mysql离开连接同一个表两次

Question

我正在尝试两次连接同一个表，但它给了我一些问题。我有两个表ee_all和ee_calendar_events，我想加入。

SELECT 
    u.first_name, 
    u.last_name, 
    u.email, 
    SUM(e1.total_vacation_hours_earned) AS vacation_hours_earned, 
    SUM(e2.absent_hours) 
FROM ee_all AS u 
LEFT JOIN ee_calendar_events AS e1 ON u.user_id = e1.sched_user_id 
LEFT JOIN ee_calendar_events AS e2 ON u.user_id = e2.sched_user_id AND e2.event_id = 2 
WHERE 
    u.user_id = 23 

的vacation_hours_earned列应该返回133，如果我拿出第二个加入的它。但是，一旦我添加它，查询将永远存在，并且vacation_hours_earned的值为2000或其他值（这是错误的）。我的猜测是，当我添加第二个连接时，它再次总结了这一行，但我不想那么做。我一直在尝试几个小时，但无法找到解决办法，希望有任何帮助。

Answer 1

当最右边的表（第二个连接）有多于一行对应左表表达式的行时，左表表达式的行将被复制并在SUM中计数多次。改用子查询。

SELECT 
    u.first_name, 
    u.last_name, 
    u.email, 
    (
     SELECT 
      SUM(e1.total_vacation_hours_earned) 
     FROM 
      ee_calendar_events AS e1 
     WHERE 
      u.user_id = e1.sched_user_id 
    ) AS vacation_hours_earned, 
    (similar) AS absent_hours 
FROM 
    ee_all AS u 
WHERE 
    u.user_id = 23 

Answer 2

SELECT 
    u.first_name, 
    u.last_name, 
    u.email, 
    SUM(e1.total_vacation_hours_earned) AS vacation_hours_earned,  
    (select SUM(e2.absent_hours) as absenthours from ee_calendar_events AS e2 where u.user_id = e2.sched_user_id AND e2.event_id = 2)  
    FROM ee_all AS u 
    LEFT JOIN ee_calendar_events AS e1 ON u.user_id = e1.sched_user_id 
    WHERE 
    u.user_id = 23 

Answer 3

使用一些MySQL的语法，你可以消除第二个左连接和简化查询;

SELECT 
    u.first_name, 
    u.last_name, 
    u.email, 
    SUM(e1.total_vacation_hours_earned) AS vacation_hours_earned, 
    SUM(e1.absent_hours * (event_id=2)) 
FROM ee_all AS u 
LEFT JOIN ee_calendar_events AS e1 ON u.user_id = e1.sched_user_id 
WHERE 
    u.user_id = 23 

演示here。