使用多个内部联接时选择所有字段

Question

我有一张像这样的表

participant_coach 表

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

教练桌

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

通过Inner Join，我想获取每位教练的完整数据。

这是我的当前查询

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，它不知道字段c1。但是，当我只想从我的 c1 表中选择 1 个属性时，它确实有效。

例如，这是可行的

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

返回：

mental_coach: 3
movement_coach: 1
nutrition_coach: 2

我是否可以从c1中选择所有字段而不是只选择1个？

这是我想要实现的目标

回复：

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

Answer 1

身份证姓名电子邮件

您的查询很好，除了select。 MySQL 没有代表表中整行的“记录”的概念。您需要明确列出列：

SELECT c1.id AS nutrition_coach_id,
       c1.name AS nutrition_coach_name,
       c1.email AS nutrition_coach_email,       
       c2.id AS movement_coach_id,
       c2.name AS movement_coach_name,
       c2.email AS movement_coach_email,
       c3.id AS mental_coach_id,
       c3.name AS mental_coach_name,
       c3.email AS mental_coach_email

Answer 2

您的 select 语句仅引用代表 表 的别名，但实际上并未说明您要使用哪一列。大概您想报告三种不同类型教练的名称，其中应该起作用：

SELECT
    c1.name AS nutrition_coach,
    c1.email AS nutrition_coach_email, -- etc. for other fields
    c2.name AS movement_coach,
    c3.name 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>