javascript - 异步获取的数据怎么return出去

Question

如图，用ajax异步获取到了数据，怎么把数据返回到上一层。。。如图所示，怎么才能让第二个return的值是异步获取返回的数据

Answer 1

The person above is right, use promise

get:function(){
    return new Promise(function(resolve,reject){
        //ajax...
        $.post("test.php",function(response){
            resolve(response)
        })
        //如果有错的话就reject
    })
}

Use

get().then(function(response){
    //response
}).catch(function(err){
    //错误处理
})

Answer 2

Either change it to synchronous or use callback, your return is useless

get:function(callback){
    $.post(.....,function(res){
        callback(res)
    })
}

get(function(res){
    console.log(res);
})

Answer 3

Make it a Promise

return Promise.resolve($.post(url,data));