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Home/ Questions/Q 7997719
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T15:01:43+00:00 2026-06-04T15:01:43+00:00

When a user registers with my website they are then required to enter personal

  • 0

When a user registers with my website they are then required to enter personal details. The entered details will be added to the personaldetails table in my datbase. I have the following code but it does not work, I cannot figure out why. Can anyone help? I am getting error:query was empty

    $myusername=$_POST['username'];
    $mypassword=$_SESSION['mypassword'];
    $firstname = $_POST['firstname'];
    $surname = $_POST['surname'];
    $dob = $_POST['dob'];
    $totalwins = $_POST['totalwins'];
    $totalloses = $_POST['totalloses'];
    $email = $_POST['email'];
    $country = $_POST['country'];
    $info = $_POST['info'];

    // Connect to server and select database.
    mysql_connect("$host", "$username", "$password")or die("cannot connect");
    mysql_select_db("$db_name")or die("cannot select DB");

    $queryreg = mysql_query("

                INSERT INTO $tbl_name VALUES('','','$myusername','$mypassword''$firstname','$surname','$totalwins','$totalloses','$email','$country','$info','$dob' )
        ");

if (!mysql_query($sql))
      {
     die('Error: ' . mysql_error());
      }
      else  {
      echo "<br><br>Your details have been successfully updated. Go back to the                 personal details page to view your updated information.";
    }
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T15:01:45+00:00Added an answer on June 4, 2026 at 3:01 pm

    First of all, you are exposed to SQL injection with your code. Please, prepare and escape your post data! This is really important to prevent SQLinjection attacks. Please read thouroughly about it.

    Then, the error occurs because of this

    if (!mysql_query($sql))
      {
     die('Error: ' . mysql_error());
      }
      else  {
      echo "<br><br>Your details have been successfully updated. Go back to the                 personal details page to view your updated information.";
    }
    

    You don’t seem to have a variable named $sql, and, you are executing two query’s!

    Please do the following and post your feedback

    $sql = "INSERT INTO $tbl_name VALUES('','','$myusername','$mypassword''$firstname','$surname','$totalwins','$totalloses','$email','$country','$info','$dob' )";
    
    if (!mysql_query($sql))
      {
     die('Error: ' . mysql_error());
      }
      else  {
      echo "<br><br>Your details have been successfully updated. Go back to the                 personal details page to view your updated information.";
    }
    
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