<form method="POST">
<input type="checkbox" id="hrm" name="hrm" />
</form>
I mean when the form is posted.
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For future reference: var_dump($_POST); :-)Christian Studer– Christian Studer2009-09-10 08:11:56 +00:00Commented Sep 10, 2009 at 8:11
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3 Answers
$_GET['hrm'] or $_POST['hrm'] (depending on your form's method attribute) will be set to 'On' if it is checked, or will not be set at all if it's unchecked. In essence, you can just check using isset($_GET['hrm']) (or _POST if that's the case) - if isset() returns true, then it was checked.
Comments
<input type="checkbox" id="hrm" name="hrm" value="yes" />
<?php
if ( isset( $_POST['hrm']) && $_POST['hrm'] === 'Yes' ) {
}
?>
2 Comments
omg
Seems isset( $_POST['hrm']) is enough,since when not checked,won't post an antry with key "hrm" at all,right?
meder omuraliev
isset() should be enough, I just like being thorough and that would explain the extra check for the value being exactly what I want.
This how:
<?PHP
if($_POST['hrm']=='ok') echo 'checked';
else echo 'not';
?>
or:
<?PHP
if(isset($_POST['hrm'])) echo 'checked';
else echo 'not';
?>
But first you have to give value for it:
<input type="checkbox" id="hrm" name="hrm" value='ok' />
2 Comments
omg
Is value neccesary? I saw from firebug that without value it'll post something like "hrm: On".
Jörg
I'd suggest to add a value since you're otherwise depending on the user's browser to add that "On" value. I don't know if you can expect every browser to do that for you. And if a user comes along without he's screwed.