Question

select * from Employee
order by 
  case left(right(Registration_no, 2), 1) 
    when '/' then 
      left(Registration_no, len(Registration_no) - 1) + '0' + right(Registration_no, 1)
    else Registration_no
  end

好的，我可以得到id string id2 1 a 1 2 b 1 3 a 2 4 c 2 5 d 3：

SELECT * FROM table WHERE string NOT IN ('a','c')

但是我也不想得到包含具有相同id2的a或c的其余行。结果：

2        b         1
5        d         3

Answer 1

具有一个子查询，该子查询返回具有a或c的id2值。进行NOT IN个id2值。

SELECT * FROM table
WHERE id2 not in (select id2 from table where string IN ('a','c'))

如果id2可以包含空值，请改为使用NOT EXISTS：

SELECT * FROM table t1
WHERE NOT EXISTS (select 1 from table t2
                  where t2.string IN ('a','c')
                    and t1.id2 = t2.id2))

Answer 2

您可以使用not exists：

select t.*
from t
where not exists (select 1
                  from t t2
                  where t2.id2 = t.id2 and t2.string in ('A', 'C')
                 );

如果只希望使用id2值，则可能会发现聚合很方便：

select id2
from t
group by id2
having sum(case when string in ('A', 'C') then 1 else 0 end) = 0;

如何查询以获得该结果？

2 个答案: