javascript – NodeJS-Funktion gibt undefiniert zurück, ich verstehe nicht ganz, warum.

Question

Endlich wird das Versprechen verwendet, das Problem zu lösen. Der Code lautet wie folgt:
exports.selectByUsername = selectByUsername;
function selectByUsername(username){

var promise = new Promise(function(resolve){
    var sql = "SELECT COUNT(*) count FROM wx_user WHERE username = ?";
    var sqlParams = [username];
    var count;
    co.query(sql,sqlParams,function(err,result){
        if(err){
            return console.log(err.message);
        }
        console.log("------------------------开始查询---------------------");
        console.log(result);
        var str = JSON.stringify(result);
        var json = JSON.parse(str);
        count = json[0].count;
        console.log(count);
        console.log("------------------------查询结束---------------------");
        resolve(count);
    });
});
promise.then(function(value){
    // console.log(value);
    return value;
});
return promise;

}

app.post('/ajax',urlencodedParser,function(req,res){

username = req.body.name;
console.log(username);
var promise = s.selectByUsername(username);
promise.then(function(value){
    console.log(value);
    if(value!==1){
        res.send("用户名不存在");
    }
});

});

Referenzdokument: http://liubin.org/promises-book/

Answer 1

query里面的function在query完成后才会执行到，而这个时候外部函数已经返回了，所以count不会赋到值，仍然为undefined

Answer 2

写在query里也不会返回count的，因为query这个方法是异步的

Answer 3

简而言之，把return count 写在query 里面就好了。