SELECT du_group_id,SUM(count) as count from ((SELECT du_group_id,COUNT(du_group_id) as count from doctor_user where doctor_id=10937 GROUP BY du_group_id ) UNION (SELECT du_group_id,0 as count FROM doctor_user_group dug ))
as duu GROUP BY duu.du_group_id
查询分组 对应记录值
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转载自1035054540-qq-com.iteye.com/blog/2215912
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