How do I post disabled input

Question

Hello I have some input but one of them is disabled ( yes and i need it for my time sheet )but how do I send it autocomplete.php to insert.php I've this error Undefined index: client1 in C:\wamp\www\testlp\insert.php on line 30

Here my code autocomplete.php

<form action = 'insert.php' method="post"  >

    <input type="text" name="client1" class = "client" size="12" id ="client1" disabled />

        </form>

here my code insert.php

    session_start(); 
    $date = $_POST['data'] ;
    $client1 = $_POST['client1'] ;

    echo($client1);
    echo($date);

EDIT I tried this :

<input type="text" name="client1" class = "client" size="12" id ="client1"readonly />

here the error : Notice: Undefined index: client1 in C:\wamp\www\testlp\insert.php on line 12

Why does it need to be disabled? I think you can't get that to work with `disabled` in it — putvande, Apr 10 '14 at 13:45
You could try using [`readonly`](https://developer.mozilla.org/en-US/docs/Web/HTML/Element/Input#attr-readonly) instead of `disabled` — Lix, Apr 10 '14 at 13:46
Ditto to @Lix see http://stackoverflow.com/questions/4727974/how-to-post-submit-an-input-checkbox-that-is-disabled for more info — user1477388, Apr 10 '14 at 13:46
i can try readonly , but i need to one can write in this input but i need t osend it to my insert. Finaly how I do with my error? — Thephpdoge, Apr 10 '14 at 13:47
But why, if you want your user to be able to do something with it, does it need to be disabled or read only? — putvande, Apr 10 '14 at 13:48
so i'll explain I have 3 input number, client, description. When I enter something in number and press enter my field client and description is complete . So that why client and description need to be lock. And after when I press end I need to send client, description and number to insert page. — Thephpdoge, Apr 10 '14 at 13:50
Try adding `value=""` or `value=""` that should do the trick; worked for me. @Thephpdoge — Funk Forty Niner, Apr 10 '14 at 14:16
You need to post the most relevant code in your question and how you're getting the data from to be inserted into your readonly field. I've tested what I could with the little information you've given. There's not much for me to go on to help you. — Funk Forty Niner, Apr 10 '14 at 14:28
If you accepted an answer "just because" and didn't reflect the actual solution, you should have posted an answer yourself and accepted it once SO lets you. In doing what you did, doesn't provide nor will it help anyone else visiting this question in the future. @Thephpdoge — Funk Forty Niner, Apr 10 '14 at 15:14
I click on Richard wecker cause its actually help my code and this is how I solved — Thephpdoge, Apr 10 '14 at 15:16
You said in a comment *"I success with a isset cause my form is not correct"* @Thephpdoge what is it that helped solve the problem then? — Funk Forty Niner, Apr 10 '14 at 15:21

cypherabe · Answer 1 · 2015-10-07T11:13:35.927

104

use the attribute readonly instead of disabled.

readonly: input can't be modified
disabled: input has no form function
(and the related third option: input type=hidden: input is not visible, but the value is submitted)

you get an error because an disabled element is not sent when the form is submitted and thus is not present in $_POST (there simply is no $_POST['client1'] in your case)

edit edited: the examples were not complete - as the accepted answer states, the name attribute must be present, too

 <input type="text" name="client1" class = "client" size="12" id ="client1" value="something" readonly />

or

 <input type="text" name="client1" class = "client" size="12" id ="client1" value="something" readonly="readonly" />

if you want to have a more xml-like syntax.

edited Oct 07 '15 at 11:13

answered Apr 10 '14 at 13:50

cypherabe

2,562
1
20
35

2

There must be something else wrong with your code. `readonly` should work that way. – putvande Apr 10 '14 at 13:58
what exactly is not working? - are you sure you spelled the name attribute and the $_POST index the same? did you reload your page? – cypherabe Apr 10 '14 at 14:01
did you see the error in my edit and my code and yes i reload – Thephpdoge Apr 10 '14 at 14:04
If you want the look of a disabled field, which is nice for the user, add a CSS class called .readonlyinput and set the color to #999999 or similar, then apply that to the input. – KingOfAllTrades Aug 17 '20 at 21:33
your code is wrong. It is readOnly (not readonly) attribute, and values are true/false (not "readonly") – qkx Aug 30 '20 at 11:23
@qkx No it's not wrong. – matronator Nov 19 '21 at 22:03
just write readonly instead of disabled . its working – PHP Laravel Developer Jul 18 '22 at 09:01
readonly not working with select controll – integrater Jul 26 '22 at 13:15

Timetrax · Answer 2 · 2019-09-12T21:06:47.213

9

Here is an idea of how you can solve this

<form action = 'insert.php' method="post"  >
  <input type="text" name="client1" class="client" size="12" id="client1" disabled />
  <input hidden name="client1" value="inserted_value_of_client1"/>
</form>

You can even remove name from the first input.
With this, your disabled input will still be displayed but php will post the value in your hidden input field.

You can use <?php echo !empty($text)?$text:'';?> to populate the value fields as shown in some answers here

TLDR;

<form action="index.php" method="post">
  <input type="text" disabled  value="my_value"/>
  <input hidden name="client" value="my_value"/>
</form>

edited Sep 12 '19 at 21:06

answered Jan 16 '19 at 04:14

Timetrax

1,373
13
15

here " value="inserted_value_of_client1"/> " means value ="" right? – User0434 Apr 16 '19 at 10:25
depends. value=6, value=tokenvalue, whatever and however you populate this value is up to the developer. so value can come from php (serverside) value can come from javascript, value can be hardcoded. The question here is how to move this value **FROM THE CLIENT to the server**. doing $_POST['client1'] means that you are moving the value **FROM THE SERVER** (php)/backend to the client(ui)/html/javascriipt. At the time of the question: we don't care how the value reached the client. We are only concerned with getting it from the client to the server – Timetrax Jun 03 '19 at 18:39
1

simple and efficient :) – vieroli Sep 18 '19 at 15:16

score -2 · Answer 3 · edited Nov 07 '16 at 18:52

If you want it disabled so it does not change in the DB, then you do not have to POST it. Use the SELECT to populate the <input> and add the attribute "disabled".

<?php
if ( !empty($_POST)) {
$other_inputs= $_POST['other'];

$valid = true;
if (empty($text)) {
    $valid = false;
}

if ($valid) {
    $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = "UPDATE table set text = ? WHERE id = ?";
    $q = $pdo->prepare($sql);
    $q->execute(array($other_inputs,$id);
}
} else {
    $pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $sql = "SELECT * FROM table where id = ?";
    $q = $pdo->prepare($sql);
    $q->execute(array($id));
    $data = $q->fetch(PDO::FETCH_ASSOC);
    $text = $data['client1'];
}
?>
<form action = 'insert.php' method="post"  >
    <input type="text" name="client1" class = "client" size="12" id ="client1" disabled vlaue="<?php echo !empty($text)?$text:'';?>" />
</form>

But they do want to POST it. – Mark Fisher Aug 27 '19 at 13:38 — Mark Fisher, Aug 27 '19 at 13:38
I do not see how this addresses the question. – Mark Fisher Aug 28 '19 at 13:14 — Mark Fisher, Aug 28 '19 at 13:14

How do I post disabled input

3 Answers3

Linked

Related