How to get info from form to php of a checkbox

Question

I got this form:

<form name="formx" id="formx" action="var.php" method="POST">
  <input type="checkbox" name="f_check1"> Check 1
  <input type="checkbox" name="f_check2">Check 2
  <input name="f_register" value="Register" type="submit">
</form>

And on the var.php file I have:

   <?php
   if($_POST['f_register'] == "Register") {
     $check1 = $_POST['f_check1'];
     $check2 = $_POST['f_check2'];
     }

 echo $check1. "<br>" ;
 echo $check2;
?>

And when i fill the form and go to the var.php, i only get results if the checkbox is on, and i want it to say "true" if checked and "false" if not.

P.S: I'm using XAMPP to run the website.

could use a shorthand syntax such as a ternary on `isset()`. — Funk Forty Niner, Mar 31 '16 at 16:33
Like use `if (isset($_POST["f_check1"])) { $check 1 = "true"; }else{ $check 1 = "false"; }` ? — Berna, Mar 31 '16 at 16:38
@Berna Shorthand means this syntax: `$var = $condition ? $val_if_true : $val_if_false;` — dokgu, Mar 31 '16 at 16:43

score 1 · Answer 1 · answered Mar 31 '16 at 16:36

1

$check1 = isset($_POST['f_check1']);
$check2 = isset($_POST['f_check2']);

answered Mar 31 '16 at 16:36

dokgu

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Um... this won't work. It's always going to be considered as being set. – Funk Forty Niner Mar 31 '16 at 16:40
@Fred-ii- This should work. Checkboxes send the value `on` if they're checked. Otherwise they don't get sent to the server at all. – dokgu Mar 31 '16 at 16:45
http://stackoverflow.com/questions/11424037/does-input-type-checkbox-only-post-data-if-its-checked – dokgu Mar 31 '16 at 16:46
That's the thing, i want then to "send" even if they are not checked :) – Berna Mar 31 '16 at 16:47
@Berna So that's why my answer tests that. If it is set, you get a value of `true`, otherwise you get a value of `false`. – dokgu Mar 31 '16 at 16:49

score 1 · Answer 2 · answered Mar 31 '16 at 16:36

1

Unchecked checkboxes aren't sent to the server. So you can account for that with:

 $check1 = isset($_POST['f_check1']) ? true:false;
 $check2 = isset($_POST['f_check2']) ? true:false;

answered Mar 31 '16 at 16:36

j08691

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What about it doesn't work? Are you expecting to see something when the value is false? PHP doesn't echo anything for false. – j08691 Mar 31 '16 at 16:45
1

No need to use ternary operators when `isset` already returns a boolean. – GGG Mar 31 '16 at 16:48
@GGG - totally true, however for a novice PHP developer this is more clear and the end result is the same. – j08691 Mar 31 '16 at 16:49
`isset()` will return false, because the value of the checkbox evaluates to false (NULL). You need to set a value in the HTML code for the checkbox - ` Check 1` – omnichad Mar 31 '16 at 16:50

Jhonny Mesquita · Accepted Answer · 2016-03-31T16:53:07.607

1

Use this block:

 <?php
   if($_POST['f_register'] == "Register") {
     $check1 = isset($_POST['f_check1']);
     $check2 = isset($_POST['f_check2']);
     }

if($check1) echo '<br>check1 true';
else echo 'check1 false';
if($check2) echo '<br>check2 true';
else echo '<br>check2 false';
?>

edited Mar 31 '16 at 16:53

answered Mar 31 '16 at 16:41

Jhonny Mesquita

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No need to use ternary operators when `isset` already returns a boolean. – GGG Mar 31 '16 at 16:49

score 1 · Answer 4 · 2016-03-31T17:07:11.747

1

Try this?

<form name="formx" id="formx" action="var.php" method="POST">
<input type="checkbox" name="f_check1"> Check 1
<input type="hidden" name="f_check1" value="0" />
<input type="checkbox" name="f_check2">Check 2
<input type="hidden" name="f_check2" value="0" />
<input name="f_register" value="Register" type="submit">
</form>

Edited as requested:

The hidden field with the same name will be passed if the checkbox is not checked.

edited Mar 31 '16 at 17:07

answered Mar 31 '16 at 16:48

1

This would be the most useful answer, if it had an explanation of WHY it works as expected. Please add an explanation of WHY having these fields performs a fallback when the checkbox is not set. – Luis Masuelli Mar 31 '16 at 17:04

How to get info from form to php of a checkbox

4 Answers4