mysql 查询中的分组记录给出错误

Question

我有一个查询应该读取 leave_requests_2014 table 中的记录，并对每个员工的所有记录的已接受休假请求执行 sum()！

SELECT lr.emp_id employee,
ifnull(MAX(CASE WHEN lr.leave_type =2 THEN (sum((TO_DAYS(lr.end_date) - TO_DAYS(lr.start_date)+1))) END ) ,0) annual_leave,
ifnull(MAX(CASE WHEN lr.leave_type =3 THEN (sum((TO_DAYS(lr.end_date) - TO_DAYS(lr.start_date)+1))) END ),0) sick_leave
FROM leave_requests_2014 lr
WHERE lr.status =5
GROUP BY lr.emp_id

查询表明returns出现如下错误：

#1111 - Invalid use of group function

我真的不知道应该对什么进行分组才能获得所需的输出！

请看一下sqlfiddle这是我真正需要的结果

employee    annual_leave    sick_leave
   5             7              2
   6             4              1

Answer 1

一个更简单（且正确）的查询版本：

SELECT lr.emp_id AS employee,
   SUM(IF(lr.leave_type = 2, DATEDIFF(lr.end_date, lr.start_date) + 1, 0)) AS annual_leave,
   SUM(IF(lr.leave_type = 3, DATEDIFF(lr.end_date, lr.start_date) + 1, 0)) AS sick_leave,
FROM leave_requests_2014 lr
WHERE lr.status = 5
GROUP BY lr.emp_id

另一个版本，更简单，它在不同的行上生成年假和病假天数（如果需要）：

SELECT lr.emp_id AS employee, lr.leave_type,
   SUM(DATEDIFF(lr.end_date, lr.start_date) + 1) AS leave_days,
FROM leave_requests_2014 lr
WHERE lr.status = 5 AND lr.leave_type IN (2, 3)
GROUP BY lr.emp_id, lr.leave_type

这需要对客户端代码进行一些处理：检查 leave_type 的值以了解 leave_days 中的值是 "annual_leave" 还是 "sick_leave".如果查询未返回其他类型的休假天数，请使用 0

Answer 2

"Group by" MAX() and/or SUM() 等函数不能在同一语句中递归执行。一种解决方案是在子查询中执行第一个聚合。例如：

SELECT employee, 
ifnull(MAX(CASE WHEN leave_type =2 THEN annual_leave END),0) annual_leave,
ifnull(MAX(CASE WHEN leave_type =3 THEN sick_leave END),0) sick_leave
FROM
    (
        SELECT lr.emp_id employee, lr.leave_type 
        (CASE WHEN lr.leave_type =2 THEN (sum((TO_DAYS(lr.end_date) - TO_DAYS(lr.start_date)+1))) END ) annual_leave,
        (CASE WHEN lr.leave_type =3 THEN (sum((TO_DAYS(lr.end_date) - TO_DAYS(lr.start_date)+1))) END ) sick_leave
        FROM leave_requests_2014 lr
        WHERE lr.status =5
        GROUP BY lr.emp_id, lr.leave_type
    ) firstAggregation
group by employee

mysql 查询中的分组记录给出错误

Grouping records in mysql query gives error

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