Question

我想使用With by删除。

所以我尝试执行以下语句：

;WITH cte As
(
 select  emp_num, [from_date],[to_date],[ req_ser], [ req_year] 
from empmission
group by emp_num, [from_date],[to_date],[ req_ser], [ req_year]
having count(*) >2
)

DELETE 
FROM    cte

但我得到以下例外：

无法更新视图或函数'cte'，因为它包含聚合，或DISTINCT或GROUP BY子句，或PIVOT或UNPIVOT 操作

Answer 1

内部联接回原始表并删除它。

;WITH cte As
(
 select   emp_num, [from_date],[to_date],[ req_ser], [ req_year] 
 from     empmission
 group by emp_num, [from_date],[to_date],[ req_ser], [ req_year]
 having   count(*) >2
)    
DELETE E
FROM   cte C
JOIN   empmission E   ON  C.emp_num  = E.emp_num
                      AND C.from_date = E.from_date
                      AND C.to_date   = E.to_date
                      AND C.req_ser   = E.req_ser
                      AND C.req_year  = E.req_year

Answer 2

使用窗口功能：

WITH todelete As (
      select em.*,
             row_number() over (partition by emp_num, [from_date],[to_date],[ req_ser], [ req_year]
                                order by (select null)) as cnt
      from empmission em
)

DELETE FROM todelete
WHERE cnt > 2;

请注意，这将删除具有重复值的所有行。通常，您希望保留其中一个值。如果是这种情况，请提出另一个问题。

Answer 3

另一种方法，跳过cte，改为EXISTS：

delete from empmission e1
where exists (select 1
              from empmission e2
              where e1.emp_num = e1.emp_num
                and e1.[from_date] = e2.[from_date]
                and e1.[to_date] = e2.[to_date]
                and e1.[ req_ser] = e2.[ req_ser]
                and e1.[ req_year] = e2.[ req_year]
              group by emp_num, [from_date],[to_date],[ req_ser], [ req_year]
              having count(*) > 2)

无法更新视图或函数'cte'，因为它包含聚合，或DISTINCT或GROUP BY子句，或PIVOT或UNPIVOT运算符

3 个答案: