当前数据
ID | Name1 | Name2
<guid1> | XMind | MindNode
<guid2> | MindNode | XMind
<guid3> | avast | Hitman Pro
<guid4> | Hitman Pro | avast
<guid5> | PPLive | Hola!
<guid6> | ZenMate | Hola!
<guid7> | Hola! | PPLive
<guid8> | Hola! | ZenMate
要求的输出
ID1 | ID2 | Name1 | Name2
<guid1> | <guid2> | XMind | MindNode
<guid3> | <guid4> | avast | Hitman Pro
<guid5> | <guid7> | PPLive | Hola!
<guid6> | <guid8> | Hola! | ZenMate
这些是应用之间的关系.我想证明阿瓦斯特和希特曼有一种关系,但在这种观点中,我不需要表明他们有什么"方向"关系.在这种观点中,这种关系是双向的.
编辑:似乎我的例子很简单.该解决方案不适用于更多数据.
DECLARE @a TABLE (ID INT, Name1 VARCHAR(50), Name2 VARCHAR(50))
INSERT INTO @a VALUES ( 1, 'XMind', 'MindNode' )
INSERT INTO @a VALUES ( 2, 'MindNode', 'XMind' )
INSERT INTO @a VALUES ( 3, 'avast', 'Hitman Pro' )
INSERT INTO @a VALUES ( 4, 'Hitman Pro', 'avast' )
INSERT INTO @a VALUES ( 5, 'PPLive Video Accelerator', 'Hola! Better Internet' )
INSERT INTO @a VALUES ( 6, 'ZenMate', 'Hola! Better Internet' )
INSERT INTO @a VALUES ( 7, 'Hola! Better Internet', 'PPLive Video Accelerator' )
INSERT INTO @a VALUES ( 8, 'Hola! Better Internet', 'ZenMate' )
SELECT a1.ID AS ID1 ,
a2.ID AS ID2 ,
a1.Name1 ,
a2.Name1 AS Name2
FROM @a a1
JOIN @a a2 ON a1.Name1 = a2.Name2
AND a1.ID < a2.ID -- avoid duplicates
然而这是有效的,所以我猜是Guid正在弄乱我.
再次编辑:
我有一段时间没看过这个,我认为它有效,但我才意识到它没有.我整个上午都在苦苦挣扎,但我必须承认SQL并不是我的强大套件.事情是这样的.
DECLARE @a TABLE (ID int, Name1 VARCHAR(50), Name2 VARCHAR(50))
INSERT INTO @a VALUES ( 1, 'XMind', 'MindNode' )
INSERT INTO @a VALUES ( 2, 'MindNode', 'XMind' )
INSERT INTO @a VALUES ( 3, 'avast', 'Hitman Pro' )
INSERT INTO @a VALUES ( 4, 'PPLive Video Accelerator', 'Hola! Better Internet' )
INSERT INTO @a VALUES ( 5, 'ZenMate', 'Hola! Better Internet' )
INSERT INTO @a VALUES ( 6, 'Hitman Pro', 'avast' )
INSERT INTO @a VALUES ( 7, 'Hola! Better Internet', 'PPLive Video Accelerator' )
INSERT INTO @a VALUES ( 8, 'Hola! Better Internet', 'ZenMate' )
INSERT INTO @a VALUES ( 9, 'XX', 'A' )
INSERT INTO @a VALUES ( 10, 'XX', 'BB' )
INSERT INTO @a VALUES ( 11, 'BB', 'XX' )
INSERT INTO @a VALUES ( 12, 'A', 'XX' )
INSERT INTO @a VALUES ( 13, 'XX', 'CC' )
INSERT INTO @a VALUES ( 14, 'CC', 'XX' )
;With CTE as
(
SELECT a1.ID AS ID1 ,
a2.ID AS ID2 ,
a1.Name1 ,
a2.Name1 AS Name2,
CheckSum(Case when a1.Name1>a2.Name1 then a2.Name1+a1.Name1 else a1.Name1+a2.Name1 end) ck, -- just for display
Row_Number() over (Partition by CheckSum(Case when a1.Name1>a2.Name1 then a2.Name1+a1.Name1 else a1.Name1+a2.Name1 end)
order by CheckSum(Case when a1.Name1>a2.Name1 then a2.Name1+a1.Name1 else a1.Name1+a2.Name1 end)) as rn
FROM @a a1
JOIN @a a2 ON a1.Name1 = a2.Name2
)
Select ID1, ID2,Name1, Name2
from CTE C1
where rn=1
当我使用此代码时,它确实可以正常使用名称,但它与ID无法正确匹配.
结果是
ID1 | ID2 | Name1 | Name2
12 | 9 | A | X (Correct)
7 | 5 | Hola! | ZenMate (Not Correct)
[..]
我整个早上都拔了头发,但我想不出来.我仍然使用Guid作为ID,并在这里使用Int来使它更具可读性.
