SQL:是否获取链接到一个项目的多个行条目？

我有一张桌子：

ID | ITEMID | STATUS | TYPE
1  | 123    | 5      | 1
2  | 123    | 4      | 2
3  | 123    | 5      | 3
4  | 125    | 3      | 1
5  | 125    | 5      | 3

此表中的任何项都可以有0到多个条目。我需要一个查询，告诉我一个项目的所有条目是否处于5或4的状态。例如，在上面的示例中，我希望得到以下结果：

ITEMID | REQUIREMENTS_MET
123    | TRUE    --> true because all statuses are either 5 or 4
125    | FALSE   --> false because it has a status of 3 and a status of 5. 
                     If the 3 was a 4 or 5, then this would be true

更好的办法是这样：

ITEMID | MET_REQUIREMENTS | NOT_MET_REQUIREMENTS
123    | 3                | 0
125    | 1                | 1

知道如何为此编写查询吗？

简单的一个：

select
    "ITEMID",
    case
        when min("STATUS") in (4, 5) and max("STATUS") in (4, 5) then 'True'
        else 'False'
    end as requirements_met
from table1
group by "ITEMID"

更好的一个：

select
    "ITEMID",
    sum(case when "STATUS" in (4, 5) then 1 else 0 end) as MET_REQUIREMENTS,
    sum(case when "STATUS" in (4, 5) then 0 else 1 end) as NOT_MET_REQUIREMENTS
from table1
group by "ITEMID";

---

一种方法是

SELECT t1.itemid, NOT EXISTS(SELECT 1
                             FROM mytable t2
                             WHERE itemid=t1.itemid
                             AND status NOT IN (4, 5)) AS requirements_met
FROM mytable t1
GROUP BY t1.itemid

更新：对于更新后的需求，您可以有如下内容：

SELECT itemid,
       sum(CASE WHEN status IN (4, 5) THEN 1 ELSE 0 END) as met_requirements,
       sum(CASE WHEN status IN (4, 5) THEN 0 ELSE 1 END) as not_met_requirements
FROM mytable
GROUP BY itemid

---

没关系，这其实很容易做到：

select  ITEM_ID , 
  sum (case when STATUS >= 3 then 1 else 0 end ) as met_requirements, 
  sum (case when STATUS < 3 then 1 else 0 end ) as not_met_requirements
from TABLE as d
group by ITEM_ID

也可以通过外部联接完成：

WITH    yes AS ( SELECT item_id, COUNT(*) AS good_count FROM items WHERE status IN (4,5) GROUP BY item_id)
,       no AS ( SELECT item_id, COUNT(*) AS bad_count FROM items WHERE status NOT IN (4,5) GROUP BY item_id)
SELECT COALESCE(y.item_id, n.item_id) AS item_id
    , COALESCE(y.good_count,0) AS good_count
    , COALESCE(n.bad_count,0) AS bad_count
FROM yes y
FULL JOIN no n ON n.item_id = y.item_id
    ;

快速、简短、简单：

SELECT itemid
      ,count(status = 4 OR status = 5 OR NULL) AS met_requirements
      ,count(status < 4 OR status > 5 OR NULL) AS not_met_requirements
FROM   tbl
GROUP  BY itemid
ORDER  BY itemid;

假设所有列都是整数而不是NULL

建立在： TRUE或NULL将生成TRUE FALSE或NULL产生NULL

并且NULL不按计数计数

---

你说得对，Aleks G，我只想展示一种可能的嵌套selectsMissing方法：表定义，Postgres版本。任何列都可以为空吗？

WITH    yes AS ( SELECT item_id, COUNT(*) AS good_count FROM items WHERE status IN (4,5) GROUP BY item_id)
,       no AS ( SELECT item_id, COUNT(*) AS bad_count FROM items WHERE status NOT IN (4,5) GROUP BY item_id)
SELECT COALESCE(y.item_id, n.item_id) AS item_id
    , COALESCE(y.good_count,0) AS good_count
    , COALESCE(n.bad_count,0) AS bad_count
FROM yes y
FULL JOIN no n ON n.item_id = y.item_id
    ;

SELECT itemid
      ,count(status = 4 OR status = 5 OR NULL) AS met_requirements
      ,count(status < 4 OR status > 5 OR NULL) AS not_met_requirements
FROM   tbl
GROUP  BY itemid
ORDER  BY itemid;