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Editorial Team
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Asked: 2026-05-27T04:05:57+00:00

The INSERT into staffprofiles mysql query (near the bottom of the code) is getting

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The INSERT into staffprofiles mysql query (near the bottom of the code) is getting the “or die” statement triggered. Can anyone see why? The first one into staffusers is fine. Am I breaking a rule by having 2 insert commands or something? If so, is there another way to code what I am trying to do? Thank you in advance for any help.

Also, I will be address SQL injection and other items shortly. I just want to get this figured out first. Thanks.

<html>
<head>
  <title>XXXXXXX</title>
  <link rel="stylesheet" type="text/css" href="style.css" />
</head>

      <body>
         <div id="back">
         </div>


        <div id="wrapper">

                <div id="header">
                 <div id="logout">
                <?php
              include 'header.php';
                ?>


                 </div>
                  <span><b>XXXXXXX</b>XXXXXX</span>
                </div>

                <?php
                include 'menu.php';

if (isset ($_POST['submit']))

{
}

else

{
echo "not set";
}


$user=$_POST['username'];
$pass1=$_POST['pass1'];
$pass2=$_POST['pass2'];
$email=$_POST['email'];
$user=strip_tags($user);
$email=strip_tags($email);
$pass1=strip_tags($pass1);

$isemail = "SELECT * from staffusers where email ='$email'";
$isemail2 = mysql_query($isemail) or die ("could not query email in table");
$isemail3 = mysql_fetch_array($isemail2);

if ($isemail3)
{
echo "That email address is already registered.<br>";
echo "<a href='forgotpassword.php'>Forgot passsword.</a><br>";
echo "<a href='staffregisterform.php'>Go back to register under a different email address.</a>";
exit;
}

if ($email == "" || $user == "" || $pass1 == ""  || $pass2 == "")
{
echo "please complete all fields before submitting<br>";
echo "<a href='staffregisterform.php'>Go Back</a>";
exit;
}

if (strlen($pass1)<6)
{
echo "Your password must be at least 6 characters in length.<br>";
echo "<a href='staffregisterform.php'>Go Back</a>";
exit;
}

if ($pass1 == $pass2)
{
$isplayer = "SELECT * from staffusers where username ='$user'";
$isplayer2 = mysql_query($isplayer) or die ("could not query user in table");
$isplayer3 = mysql_fetch_array($isplayer2);

if ($isplayer3)
{
echo "The username ''$user'' is already taken. Please try another username.<br>";
echo "<a href='staffregisterform.php'>Go Back</a>";
exit;
}

if (strlen($user)>16 || strlen($user)<2)
{
echo "Your username must be be at 2 two characters and at most, 16 characters.<br>";
echo "<a href='staffregisterform.php'>Go Back</a>";
exit;
}
}

else
{
echo "The passwords do not match.<br>";
echo "<a href='staffregisterform.php'>Go Back</a>";
exit;
}

$pass1=MD5($pass1);

$register = "INSERT INTO staffusers (username, password, email) VALUES ('$user', '$pass1', '$email')";
mysql_query($register) or die ("staffusers.");

$profile = "INSERT INTO staffprofiles (username, email) VALUES ('$user', '$email')";
mysql_query($profile) or die ("staffprofiles.");

echo "Thank you for registering, $user.<br>";
echo "<a href='index.php'>Login</a>";

?>
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1 Answer

  1. Editorial Team

    If you have valid fieldnames and table names then try;

    $profile = mysql_query("INSERT INTO staffprofiles (username, email) VALUES ('$user', '$email')");

// use some value directly to check as:
$user = "Tester";
$email= "something@test.com";
$profile = mysql_query("INSERT INTO staffprofiles (username, email) VALUES ('$user', '$email')");

//if this value gets inserted then you have problem in $user variables value

    Hope it helps

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