jQuery array() with select on change
I am trying to minimize my code by putting it into an array but nothing happens. I can't figure out what I am doing wrong. Here's the code
<html>
<head>
<title>test</title>
<!-- JavaScript -->
<script src="js/jquery-1.5.2.js" type="text/javascript"></script>
<script type="text/javascript">
var phpf开发者_如何学Goile = new Object();
phpfile["testselect"] = "zoomchange.php";
var elementID = new Object();
elementID["testselect"] = "#testdiv";
$(document).ready(function(){
$("select").change(function() {
$.post(
phpfile[$(this).id()],
$(this).serialize(),
function(data) {
$(elementID[$(this).id()]).html(data)
}
);
});
});
</script>
</head>
<body>
<select id="testselect">
<option value="1">1</option>
<option value="2">2</option>
</select>
<div id="testdiv"></div>
</body>
</html>
here is the zoomchange.php:
<?PHP echo $_REQUEST['testselect'] ; ?>
Your initializers shouldn't look like this:
var phpfile = new Array();
phpfile["testselect"] = "zoomchange.php";
var elementID = new Array();
elementID["testselect"] = "#testdiv";
A JavaScript Array is indexed by numbers, not strings. You want simple object literals:
var phpfile = { testselect: 'zoomchange.php' };
var elementED = { testselect: '#testdiv' };
Then, your POST callback is confused:
function(data) {
$(elementID[$(this).id()]).html(data)
}
this isn't what you think it is when that function is called. You want something more like this:
$("select").change(function() {
var that = this;
$.post(
phpfile[that.id],
$(this).serialize(),
function(data) {
$(elementID[that.id]).html(data);
}
);
});
This
function(data)
{
$(elementID[$(this).id()]).html(data);
}
instead of this
function(data)
{
$(elementID[$(this).id()]).html(data)
}
Is this the error ?
You should do new Object() instead of new Array().
Edit: There are other mistakes, your js code should be this:
<script type="text/javascript">
var phpfile = {};
phpfile["testselect"] = "zoomchange.php";
var elementID = {};
elementID["testselect"] = "#testdiv";
$(document).ready(function(){
$("select").change(function() {
var $select = $(this);
$.post(
phpfile[$select.attr("id")],
$select.serialize(),
function(data) {
$(elementID[$select.attr("id")]).html(data)
}
);
});
});
</script>
加载中,请稍侯......
精彩评论