我试图用以下信息创建一个json对象
table
电影：主键电影ID
参与者：主键actorid
acted in：电影和演员的m到n关系的中间表主键=movieid和actorid。
这就是我要做的。
选择actorid=1的参与者
查询：

SELECT * FROM actors WHERE ActorID=1.

选择演员演过的所有电影。
查询：

SELECT * 
  from movies 
  LEFT 
  JOIN 
    ( SELECT *
       from actedin 
  WHERE ActorID = '+actorID+'
  ) AS actors_temp 
    ON movies.MovieID=actors_temp.MovieID AS movies_actor

求列的平均值。
query:'从电影中选择平均（rtrating）'；
求列平均值。
query:'从电影中选择平均值（mcrating）'；
这是我的nodejs代码。
假设app=express（），con=mysql连接。我将查询结果附加到actors变量，并通过res.send将它们发送回。

//#4 ActorID-> actor data, movies data, average rating 
app.get('/actor',function(req,res){
    let query_actorID = con.escape(req.query.actorID);
    let selectActor = 'SELECT * from actors WHERE ActorID = '+query_actorID;
        con.query(selectActor, function(err,result){
        if(err) throw err;
    console.log('retrieved actor info');
    console.log(result);
    var actor = result[0];
    var actorID = con.escape(actor.ActorID);
    //Retrieve Actor information
    let movie_query = 'SELECT * from movies LEFT JOIN (SELECT * from actedin WHERE ActorID = '+actorID+') AS actors_temp ON movies.MovieID=actors_temp.MovieID AS movies_actor'; 
    con.query(movie_query, function(err,movies){
    if(err) throw err;
        console.log('retrieved movies info');
        console.log(movies);
        actor.movies = movies;
        let RTR_query = 'SELECT AVG(RTRating) From movies_actor';
        con.query(RTR_query, function(err,average){
        if(err) throw err;
            console.log('retrieved movies info');
            console.log(average);
            actor.averageRTRating = average;
            });
        let MCR_query = 'SELECT AVG(MCRating) From movies_actor';
        con.query(RTR_query, function(err,average){
        if(err) throw err;
            console.log('retrieved movies info');
            console.log(average);
            actor.average_MCRating = average;
            });
        });

    res.send(JSON.stringify(actor));
    });
});

我的语法显然错了。我想我用错了。