GROUP BY 没有按预期工作

Question

这是我的样本 table，只有一点信息。

select * from juniper_fpc';       
    id     |  router   |     part_name      
-----------+-----------+--------------------
 722830939 | BBBB-ZZZ1 | MPC-3D-16XGE-SFPP 
 722830940 | BBBB-ZZZ1 | MPC-3D-16XGE-SFPP 
 723103163 | AAAA-ZZZ1 | DPCE-R-40GE-SFP   
 723103164 | AAAA-ZZZ1 | MPC-3D-16XGE-SFPP 
 723103172 | AAAA-ZZZ1 | DPCE-R-40GE-SFP   
 722830941 | BBBB-ZZZ1 | MPC-3D-16XGE-SFPP

我想做的是从路由器列中识别出只有 part_name 条目以 MPC 开头的元素。我想出的是这个，但它是错误的，因为它列出了上面的两个元素。

SELECT   router
FROM     juniper_fpc
WHERE    part_name LIKE 'MPC%'
GROUP BY router
ORDER BY router;
  router   
-----------
 AAAA-ZZZ1
 BBBB-ZZZ1

Answer 1

如果您的意思是 select 只有那些路由器，它们的所有部件名称都以 MPC 开头，那么您的查询应该是：

SELECT   s.router
FROM     juniper_fpc s
WHERE    NOT EXISTS(select distinct id from juniper_fpc t
                    where t.id = s.id part_name NOT LIKE 'MPC%')
GROUP BY s.router
ORDER BY s.router;

Answer 2

假设您想要只有 part_name 的路由器，例如 'MPC%'，您可以使用条件计数：

select * from (
  select router, 
    count(case when part_name like 'MPC%' then 1 else null end) as cnt_mpc,
    count(*) as cnt_overall
  from juniper_fpc  
  group by router) v_inner
where cnt_mpc = cnt_overall

这可以写得更紧凑（尽管可读性稍差）如

  select router
  from juniper_fpc  
  group by router
  having count(case when part_name like 'MPC%' then 1 else null end)  = count(*)

SQL Fiddle

Answer 3

这应该表现良好：

SELECT j1.router
FROM  (
   SELECT   router
   FROM     juniper_fpc
   WHERE    part_name LIKE 'MPC%'
   GROUP    BY router
   ) j1
LEFT   JOIN juniper_fpc j2 ON j2.router = j1.router
                          AND j2.part_name NOT LIKE 'MPC%'
WHERE  j2.router IS NULL
ORDER  BY j1.router;

和 NOT EXISTS 也可以，如果你做对了：

SELECT router
FROM   juniper_fpc j
WHERE  NOT EXISTS (
   SELECT 1
   FROM   juniper_fpc
   WHERE  router = j.router
   AND    part_name NOT LIKE 'MPC%'
   )
GROUP  BY router
ORDER  BY router;

详情：

Select rows which are not present in other table

SQL Fiddle.

或者，使用 Postgres 9.4 或更高版本的语法：

SELECT router
FROM   juniper_fpc
GROUP  BY router
HAVING count(*) = count(*) FILTER (WHERE part_name LIKE 'MPC%')
ORDER  BY router;

最好为每个人在 (router, partname) 上建立索引。

Answer 4

你也可以 Window 这里的功能：

SELECT
    *
FROM
    (
        SELECT 
            router,
            part_name,
            COUNT(distinct part_name) OVER (PARTITION BY router) as count_of_distinct_parts
        FROM juniper_fpc
    )subqry
WHERE part_name like 'MPC%' AND count_of_distinct_parts = 1

如果此查询范围扩大，这将为更复杂的条件打开大门。

GROUP BY 没有按预期工作

GROUP BY isn't working as expected

sql

postgresql

aggregate