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Home/ Questions/Q 6708603
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T07:46:47+00:00 2026-05-26T07:46:47+00:00

I am trying to submit the page to itself but some reason the following

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I am trying to submit the page to itself but some reason the following code is not working. Also How can I get the table1 primary key ID back after inserting the data successfully? I have a child table which needs this ID. Thanks for any suggestions.

<?php
include('db_login.php');
$connection = mysql_connect( $db_host, $db_username, $db_password );

if (!$connection){
  die ("Could not connect to the database: <br />". mysql_error());
}
// Select the database
$db_select=mysql_select_db($db_database);
if (!$db_select){
  die ("Could not select the database: <br />". mysql_error());

  if ($_POST['Submit']) 
  { 
    $first = $_POST["first"];
    $first = mysql_real_escape_string(get_magic_quotes_gpc() ? stripslashes($first): $first);
    $last = $_POST["last"];
    $last = mysql_real_escape_string(get_magic_quotes_gpc() ? stripslashes($last): $last);


    $insertsql = "INSERT INTO table1(FirstName,LastName) VALUES ('".$first."', '" .$last. "')";

    $result1 = mysql_query($insertsql) or die(mysql_error());
 }
 ?>
 <form name="hotlineForm" action="<?php echo htmlentities($_SERVER['PHP_SELF']); ?>"
  method="post">
 <input id="first" type="text">
 <input id="last" type="text">
 <input type="submit" value="Submit"></form></body>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T07:46:48+00:00Added an answer on May 26, 2026 at 7:46 am

    Change your form inputs to include name attributes. Without them, your $_POST will be empty.

    <input name='first' id="first" type="text">
    <input name='last' id="last" type="text">
    <input name='Submit' type="submit" value="Submit">
    

    As mentioned in the comments, get_magic_quotes should not be used. You’ve correctly called mysql_real_escape_string() on your inputs already.

    Following your insert, get the id from mysql_insert_id():

    $result1 = mysql_query($insertsql) or die(mysql_error());
    $new_id = myqsl_insert_id();
    
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