUser = 0;

    $user = mysql_fetch_array($query);
    if ($user !== '')
    {
    $correctUser = 1;
    }

    if ($correctUser == 0)
    {
    $theEnd = <<<END
    <script language='Javascript' type='text/javascript'>
    alert("We do not have that username on file.");
    </script>
    END;
    die($theEnd);
    }

    if($pass != mysql_query("SELECT password FROM 'customers' WHERE user=$name"))
    {
    $theEnd = <<<END
    <script language='Javascript' type='text/javascript'>
    alert("The password you entered does not match the username you provided.");
    </script>
    END;
    die($theEnd);
    }

    [/code]

    But it didn't work, so I tried a longer route:

    [code]

        $resultant = mysql_query("SELECT password FROM customers WHERE user=$name",$con);
        $correctPass = 0;
       
        foreach(mysql_fetch_array($resultant) as $val)
        {
        if ($pass == $val)
        {
        $correctPass = 1;
        }
        }
       
    if ($correctPass == 0)
    {
    $theEnd = <<<END
    <script language='Javascript' type='text/javascript'>
    alert("The password you entered does not match the username you provided.");
    </script>
    END;
    die($theEnd);
    }

    if (mysql_query("SELECT id FROM customers WHERE password=$pass") != mysql_query("SELECT id FROM customers WHERE user=$name"))
    {
      $theEnd = <<<END
    <script language='Javascript' type='text/javascript'>
    alert("The password you entered does not match the username you provided.");
    </script>
    END;
    die($theEnd);
    }
    [/code]

    But it will only accept the first password in the list, no matter what the username is.

    Help!!!

    --Guy
  2. I'm sure that my problem is caused by a really stupid typo, but I can't find it.  I'm trying to grab a column from a database, and then go through it as an array in order to find a username.  However, my foreach loop only checks the first username in the column, and then stops.  I searched the web for a similar problem, and then I scoured the forums, and it appears that no one else has this problem, which is why I'm betting that it's a typo, but still, it's stumping me...

    PLEASE HELP!!!

    [code]
    <?php
    $name = $_POST["compid"];
    $pass = $_POST["pass"];

    $con = mysql_connect("localhost","root","quantico");
    if (!$con)
                {
        die('Could not connect: ' . mysql_error());
    }

    mysql_select_db("argo", $con);

    $results = mysql_fetch_array(mysql_query("SELECT user FROM customers",$con));
    $correctUser = 0;

    foreach ($results as $value)
    {
    if ($name == $value)
    {
    $correctUser = 1;
    }
    }

    if ($correctUser == 0)
    {
    $theEnd = <<<END
    <script language='Javascript' type='text/javascript'>
    alert("We do not have that username on file.");
    </script>
    END;
    die($theEnd);
    }
    [/code]

    There's more code, but I didn't think it was part of the problem.  If this part of the code isn't the problem, I'll post the rest.

    Thanks in advance,
    Guy
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