Using Variable Inside $_POST[ ]

Question

How to use variable inside $_POST as i have used the following code but it says undefined offset in second line. How can i solve it?

    $p = $_GET['cii'];
$selectOption = $_POST[$p];

Below is the code:

echo'<form method ="POST">
            <select name="'.$abc[3].'">
                <option value="slow">slow</option>
                <option value="medium">medium</option>
                <option value="fast">fast</option>
                </select>
                <br>
                <a href ="?change&&cii='.$abc[3].'">Click to change</a>
            </form>';

$abc[] has some numbers and the select box is made with same name as the number.

then check if $_post have the value of same variable. try var_dump($_POST), see if that index exits or not — reza
– reza, Commented Jan 14, 2017 at 7:34
then you are not sending any value with post method.... show us your html form where you are setting the values — reza
– reza, Commented Jan 14, 2017 at 7:48

Craig · Accepted Answer · 2017-01-14 07:27:58Z

1

Almost there:

    $p = 'cii';
    $selectOption = $_POST[$p];

answered Jan 14, 2017 at 7:27

Craig

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Comments

Niklesh Raut · Accepted Answer · 2017-01-14 08:08:26Z

0

you are accessing same page when you come first time you get undefined offset

So first check existence

 if(isset($_GET['cii'])){
   $p = $_GET['cii'];
   $selectOption = $_POST[$p];
 }

Also no need to add double && just use single & here

<a href ="?change&cii='.$abc[3].'">Click to change</a>

Assuming variable $abc[3] gives you right value.

answered Jan 14, 2017 at 8:08

Niklesh Raut

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