Question

我有两张桌子，T1和T2如下：

CATEGORY     ID
1           1100
1           1200
1           1300
1           1500 
2           2000
2           2100
2           2300
2           2500

我需要知道：

T1和T2之间有多少行相似（相同的CATEGORY和ID）
T2中有多少行不在T1中
T1中有多少行不在T2

从今天早晨开始，我就紧紧抓住它，并尝试这样做以获得类似的行：

select count(*) from T1, T2 WHERE 
T1.CATEGORY = T2.CATEGORY AND T1.ID = T2.ID;

但我无法弄清楚如何获得唯一的行（仅在T1或T2中）。

Answer 1

问题1

SELECT  COUNT(*) totalCount
FROM    T1 a 
        INNER JOIN T2 b
            ON a.Category = b.Category AND
                a.ID = b.ID

问题2（使用LEFT JOIN ）

SELECT  COUNT(*) totalCount
FROM    T2 a 
        LEFT JOIN T1 b
            ON a.Category = b.Category AND
                a.ID = b.ID
WHERE   b.Category IS NULL

问题3（使用LEFT JOIN ）

SELECT  COUNT(*) totalCount
FROM    T1 a 
        LEFT JOIN T2 b
            ON a.Category = b.Category AND
                a.ID = b.ID
WHERE   b.Category IS NULL

Answer 2

DROP SCHEMA tmp CASCADE;
CREATE SCHEMA tmp ;
SET search_path=tmp;

CREATE TABLE lutser
        ( id INTEGER NOT NULL
        , category INTEGER NOT NULL
        );
INSERT INTO lutser(category, id) VALUES
(1,1100) ,(1,1200) ,(1,1300) ,(1,1500)
,(2,2000) ,(2,2100) ,(2,2300) ,(2,2500)
,(1,3500) -- added these
,(2,3500)
        ;

这些查询为类别== 1,2构建“位掩码”1，为类别== 2构建2，并将它们相加。因此，当id存在于两个集合中时，掩码为3，仅当在第一个集合中时为1，而仅在第二个集合中时为2。外连接+合并在这里做了诀窍。

        --
        -- CTE version
        --
WITH flags AS (
        WITH one AS ( SELECT category AS flag , id FROM lutser WHERE category = 1)
        , two AS ( SELECT category AS flag , id FROM lutser WHERE category = 2)
        SELECT COALESCE(one.flag, 0) + COALESCE(two.flag, 0) AS flag
        FROM one
        FULL OUTER JOIN two ON two.id = one.id
        )
SELECT flag, COUNT(*)
FROM flags
GROUP BY flag;

        --
        -- Non-CTE version
        --
SELECT COALESCE(one.flag, 0) + COALESCE(two.flag, 0) AS flags
        , COUNT(*)
FROM (
        SELECT category AS flag , id
        FROM lutser WHERE category = 1
        ) one
FULL OUTER JOIN (
        SELECT category AS flag , id
        FROM lutser WHERE category = 2
        ) two ON two.id = one.id
GROUP BY flags;

结果（对于两个查询; - ）：

 flags | count 
-------+-------
     1 |     4
     2 |     4
     3 |     1

Answer 3

如果您不能假设行是不同的，那么您需要采取稍微不同的方法。这是一个同时回答所有三个问题的方法，考虑到重复的行：

select (case when isT1 = 1 and isT2 = 0 then 'BOTH'
             when isT1 = 1 then 'T1-Only'
             else 'T2-Only'
        end) as WhereRow,
       count(*) as NumDistinctRows,
       sum(cnt) as NumTotalRows
from ((select category, id, count(*) as cnt, 1 as isT1, 0 as isT2
       from t1
       group by category, id
      ) union all
      (select category, id, count(*) as cnt, 0 as isT1, 1 as isT2
       from t2
       group by category, id
      )
     ) t
group by isT1, isT2

从两个表中选择计算不同行和公共行

3 个答案: