Jquery构建Form表单Post提交数据的简单方法

PostForm

$.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\' })