在ajax成功回调函数里把一个对象return出去，在外面console.log却是undefined

Question

function getSysTime(){
    var time = {}
    $.ajax({
        type: "HEAD",
        url: "/",
        complete: function(x) {
            //console.log(x.getResponseHeader("Date"));
            var date = x.getResponseHeader("Date").substring(17,22).split(":")
            return time = {
                hours:parseInt(date[0])+8,
                minutes:parseInt(date[1])
            }
            
        }
    })
}
var s = getSysTime()
console.log(s)

Answer 1

ajax 默认属于 异步回调函数. 执行了getSysTime() 函数后 实际上该函数没有任何显示 return操作, 默认返回 undefined.

如何修改:

function getSysTime(){
    var time = {}
    $.ajax({
        type: "HEAD",
        url: "/",
        complete: function(x) {
            //console.log(x.getResponseHeader("Date"));
            var date = x.getResponseHeader("Date").substring(17,22).split(":")
            ajaxSuccess({
                hours:parseInt(date[0])+8,
                minutes:parseInt(date[1])
            });
            
        }
    })
}
function ajaxSuccess (data) {
  console.log(data);
}
getSysTime();