Question

我有一些以下格式的传入行。

| Col1 | Col2 | Col3 |
| 1    | A    | 1    |
| 1    | A    | 1,2  |
| 1    | A    | 1,3  |
| 1    | A    | 2,4  |

所需的输出是

| Col1 | Col2 | Col3    |
| 1    | A    | 1,2,3,4 |

基本上，根据Col1和Col2对所有行进行分组，然后从Col3连接并删除重复项。

SELECT COL1, COL2, {?????}
FROM TABLEA
GROUP BY COL1, COL2;

此刻我想不出多少。任何指针都将非常感激。我倾向于使用WX2数据库，但任何符合ANSI标准的代码段都会有所帮助。

Answer 1

对于SQL Server：首先使用STUFF方法和INSERT INTO CTE表连接所有col3值。基于此CTE表，基于CTE表将所有行分离为单个列。最后在STUFF的帮助下连接所有DISTINCT字符串。< / p>

 CREATE TABLE #table ( Col1 INT ,   Col2 VARCHAR(10) , Col3 VARCHAR(10))
 INSERT INTO #table ( Col1  ,   Col2  , Col3 )
 SELECT  1   , 'A'   , '1'  UNION ALL 
 SELECT  1   , 'A'   , '1,2' UNION ALL
 SELECT  1   , 'A'   , '1,3'  UNION ALL
 SELECT  1   , 'A'   , '2,4' 

 ;WITH CTEValues ( Colval ) AS
 (
   SELECT STUFF ( ( SELECT ',' +  Col3 FROM #table T2 WHERE T2.Col2 =  
                T1.col2 FOR XML PATH('') ),1,1,'') 
   FROM #table T1 
   GROUP BY Col2
 )

 SELECT * INTO #CTEValues
 FROM CTEValues

 ;WITH CTEDistinct ( SplitValues , SplitRemain ) AS
 (
  SELECT SUBSTRING(Colval,0,CHARINDEX(',',Colval)),    
         SUBSTRING(Colval,CHARINDEX(',',Colval)+1,LEN(Colval))
  FROM #CTEValues
  UNION ALL
  SELECT CASE WHEN CHARINDEX(',',SplitRemain) = 0 THEN SplitRemain ELSE  
                   SUBSTRING(SplitRemain,0,CHARINDEX(',',SplitRemain)) END, 
         CASE WHEN CHARINDEX(',',SplitRemain) = 0 THEN '' ELSE                               
        SUBSTRING(SplitRemain,CHARINDEX(',',SplitRemain)+1,LEN(SplitRemain))          
        END
  FROM CTEDistinct
  WHERE SplitRemain <> ''
 )

  SELECT STUFF ( ( SELECT DISTINCT ',' +  SplitValues FROM CTEDistinct T2  
  FOR XML PATH('') ),1,1,'')

Answer 2

您可以尝试使用转置或连接功能。困难来自于col3是varchar并且需要转换才能获得不同的值。使用MySQL：

SELECT col1, col2, GROUP_CONCAT(DISTINCT col3) AS col3 FROM
(SELECT col1, col2, CONVERT(SUBSTR(col3, 1), UNSIGNED INTEGER) AS col3 FROM (
SELECT 1 AS col1, 'A' AS col2, '1' AS col3 UNION ALL
SELECT 1 AS col1, 'A' AS col2, '1,2' AS col3 UNION ALL
SELECT 1 AS col1, 'A' AS col2, '1,3' AS col3 UNION ALL  
SELECT 1 AS col1, 'A' AS col2, '2,4' AS col3
) AS t
UNION ALL
SELECT col1, col2, CONVERT(SUBSTR(col3, 3), UNSIGNED INTEGER) AS col3 FROM (
SELECT 1 AS col1, 'A' AS col2, '1' AS col3 UNION ALL
SELECT 1 AS col1, 'A' AS col2, '1,2' AS col3 UNION ALL
SELECT 1 AS col1, 'A' AS col2, '1,3' AS col3 UNION ALL  
SELECT 1 AS col1, 'A' AS col2, '2,4' AS col3
) AS t1
) AS t2
WHERE col3 <> 0

结果：

col1 | col2 | col3 
1    |   A  | 1,2,3,4

Answer 3

对于Postgres使用此：

ip route get 8.8.8.8 2> /dev/null | awk 'NR==1 {print $NF}'

除select col1, col2, string_agg(distinct col3, ',') as col3 from ( select col1, col2, x.col3 from tablea, unnest(string_to_array(col3, ',')) as x(col3) ) t group by col1, col2;和string_to_array()函数外，这主要符合ANSI标准。

连接和重复多行

3 个答案: