挺基础的JavaScript问题，关于函数内向外传参

Question

var request=new XMLHttpRequest();
    request.open("get","url.json",false);
    request.send(null);
    request.onload = function (){
        if (request.status == 200) {
            menu = JSON.parse(request.responseText);
        }
    };

请问这里的menu怎么传出去供其他函数使用

本人小白，见笑了

Answer 1

var menu = "";
var request=new XMLHttpRequest();
    request.open("get","url.json",false);
    request.onload = function (){
        if (request.status == 200) {
            menu = JSON.parse(request.responseText);
        }
    };
    request.send(null);
console.log(menu);