PHP form checkbox always returning value

Question

Even when the checkbox is unselected it is still returning the Value. I am passing the form through an Ajax request and printing the returned value, but it is the same result checked or unchecked. Isset not working properly either.

<input type="checkbox" value="Agree" id="siteAgreement">

if(!isset($siteAgreement) || !$siteAgreement || $siteAgreement != "Agree"){
//////Unchecked
}

I am posting the data via javascript: "siteAgreement = $("siteAgreement").value;" So I think the problem is that is it always going to get the value from checkbox. Is there a way to change the value oncheck? — Ajezior
– Ajezior, Commented Jan 11, 2013 at 19:06

Jan Tojnar · Accepted Answer · 2013-01-11 19:26:49Z

3

Change

siteAgreement = $("#siteAgreement").value;

to something like this:

siteAgreement = $("#siteAgreement").is(':checked') ? $("#siteAgreement").val() : null;

But you should also add the name attribute so browsers with javascript turned off can also use your site.

Also you can use some code that generates request automatically from form (something like jquery.form), so you don't have to update javascript whenever you change the form.

Example here: http://jsfiddle.net/RhasK/

edited Jan 11, 2013 at 19:26

answered Jan 11, 2013 at 19:07

Jan Tojnar

5,5443 gold badges32 silver badges53 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Ajezior Over a year ago

Although, now when it is checked it wont execute the Javascript

Jan Tojnar Over a year ago

Won't execute at all? Do you have current version of jQuery?

user1909426 · Accepted Answer · 2013-01-11 18:09:04Z

0

Here try this:

 if(!isset($_POST['siteAgreement']) || !$_POST['siteAgreement'] || $_POST['siteAgreement'] != "Agree"){
 //////Unchecked
 }

Unless you're extracting the variable before, $siteAgreement won't be set. You can use $_POST['nameofelement'] to access the data.

answered Jan 11, 2013 at 18:09

user1909426

1,6668 silver badges20 bronze badges

Comments

Stefan · Accepted Answer · 2013-01-11 18:10:54Z

0

if($_POST["siteAgreement"] !== "Agree") {
//not checked
}

will work fine

answered Jan 11, 2013 at 18:10

Stefan

2,2141 gold badge25 silver badges40 bronze badges

1 Comment

Ajezior Over a year ago

But the problem is when I submit even unchecked it's still returning checked

Boundless · Accepted Answer · 2013-01-11 19:04:01Z

0

you should only need to check if it is set or not.

if(!isset($_POST['siteAgreement'])){
//then it is unchecked
}

Also, you didn't give your input a name. You need to put...

<input type="checkbox" name="siteAgreement" value="Agree" id="siteAgreement">

You need a name to access it in PHP.

edited Jan 11, 2013 at 19:04

answered Jan 11, 2013 at 18:12

Boundless

2,4642 gold badges26 silver badges40 bronze badges

2 Comments

Ajezior Over a year ago

Name didn't help. I pull the data via Javascript/ Ajax and pull element values by id

Boundless Over a year ago

The only way you can get form values from PHP is by setting the name attribute of your input. Did you also try the php code that I put above? Are you sure that the input tag is inside of the form that you're submitting? This is the way that I use check boxes, and it works on my sites. Also, note that where you have value='Agree' in your input, that doesn't get used. If you want to check an input box you write... checked="yes"

Collectives™ on Stack Overflow

PHP form checkbox always returning value

4 Answers 4

2 Comments

Comments

1 Comment

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

2 Comments

Comments

1 Comment

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related