我有两个桌子
餐桌假
ID | TYPE
1 Annual
2 Sick
3 Unpaid
4 Marriage
表离开数据
IDLEAVEDATA | LEAVETYPE*
1 1
2 1
3 2
4 2
LEAVETYPE是外键(请参阅表假中的ID)
如何计算ID在表LeaveData中的出现?
输出示例:
TYPE | COUNT
Annual 2
Sick 2
Unpaid 0 or null
Marriage 0 or null
最佳答案
尝试,
SELECT l.TYPE , COUNT(ld.LEAVETYPE) as COUNT
FROM Leave AS l
LEFT JOIN LeaveData AS ld ON ld.LEAVETYPE = L.ID
GROUP BY ld.LEAVETYPE
关于mysql - mysql连接数,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/3798456/