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Home/ Questions/Q 6612615
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T20:07:10+00:00 2026-05-25T20:07:10+00:00

I have a form that inserts data into mySQL. It works fine. After the

  • 0

I have a form that inserts data into mySQL. It works fine.

After the submission, I have a success page that displays ‘part_no’. I need to also show ‘add_qty’, as well.

How do i alter my post script to show two field data’s on the success page?

Here is part of my code (that already works):

$part_no = $_REQUEST['part_no'] ;

header("location: inv_fc_add_success.php?part_no=" . urlencode($part_no));
}
else {
header("location: inv_fc_add_fail.php");
}
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T20:07:11+00:00Added an answer on May 25, 2026 at 8:07 pm

    You need 3 things.

    First, you’ll need:

    $add_qty = $_REQUEST['add_qty'];
    

    Alternatively you can simply use $_REQUEST['add_qty'] directly instead of assigning it to a new variable ($add_qty).

    Then instead of:

    header("location: inv_fc_add_success.php?part_no=" . urlencode($part_no));
    

    you need:

    header("location: inv_fc_add_success.php?part_no=" . urlencode($part_no) . "&add_qty=" . urlencode($add_qty));
    

    Then, on the page inv_fc_add_success.php (which you will need to edit), you display the variable the same way that you display $part_no. As always, you can either assign it to a variable, like you already do at the top, or you can just use $_REQUEST['add_qty'] directory.

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