我有 3 个这样的表:
表事件
+------+----------+----------------------+
| ID | Title | Employees |
+------+----------+----------------------+
| 1 | Event1 | john,james |
+------+----------+----------------------+
| 2 | Event2 | sarah,jessica |
+------+----------+----------------------+
表 checkin
+------+----------+----------+---------------------+
| ID | Time | EventID | By |
+------+----------+----------+---------------------+
| 1 | 08:30 | 1 | john |
+------+----------+----------+---------------------+
| 2 | 08:30 | 1 | james |
+------+----------+----------+---------------------+
| 3 | 09:30 | 1 | john |
+------+----------+----------+---------------------+
| 4 | 10:30 | 2 | sarah |
+------+----------+----------+---------------------+
| 5 | 10:35 | 2 | sarah |
+------+----------+----------+---------------------+
表格问题
+------+----------------+----------+---------------------+
| ID | Comment | EventID | By |
+------+----------------+----------+---------------------+
| 1 | Broken door | 1 | john |
+------+----------------+----------+---------------------+
| 2 | Slippery floor | 1 | john |
+------+----------------+----------+---------------------+
| 3 | Leaking tap | 1 | john |
+------+----------------+----------+---------------------+
| 4 | Broken window | 2 | jessica |
+------+----------------+----------+---------------------+
| 5 | Broken glass | 2 | jessica |
+------+----------------+----------+---------------------+
我想打印这样的东西:
+------+----------+---------------+-------------------+-------------------+
| ID | Title | Employees | Count_Check_In | Count_Problems |
+------+----------+---------------+-------------------+-------------------+
| 1 | Event1 | john,james | john:2,james:1 | john:3,james:0 |
+------+----------+---------------+-------------------+-------------------+
| 2 | Event2 | sarah,jessica | sarah:2,jessica:0 | sarah:0,jessica:2 |
+------+----------+---------------+-------------------+-------------------+
我知道如果数据库设计得当，这个问题会很微不足道，但我们目前没有重写应用程序的奢侈。
最佳答案
您首先需要使用 union 从签到表和问题表中获取每个事件 ID 的所有员工。
然后 left join 将每个签到表和问题表的计数与之前的结果相加，也得到 0 计数。
最后使用 group_concat 为每个事件 ID 在一行中获取结果。
select te.id,te.title,te.employees
,group_concat(concat(t.`By`,':',coalesce(tccnt.cnt,0))) count_check_in
,group_concat(concat(t.`By`,':',coalesce(tpcnt.cnt,0))) count_problems
from table_events te
left join (select eventid,`By` from table_check_in
union
select eventid,`By`from table_problems) t on te.id = t.eventid
left join (select eventid,`By`,count(*) cnt from table_check_in group by eventid,`By`) tccnt on tccnt.eventid = t.eventid and tccnt.`By`=t.`By`
left join (select eventid,`By`,count(*) cnt from table_problems group by eventid,`By`) tpcnt on tpcnt.eventid = t.eventid and tpcnt.`By`=t.`By`
group by te.id,te.title,te.employees
Sample Demo (感谢@valex 设置架构)
关于MySQL 计数逗号分隔，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/41165507/