Is it possible to get past undefined variable using dynamic input fields in PHP?

Question

Got an error of

Undefined index: mytext

because I've got a form wherein it has a button wherein it adds an input field dynamically! But what if I will not add an input field? Can it be possible that it will get passed the variable in the php code.

Here is my declaration of mytext variable

if ($_POST['mytext']) {
    foreach ( $_POST['mytext'] as $key=>$value ) {
    $values = mysqli_real_escape_string($conn, $value);
    $query = mysqli_query($conn,"INSERT INTO tblalumni_past_o (pastoccu, alum_id) VALUES ('$values', '$id')");
        }
    }

So can it get passed if the input field mytext is not in the form?

You should check with isset or array_key_exists to avoid "undefined index", if you really want to ignore those notices (not recommended) you could specify error_level as warning, or use @ operator in front of the variable — Dan Ionescu, Mar 04 '17 at 12:55
@DanIonescu Thank you for understanding my question. ! Thanks — sauce, Mar 04 '17 at 12:56
Possible duplicate of [PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-notice-undefined-index-and-notice-undef) — user3942918, Mar 04 '17 at 17:23

Shailesh Chauhan · Answer 1 · 2017-03-04T13:01:23.983

0

Just change you condition if(isset($_POST['mytext']) && is_array($_POST['mytext']))

edited Mar 04 '17 at 13:01

answered Mar 04 '17 at 12:54

Shailesh Chauhan

643
5
18

what? I will replace the same? – sauce Mar 04 '17 at 12:57

score 0 · Answer 2 · answered Mar 04 '17 at 12:57

0

Replace if( isset($_POST['mytext']) && is_array($_POST['mytext']) ) with if ($_POST['mytext'])

answered Mar 04 '17 at 12:57

Vineet1982

7,552
4
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what? you are like chauhan! maybe the other way around? – sauce Mar 04 '17 at 13:00

score 0 · Accepted Answer · answered Mar 04 '17 at 13:01

To avoid "undefined index" notices the right way you can use one of the following:

if (isset($_POST['mytext'])) {
    // your code
}

or

if (array_key_exists('mytext', $_POST)) {
    // your code
}

You could also choose to ignore them using the @ operator or by setting error_reporting above notice but that is not recommended.

Is it possible to get past undefined variable using dynamic input fields in PHP?

3 Answers3