MySQL子查询替代方案

Question

我有一个查询，其中包含4个子查询。查询是这样的：

SELECT
  (SELECT
     COUNT(id)
   FROM timelog
   WHERE emp_id = 1
       AND am_in > GET_TIME_IN1(emp_id, DATE)) AS tardy1,
  (SELECT
     COUNT(id)
   FROM timelog
   WHERE emp_id = 1
       AND pm_in > GET_TIME_IN2(emp_id, DATE)) AS tardy2,
   (SELECT balance FROM leave_credit lc JOIN leave_type lt ON lc.leave_type_id = lt.id WHERE emp_id = 1 AND lt.active = TRUE) AS balance,
   (SELECT leave_type_id FROM leave_credit lc JOIN leave_type lt ON lc.leave_type_id = lt.id WHERE emp_id = 1 AND lt.active = TRUE) AS leave_type_id

我是这样做的，所以我只有1个查询字符串从PHP到SQL服务器并获得实例中的所有结果。我知道子查询会影响性能，但在我的情况下是否有更好的方法来解决我的问题？

示例数据： 时间日志表

离开信用表

Answer 1

以下是替代版本：

SELECT tardy1, tardy2, balance, leave_type_id
FROM (SELECT emp_id, SUM(case when GET_TIME_IN1(emp_id, DATE) then 1 else 0 end) as tardy1,
             SUM(case when pm_in > GET_TIME_IN2(emp_id, DATE) then 1 else 0 end) as tardy2
      FROM timelog
      WHERE emp_id = 1 
      group by emp_id
     ) tardy join
    (SELECT emp_id, balance, leave_type_id
     FROM leave_credit lc full outer JOIN
          leave_type lt
          ON lc.leave_type_id = lt.id
     WHERE emp_id = 1 AND lt.active = TRUE
    ) balance
    on tardy.emp_id = balance.emp_id
where tardy.emp_id = 1 

对所有员工：

SELECT tardy1, tardy2, balance, leave_type_id
FROM (SELECT emp_id, SUM(case when GET_TIME_IN1(emp_id, DATE) then 1 else 0 end) as tardy1,
             SUM(case when pm_in > GET_TIME_IN2(emp_id, DATE) then 1 else 0 end) as tardy2
      FROM timelog
      group by emp_id
     ) tardy full outer join
    (SELECT emp_id, balance, leave_type_id
     FROM leave_credit lc JOIN
          leave_type lt
          ON lc.leave_type_id = lt.id
     WHERE lt.active = TRUE
    ) balance
    on tardy.emp_id = balance.emp_id

如果您尝试组合这些子查询，则必须小心，因为timelog上的多行，以及员工在一个表中但不在另一个表中的可能性。