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Editorial Team
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Asked: 2026-05-25T21:01:13+00:00

I’m trying to learn php and mysql and its been going good but I’m

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I’m trying to learn php and mysql and its been going good but I’m trying to use the $_POST method to insert something to a database from another page, and when I check the database to see if it worked it creates another row but with no information.

First page

<html>
<body>

    <form action="InsertExternal.php" method=post>
        First Name: <input type="text" name="firstname" />
        Last Name: <input type="text" name="lastname" />
        Age: <input type="text" name="age" />
        <input type="submit" />
    </form>

</body>

Second page

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

mysql_select_db("Learning", $con);

$sql="INSERT INTO Persons (FirstName, LastName, Age)
    VALUES
        ('$_POST[firstname]','$_POST[lastname]','$_POST[age]')";

if (!mysql_query($sql,$con))
    {
        die('Error: ' . mysql_error());
    }
    echo "1 record added";

mysql_close($con)
?>

Thanks for any help. I imagine its something simple I’m missing.

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1 Answer

  1. Editorial Team

    Make sure you are not contacting this page directly.
    In this case before doing the actual insert check if there are values in the fields:

         if(isset($_POST['firstname']) && isset($_POST['lastname']) && ..

    Second thing,
    You must sanitize your input before inserting into the database:
    use:

            $firstname = mysql_real_esacpe_string($_POST['firstname']);

    You can also use or with in the statement:

          mysql_query($q)
             or die(mysql_error());
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