Call ajax each time when a checkbox is checked in php
Question
I have a page in which there are some checkboxes.Right now when ever i press submit btton, im calling ajax.but i want the ajax to be called when any of the checkboxes is checked.
My code is as below for calling ajax.
<html>
<head>
<title>Insert title here</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
</head>
<body>
<script type="text/javascript">
function get_check_value() {
var c_value = [];
$('input[name="brand"]:checked').each(function () {
c_value.push(this.value);
});
return c_value.join(',');
}
function get_store_value(){
var d_value=[];
$('input[name="store"]:checked').each(function () {
d_value.push(this.value);
});
return d_value.join(',');
}
$(document).ready(function(){
$('#btnSubmit').on('change', function (e)
{e.preventDefault();
alert("hi");
//var os = $('#originState').val();
//var c = $('#commodity').val();
//var ds = $('#destState').val();
var ser = get_check_value();
var store=get_store_value();
//var queryString = "os=" + os;
var data = "?ser=" + ser;
var queryString = "&ser=" + ser;
alert(ser);
$.ajax({
//alert("ajax");
type: "POST",
url: "sortingajax.php",
data: {ser:ser,store:store},
dataType : 'html',
success: function (b) {
// alert(a+' ok. '+b)
$('#results').html(b);
console.log(b);
}
});
});
});
</script>
brand
<input type="checkbox" name="brand" value="Sunbaby" id="check" />Sunbaby
<br/>
<input type="checkbox" name="brand" value="Advance Baby" id="check"/>Advance Baby
<br/>
store
<br/>
<input type="checkbox" name="store" value="JC Penny" id="check"/>JC Penny
<br/>
<input type="checkbox" name="store" value="Supermart" />Suoermart
<br/>
<input type="checkbox" name="store" value="Target" />Target
<br/>
<button id="btnSubmit">sort</button>
<div id="results">
</div>
</body>
</html>
Currently im calling ajax on pressing button,but i want to call ajax each time and fetch the results when ever any of the checkbox is checked.. Please tell me what can i change or add here in above code.
2 answers
solution1
0 2014-04-18 19:56:36
只需更改您的事件，然后处理您的复选框，即可：
$( "[type=checkbox]" ).change(function () { ...
solution2
0 2014-04-18 20:04:14
Use the change event for a changed checkbox and use the prop for checking if he is checked.
$("[type=checkbox]").change(function () {
if(this).prop('checked')
{
// textbox is checked
}
});
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