2017-01-03 72 views

回答

1
;WITH tb(id,id_ang,bulan)AS(
     SELECT 1,1,1 UNION 
     SELECT 2,2,1 UNION 
     SELECT 3,1,13 UNION 
     SELECT 4,1,2 
    ) 
    SELECT id_ang,STUFF(c.bulans,1,1,'') AS bulans FROM tb 
    OUTER APPLY(SELECT ','+LTRIM(bulan) FROM tb AS tt WHERE tt.id_ang=tb.id_ang ORDER BY ID FOR XML PATH('')) c(bulans) 
    GROUP BY id_ang,c.bulans 
 
id_ang bulans 
1 1,13,2 
2 1 
+0

感謝諾蘭。這是完美的工作 –

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