我有一张这样的table
参与者教练表

id----nutrition_coach----mental_coach----movement_coach
1 ----       2       ----      5     ----      4

教练台

id----name----email----
2 ----NAME---- @@@@ ----

通过 Inner Join 我想得到每个教练的全部数据。
这是我目前的问题

SELECT
    c1 AS nutrition_coach,
    c2 AS movement_coach,
    c3 AS mental_coach
FROM participant_coaches AS pc 
INNER JOIN coaches AS c1 ON pc.nutrition_coach = c1.id 
INNER JOIN coaches AS c2 ON pc.movement_coach = c2.id 
INNER JOIN coaches AS c3 ON pc.mental_coach = c3.id 
WHERE participant = + participantId

这不起作用，它给出了一个错误，它不知道该字段 c1 . 但是，当我只想从我的c1表中选择1个属性时，这就行了。
例如，这是有效的

SELECT c1.id AS nutrition_coach, 
       c2.id AS movement_coach, 
       c3.id AS mental_coach

退货：

mental_coach: 3
movement_coach: 1
nutrition_coach: 2

我是否可以从中选择所有字段 c1 而不是只有1个？
以下是我想要达到的目标
答复：

{
       mental_coach: {
          id: 1
          name: 'MYNAME'
          email '@@@@'
       }
       movement_coach: {
          ...
       }
       nutrition_coach: {
          ...
       }
   }