我有一张具有以下结构的表......
+--------+------+------+------+------+------+
| ID | colA | colB | colC | colD | colE | [...] etc.
+--------+------+------+------+------+------+
| 100100 | 15 | 100 | 90 | 80 | 10 |
+--------+------+------+------+------+------+
| 100200 | 10 | 80 | 90 | 100 | 10 |
+--------+------+------+------+------+------+
| 100300 | 100 | 90 | 10 | 10 | 80 |
+--------+------+------+------+------+------+
我需要返回列名的连接值,每行最多包含3个值...
+--------+----------------------------------+
| ID | maxCols |
+--------+----------------------------------+
| 100100 | colB,colC,colD |
+--------+------+------+------+------+------+
| 100200 | colD,colC,colB |
+--------+------+------+------+------+------+
| 100300 | colA,colB,colE |
+--------+------+------+------+------+------+
maxCol1 | maxCol2 | maxCol3
答案 0 :(得分:2)
您可以使用UNPIVOT
并为每个ID获得TOP 3
;with temp AS
(
SELECT ID, ColValue, ColName
FROM @SampleData sd
UNPIVOT
(
ColValue For ColName in ([colA], [colB], [colC], [colD], [colE])
) unp
)
SELECT sd.ID, ca.ColMax
FROM @SampleData sd
CROSS APPLY
(
SELECT STUFF(
(
SELECT TOP 3 WITH TIES
',' + t.ColName
FROM temp t
WHERE t.ID = sd.ID
ORDER BY t.ColValue DESC
FOR XML PATH('')
)
,1,1,'') AS ColMax
) ca
在此处查看演示:http://rextester.com/CZCPU51785
答案 1 :(得分:1)
以下是使用Cross Apply
和Table Valued Constructor
SELECT Id,
maxCols= Stuff(cs.maxCols, 1, 1, '')
FROM Yourtable
CROSS apply(SELECT(SELECT TOP 3 ',' + NAME
FROM (VALUES (colA,'colA'),(colB,'colB'),(colC,'colC'),
(colD,'colD'),(colE,'colE')) tc (val, NAME)
ORDER BY val DESC
FOR xml path, type).value('.[1]', 'nvarchar(max)')) cs (maxCols)
如果需要,可以使用Information_schema.Columns