SELECT * FROM task a, (SELECT id, name, addtime FROM task_step j, (SELECT id, name, max(addtime) addtime FROM task_step GROUP BY id) q WHERE j.id=q.id AND j.addtime=q.addtime) WHERE a.id=b.id
SELECT * FROM task a, (SELECT id, name, addtime FROM task_step j, (SELECT id, name, max(addtime) addtime FROM task_step GROUP BY id) q WHERE j.id=q.id AND j.addtime=q.addtime) WHERE a.id=b.id
https://www.gbtgames.com/thread-1198.htm 都是安裝包, 扔迅雷裏面就行. 都很乾淨沒廣告.
1、數據 2、全選日期(從8到22的所有日期) 3、爲全選日期新增規則 規則1(當前日期>=開始日期,當前日期<=結束日期) 公式裏面$符號去掉格式如下 格式化如下 點確定如下 規則2(超過當前日期)