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「PERSON」というテーブルがあります

GROUP_ID  PERSON_NAME
1001       ALEX
1002       MATHEW
1001       GEORGE
1002       THOMAS
1001       JAMES

以下のような出力が必要です"|"

GROUP_ID   PERSON_NAME
1001       ALEX|GEORGE|JAMES
1002       THOMAS|MATHEW
4

1 に答える 1

0
SELECT  a.[GROUP_ID], 
        SUBSTRING(d.Names,1, LEN(d.Names) - 1) NamesList
FROM
        (
            SELECT DISTINCT [GROUP_ID]
            FROM Table1
        ) a
        CROSS APPLY
        (
            SELECT [PERSON_NAME] + '|' 
            FROM table1 AS B 
            WHERE A.[GROUP_ID] = B.[GROUP_ID] 
            FOR XML PATH('')
        ) D (Names) 

また

SELECT [GROUP_ID],
       STUFF(
           (SELECT '|' + [PERSON_NAME]
            FROM Table1
            WHERE [GROUP_ID] = a.[GROUP_ID]
            FOR XML PATH (''))
            , 1, 1, '')  AS NamesList
FROM Table1 AS a
GROUP BY [GROUP_ID]
于 2012-11-26T00:01:18.077 に答える