DECLARE @a TABLE (ID INT, Name1 VARCHAR(50), Name2 VARCHAR(50))
INSERT INTO @a VALUES ( 1, 'XMind', 'MindNode' )
INSERT INTO @a VALUES ( 2, 'MindNode', 'XMind' )
INSERT INTO @a VALUES ( 3, 'avast', 'Hitman Pro' )
INSERT INTO @a VALUES ( 4, 'Hitman Pro', 'avast' )
SELECT a1.ID AS ID1 ,
a2.ID AS ID2 ,
a1.Name1 ,
a2.Name1 AS Name2
FROM @a a1
JOIN @a a2 ON a1.Name1 = a2.Name2
AND a1.ID < a2.ID -- avoid duplicates
提到你的问题的修改和扩展,需要一个更复杂的解决方案.我们在a1.Name1,a2.Name上形成一个CHECKSUM(以获得我们在大小上交换的相同内容).
使用此方法,我们使用ROW_NUMBER(Transact-SQL)生成一个数字,并仅使用数字为1的结果中的行.
DECLARE @a TABLE (ID uniqueIdentifier, Name1 VARCHAR(50), Name2 VARCHAR(50))
INSERT INTO @a VALUES ( NewID(), 'XMind', 'MindNode' )
INSERT INTO @a VALUES ( NewID(), 'MindNode', 'XMind' )
INSERT INTO @a VALUES ( NewID(), 'avast', 'Hitman Pro' )
INSERT INTO @a VALUES ( NewID(), 'Hitman Pro', 'avast' )
INSERT INTO @a VALUES ( NewID(), 'PPLive Video Accelerator', 'Hola! Better Internet' )
INSERT INTO @a VALUES ( NewID(), 'ZenMate', 'Hola! Better Internet' )
INSERT INTO @a VALUES ( NewID(), 'Hola! Better Internet', 'PPLive Video Accelerator' )
INSERT INTO @a VALUES ( NewID(), 'Hola! Better Internet', 'ZenMate' )
INSERT INTO @a VALUES ( NewID(), 'XX', 'A' )
INSERT INTO @a VALUES ( NewID(), 'A', 'XX' )
INSERT INTO @a VALUES ( NewID(), 'XX', 'BB' )
INSERT INTO @a VALUES ( NewID(), 'BB', 'XX' )
INSERT INTO @a VALUES ( NewID(), 'XX', 'CC' )
INSERT INTO @a VALUES ( NewID(), 'CC', 'XX' )
;With CTE as
(
SELECT a1.ID AS ID1 ,
a2.ID AS ID2 ,
a1.Name1 ,
a2.Name1 AS Name2,
CheckSum(Case when a1.Name1>a2.Name1 then a2.Name1+a1.Name1 else a1.Name1+a2.Name1 end) ck, -- just for display
Row_Number() over (Partition by CheckSum(Case when a1.Name1>a2.Name1 then a2.Name1+a1.Name1 else a1.Name1+a2.Name1 end)
order by CheckSum(Case when a1.Name1>a2.Name1 then a2.Name1+a1.Name1 else a1.Name1+a2.Name1 end)) as rn
FROM @a a1
JOIN @a a2 ON a1.Name1 = a2.Name2
)
Select *
from CTE C1
where rn=1
编辑:
如果您只想获得两个字段都适合的那些,那么所需的查询就是:
SELECT a1.ID AS ID1 , a2.ID AS ID2 , a1.Name1 , a2.Name1 AS Name2
FROM @a a1
JOIN @a a2 ON a1.Name1 = a2.Name2 and a1.Name2 = a2.Name1 AND a1.ID < a2.ID
如果输出应仅包含双向关系('XX' + 'A') AND ('A' + 'XX')，请尝试以下操作：
;
WITH m (ID1, ID2, Name1, Name2) AS (
SELECT ID1, ID2, Name1, Name2
FROM (
SELECT a1.ID AS ID1
,a2.ID AS ID2
,a1.Name1 AS Name1
,a2.Name1 AS Name2
,ROW_NUMBER() OVER (PARTITION BY a1.Name1, a2.Name1 ORDER BY (SELECT 1)) AS n
FROM @a AS a1
JOIN @a AS a2
ON a1.Name1 = a2.Name2
AND a1.Name2 = a2.Name1
) AS T
WHERE n = 1
)
SELECT DISTINCT *
FROM (
SELECT ID1, ID2, Name1, Name2
FROM m
WHERE ID1 <= ID2
UNION ALL
SELECT ID2, ID1, Name2, Name1
FROM m
WHERE ID1 > ID2
) AS dm
它产生如下输出：
+------+-----+--------------------------+-----------------------+
| ID1 | ID2 | Name1 | Name2 |
+------+-----+--------------------------+-----------------------+
| 1 | 2 | XMind | MindNode |
| 3 | 6 | avast | Hitman Pro |
| 4 | 7 | PPLive Video Accelerator | Hola! Better Internet |
| 5 | 8 | ZenMate | Hola! Better Internet |
| 9 | 12 | XX | A |
| 10 | 11 | XX | BB |
| 13 | 14 | XX | CC |
+------+-----+--------------------------+-----------------------+
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