Note that, you are missing the quote here name=options[cid]"
.
And you are using onChange="chkdeptCount(this.value)"
event with current value this
, this will only return one value at once.
This is very basic Example:
HTML:
<form method="post" id="formID" action="">
<input type="hidden" name="Action" value="IMPLODEARRAY">
<input type="checkbox" name="options[cid]" value='1' class="test">
<input type="checkbox" name="options[cid]" value='2' class="test">
<input type="submit" name="submit" value="Submit" id="SubmitButton">
</form>
AJAX:
<script type="text/javascript">
$(document).ready(function(){
$("#SubmitButton").click(function(){ // when submit button press
var data = $("#formID").serialize(); // get all form input in serialize()
$.ajax({
url: YourURL, // add your url here
type: "POST", // your method
data: data, // your form data
dataType: "json", // you can use json/html type
success: function(response) {
console.log(response); // your response
},
beforeSend: function()
{
// if you want to display any loading message
}
}); // JQUERY Native Ajax End
return false;
});
});
</script>
PHP:
<?php
if(count($_POST) > 0){ // if you have some value in AJAX request
if($_POST['Action'] == 'IMPLODEARRAY'){ // your condition
print_r($_POST['options']); // get all checkbox value.
}
}
?>
Few more example will help you to understand, you can you use: Submitting HTML form using Jquery AJAX |
jQuery AJAX submit form