对于这种类型的数据转换,您将需要先执行数据UNPIVOT
,然后再执行PIVOT
数据。如果您知道要转换的值,则可以使用静态数据透视对查询进行硬编码,否则可以使用动态 sql。
创建表:
CREATE TABLE Table_User
([UserId] int, [UserName] varchar(5))
;
INSERT INTO Table_User
([UserId], [UserName])
VALUES
(1, 'John'),
(2, 'Joe'),
(3, 'James')
;
CREATE TABLE Table_UserPermissions
([UserId] int, [PermissionId] int, [Rights] int)
;
INSERT INTO Table_UserPermissions
([UserId], [PermissionId], [Rights])
VALUES
(1, 10, 1),
(1, 11, 2),
(1, 12, 3),
(2, 11, 2),
(2, 12, 3),
(3, 10, 2)
;
静态枢轴:
select *
from
(
select userid,
username,
value,
col + '_'+ cast(rn as varchar(10)) col
from
(
select u.userid,
u.username,
p.permissionid,
p.rights,
row_number() over(partition by u.userid
order by p.permissionid, p.rights) rn
from table_user u
left join Table_UserPermissions p
on u.userid = p.userid
) src
unpivot
(
value
for col in (permissionid, rights)
) unpiv
) src
pivot
(
max(value)
for col in (permissionid_1, rights_1,
permissionid_2, rights_2,
permissionid_3, rights_3)
) piv
order by userid
请参阅带有演示的 SQL Fiddle
动态枢轴:
如果您有未知数量的permissionid
s 和rights
,那么您可以使用动态 sql:
DECLARE
@query AS NVARCHAR(MAX),
@colsPivot as NVARCHAR(MAX)
select @colsPivot = STUFF((SELECT ','
+ quotename(c.name +'_'+ cast(t.rn as varchar(10)))
from
(
select row_number() over(partition by u.userid
order by p.permissionid, p.rights) rn
from table_user u
left join Table_UserPermissions p
on u.userid = p.userid
) t
cross apply sys.columns as C
where C.object_id = object_id('Table_UserPermissions') and
C.name not in ('UserId')
group by c.name, t.rn
order by t.rn
FOR XML PATH(''), TYPE
).value('.', 'NVARCHAR(MAX)')
,1,1,'')
set @query
= 'select *
from
(
select userid,
username,
value,
col + ''_''+ cast(rn as varchar(10)) col
from
(
select u.userid,
u.username,
p.permissionid,
p.rights,
row_number() over(partition by u.userid
order by p.permissionid, p.rights) rn
from table_user u
left join Table_UserPermissions p
on u.userid = p.userid
) src
unpivot
(
value
for col in (permissionid, rights)
) unpiv
) x1
pivot
(
max(value)
for col in ('+ @colspivot +')
) p
order by userid'
exec(@query)
请参阅带有演示的 SQL Fiddle
两者的结果是:
| USERID | USERNAME | PERMISSIONID_1 | RIGHTS_1 | PERMISSIONID_2 | RIGHTS_2 | PERMISSIONID_3 | RIGHTS_3 |
---------------------------------------------------------------------------------------------------------
| 1 | John | 10 | 1 | 11 | 2 | 12 | 3 |
| 2 | Joe | 11 | 2 | 12 | 3 | (null) | (null) |
| 3 | James | 10 | 2 | (null) | (null) | (null) | (null) |