Jquery構建Form表單Post提交數據的簡單方法

$.extend({
    PostSubmitForm: function (url, args) {
        var body = $(document.body),
              form = $("<form method='post' style='display:none'></form>"),
           input;
        form.attr({ "action": url });
        $.each(args, function (key, value) {
            input = $("<input type='hidden'>");
            input.attr({ "name": key });
            input.val(value);
            form.append(input);
        });

        //IE低版本和火狐下
        form.appendTo(document.body);
        form.submit();
        document.body.removeChild(form[0]);
    }
});

//示例
$.PostSubmitForm('url', { data1:'1',data2:'2' })