JavaScript异步函数.then回调返回undefined问题分析与解决

调用Login函数将用户Google ID与数据库比对，判断是否存在账号，调用代码如下：

Login(DecodedGoogleCredentials.sub).then((localID)=>{//Login returns a promise with the local id for the database

             console.log(localID);//used to confirm id
             sessionStorage.setItem("CurrentUserID", localID);//set for refresh

             setID(localID);//set to recall App
             setLogin(true);//set to ensure it will be recalled

       });//end of Login().then(()=>{});

在async函数内部能获取到预期的localID，但在.then回调中执行console.log(localID)时结果为undefined。原Login函数代码如下：

async function Login(GoogleID) {//Called from App()
    
    const idQuery = query(idRef, where("id", "==", GoogleID.toString()));//build query for user's id
    const idDoc = await getDocs(idQuery);//get doc found in query. There should only be one
    
    var userFound = false;//user is not found
    
    idDoc.forEach((doc) => {//loop through docs

        userFound = true;//user is found  
        const localID = doc.data().localID;
        console.log(localID)
        try {
            return localID;
        }
        catch(e) {
            console.log(e)
        }
        
    });
    //there is no user so far
    if (userFound==false) {//create new user if user is not found
        
       return await CreateUser(GoogleID);//create the login
    }
}//end of Login(id)

此处的console.log(localID)能正常输出预期值，为何.then中会得到不同结果？

问题根源 #

问题出在forEach循环内的return操作：forEach的回调函数是独立的函数作用域，回调里的return localID只会终止当前回调的执行，并不会把值返回给外层的Login函数。当找到用户时，外层Login函数在forEach执行完毕后没有任何return语句，因此函数默认返回undefined，这就导致.then回调中拿到的localID是undefined。

修正后的代码 #

async function Login(GoogleID) {//Called from App()
    
    const idQuery = query(idRef, where("id", "==", GoogleID.toString()));//build query for user's id
    const idDoc = await getDocs(idQuery);//get doc found in query. There should only be one

    var foundDoc;
    idDoc.forEach((doc) => {//loop through docs
       foundDoc = doc
    });
    if (foundDoc != null) {
        const localID = foundDoc.data().localID;
        
        return localID;
        
    }
   else {//create new user if user is not found
        
       return CreateUser(GoogleID);//create the login
    }
}//end of Login(id)

内容的提问来源于stack exchange，提问作者ACCE