javascript - 从 Javascript 中的回调函数返回

Question

如何从 node.js/javascript 回调的值中获取返回值？

function get_logs(){
    User_Log.findOne({userId:req.user._id}, function(err, userlogs){
        if(err) throw err;
        if(userlogs){
            // logs = userlogs.logs;
            return "hello there is a logs";
        } else {
            return "there is no logs yet..."
        }
    })
}

var logs = get_logs();
console.log(logs);

Answer 1

You can't return the result from a function whose execution is asynchronous.

The simplest solution is to pass a callback :

function get_logs(cb){
    User_Log.findOne({userId:req.user._id}, function(err, userlogs){
        if(err) throw err;
        if(userlogs){
            // logs = userlogs.logs;
            cb("hello there is a logs");
        } else {
            cb("there is no logs yet...)"
        }
    })
}

get_logs(function(logs){
    console.log(logs);
});

Answer 2

You can't. You should instead pass another callback to your function. Something like this:

function get_logs(callback){
    User_Log.findOne({userId:req.user._id}, function(err, userlogs){
        if(err) throw err;
        if(userlogs){
            callback("hello there is a logs");
        } else {
            callback("there is no logs yet...");
        }
    })
}

get_logs(function(arg1) {
   console.log(arg1);
});

Answer 3

在 node.js 中，几乎所有回调都在函数返回后运行，所以你可以做这样的事情

function get_logs(){
    User_Log.findOne({userId:req.user._id}, function(err, userlogs){
        if(err) throw err;
        if(userlogs){
            // logs = userlogs.logs;
               do_something(logs)
        } else {
            console.log('No logs')
        }
    })
}

Answer 4

function get_logs(callback) {
    User_Log.findOne({
        userId: req.user._id
    }, function (err, userlogs) {
        if (err) throw err;
        if (userlogs) {
            // logs = userlogs.logs;
            callback("hello there is a logs");
        } else {
            callback("there is no logs yet...");
        }
    })
}

get_logs(function (data) {
    console.log(data);
});

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