[英]MySQL Subquery Alternative
我有一个包含4个子查询的查询。 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
我这样做是为了让我只有1个查询字符串(从PHP到SQL服务器),并将所有结果存储在一个实例中。 我知道子查询会影响性能,但是在我的情况下,有没有更好的方法来解决我的问题?
样本数据:Timelog表
离开信用表
这是替代版本:
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
对于所有员工:
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
如果尝试合并这些子查询,则必须小心,因为timelog
上有多行,而且雇员可能在一个表中,而不在另一个表中。
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