由於您使用的是SQL Server,因此您可以通過幾種不同的方式獲得結果,但所有這些方法都涉及使用row_number()
。
可以使用聚合函數CASE表達式:
select visitnumber,
max(case when seq=1 then id end) ID1,
max(case when seq=2 then id end) ID2,
max(case when seq=3 then id end) ID3,
max(case when seq=4 then id end) ID4,
max(case when seq=5 then id end) ID5
from
(
select visitnumber, id,
row_number() over(partition by visitnumber
order by id) seq
from yourtable
) d
group by visitnumber;
見SQL Fiddle with Demo
可以使用旋轉功能:
select visitnumber, ID1, ID2, ID3, ID4, ID5
from
(
select visitnumber, id,
'ID'+cast(row_number() over(partition by visitnumber
order by id) as varchar(10)) seq
from yourtable
) d
pivot
(
max(id)
for seq in (ID1, ID2, ID3, ID4, ID5)
) piv;
見SQL Fiddle with Demo。你說,你只會有一個最大的5分的,但如果你有一個未知的數字,那麼你可以使用動態SQL得到的結果:
DECLARE @cols AS NVARCHAR(MAX),
@query AS NVARCHAR(MAX)
select @cols = STUFF((SELECT distinct ',' + QUOTENAME('ID'+cast(row_number() over(partition by visitnumber
order by id) as varchar(10)))
from yourtable
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query = 'SELECT visitnumber,' + @cols + '
from
(
select visitnumber, id,
''ID''+cast(row_number() over(partition by visitnumber
order by id) as varchar(10)) seq
from yourtable
) x
pivot
(
max(id)
for seq in (' + @cols + ')
) p '
execute sp_executesql @query;
見SQL Fiddle with Demo。所有版本會給結果:
| VISITNUMBER | ID1 | ID2 | ID3 | ID4 | ID5 |
------------------------------------------------------------
| 39332 | 759666 | 769445 | 775795 | 783781 | 861130 |
| 40329 | 762595 | 769447 | 775796 | 783782 | (null) |
你使用MSSQL服務器? – valex
是的,我正在使用MSSQL服務器。 – Kenry87
你是否需要在單獨的列或ID列表中的ID? VisitNumber可以有多於5個ID嗎? – BWS