GROUP_CONCAT 不考虑过滤器

Question

我有一个名为 attendance

的示例表

+------+--------------+--------------+-------------+--------------------+
| id   | employee_id  | date         | total_work  | is_regular_holiday |
+------+--------------+--------------+-------------+--------------------+
| 1    | e10991       |  11-01-2020  | 28800       | 1                  |
+------+--------------+--------------+-------------+--------------------+
| 2    | e10991       |  11-02-2020  | 28800       | 0                  |
+------+--------------+--------------+-------------+--------------------+
| 3    | e10992       |  11-01-2020  | 28800       | 1                  |
+------+--------------+--------------+-------------+--------------------+
| 4    | e10992       |  11-02-2020  | 28800       | 0                  |
+------+--------------+--------------+-------------+--------------------+
| 5    | e10993       |  11-02-2020  | 28800       | 0                  |
+------+--------------+--------------+-------------+--------------------+

现在我有一个问题

select 
  employee_id, 
  sum(total_work) as `total_working_hours`
from attendance
group by employee_id;

这个查询有效

现在我想得到 total_work 的所有总和（不管 is_regular_holiday = 1 与否），假期工作（如果 is_regular_holiday = 1）连接到一个字符串结果，

我尝试使用下面的查询

select 
  employee_id, 
  sum(total_work) as `total_working_hours`, 
  group_concat(select date from attendance_2 where is_holiday = 1) as holidays_worked  
from attendance
group by employee_id;

它返回一个语法错误

我在查询中遗漏了什么方法，或者可以做些什么来获得 group_concat holiday_worked？

任何帮助都会很棒！

Answer 1

考虑group_concat() 中的case 表达式：

select 
    employee_id, 
    sum(total_work) as total_working_hours, 
    group_concat(case when is_regular_holiday then date end order by date) as holidays_worked  
from attendance
group by employee_id;