jquery ajax promise

$request = $.getJSON('test.php');

    $request.done(process1);
    $request.done(process2);
    $request.always(process3);

    function process1() {
        console.log('process1:');
    }
    function process2() {
        console.log('process2');
    }
    function process3() {
        console.log('process3');
    }

$promise = $.Deffered()

 

posted @ 2013-12-20 11:25  sexy_girl  阅读(268)  评论(0编辑  收藏  举报