Question

所以，一旦我解决了我之前的问题的承诺问题，我试图在登录时从数据库中获取玩家账号X，Y坐标（这样他们就不会被放在1,1，而是放在他们的最后跟踪坐标）

经过一些调试后，我来到了这里：

var x; //note I also tried to define these directly above the getX/Y() and it didn't work
var y;
return con.getConnectionAsync().then(function(connection) {
    return connection.queryAsync('SELECT password,id FROM player WHERE name='+mysql.escape(req.body.user))
        .spread(function(rows, fields) {
        if (hash.verify(req.body.pass,rows[0].password)) {
            req.session.loggedIn = true;
            req.session.user = rows[0].id;
            getX(rows[0].id,con,mysql).then(function(data) {
                x = data;
            });
            getY(rows[0].id,con,mysql).then(function(data) {
                y = data;
            });
            console.log(x,y,"line77");
            ref = new P(rows[0].id,x,y);
            res.send({
                "msg":"You have logged in!",
                "flag":false,
                "title":": Logged In"
            });
            return ref;
        } else {
            res.send({
                "msg":"Your username and or password was incorrect.",
                "flag":true,
                "title":": Login Failed"
            });
        }
    }).finally(function() {
        connection.release();
    });
});

这就是整个功能 - 以防某些范围缺失。但问题在于：

getX(rows[0].id,con,mysql).then(function(data) {
    x = data; //x logs the return 7 from the db
});
getY(rows[0].id,con,mysql).then(function(data) {
    y = data; //y logs 45 from the db
});
console.log(x,y,"line77"); //logs "undefined undefined line77"
ref = new P(rows[0].id,x,y);

我的印象Promise会在我的查询完成之前解决此问题触发问题，但我猜不会。

为什么我的函数在设置X，Y变量之前返回？

注意：NEXT步骤是将我的关注点与功能区分开来，所以请忽略我在比赛中像接力棒一样传递con和mysql。谢谢！

Answer 1

为什么我的函数在设置X，Y变量之前返回？

Because JavaScript I/O is asynchronous。

如果你想等待两个承诺 - 你需要挂钩承诺的完成。幸运的是，承诺通过Promise.all和Promise.spread使您变得非常简单。

Promise.all(getX(rows[0].id,con,mysql),getY(rows[0].id,con,mysql)).spread(function(x,y){
     console.log(x,y);//should work;
});

Answer 2

承诺不会改变语言，它只是一个图书馆。它无法改变函数返回的工作方式。但是，你想要这样的东西：

return con.getConnectionAsync().then(function(connection) {
    var x;
    var y;
    return connection.queryAsync('SELECT password,id FROM player WHERE name='+mysql.escape(req.body.user))
        .spread(function(rows, fields) {
        if (hash.verify(req.body.pass,rows[0].password)) {
            req.session.loggedIn = true;
            req.session.user = rows[0].id;
            ref = new P(rows[0].id,x,y);
            res.send({
                "msg":"You have logged in!",
                "flag":false,
                "title":": Logged In"
            });
            return Promise.all([
                getX(rows[0].id,con,mysql),
                getY(rows[0].id,con,mysql)
            ]).then(function(xy) {
                x = xy[0];
                y = xy[1];
            }).return(ref);
        } else {
            // Should probably throw a LoginError here or something
            // because if we get here, we don't define x and y
            // and that will require an annoying check later
            res.send({
                "msg":"Your username and or password was incorrect.",
                "flag":true,
                "title":": Login Failed"
            });
        }
    }).then(function() {
        // If the login succeeded x and y are defined here.
        // However, in the else branch you don't define
        // x and y so you will need to check here.
        // Had you thrown an error in the else branch
        // you would know that x and y are always defined here.
        use(x, y);
    }).finally(function() {
        connection.release();
    });
});

Answer 3

函数getX和getY是异步函数。 Promise解决了嵌套匿名函数的问题，但它不会使函数同步和阻塞。

因此，在您创建x对象并返回此对象时，y和ref未设置。

尝试这样的事情：

getX(rows[0].id,con,mysql).then(function(data) {
    x = data; //x logs the return 7 from the db
    getY(rows[0].id,con,mysql).then(function(data) {
        y = data; //y logs 45 from the db
        console.log(x,y,"line77"); //logs "undefined undefined line77"
        ref = new P(rows[0].id,x,y);
    });
});

另外，因为您的整个函数都是异步的，所以您必须使用回调或返回一个承诺，您可以在其中访问ref对象。

失去可变范围

3 个答案: