ajax请求捕获异常

 

$.ajax({
        url: "/es/find",
        type: "POST",
        async: false,
        success: function(data) {
            alert("发生错误的信息如下");
    },
        error: function (response, ajaxOptions, thrownError) {
        alert("error"); 
    }
});    

 

posted @ 2019-11-19 17:49  闫超辉  阅读(930)  评论(0编辑  收藏  举报