我有一个数据库表,其中包含一个堆叠数据的列,其中包含两个级别,其中包含一个我要拆分的列。这是数据的例子(改变数据以保护无辜者:):
ID = varchar(100)
CarData = varchar(1000)
ID CarData
1 Nissan:blue:20000,Ford:green:10000
2 Nissan:steel:20001,Ford:blue:10001,Chevy:blue:10000,Ford:olive:10000
** Note that cardata can is not fixed, and can have many cars in it
ID Manufacture Color Cost
1 Nissan Blue 20000
1 Ford green 10000
2 Nissan steel 20001
... and on
所以说明白我需要打破第一个堆叠字段,这是一个逗号并为其创建一行,然后将第二个堆叠字段分解为列。
非常感谢任何帮助。
答案 0 :(得分:6)
-- Sample data
declare @T table(ID int, CarData varchar(100))
insert into @T values
(1, 'Nissan:blue:20000,Ford:green:10000'),
(2, 'Nissan:steel:20001,Ford:blue:10001,Chevy:blue:10000,Ford:olive:10000')
-- Recursice CTE to get one row for each car
;with cte(ID, Car, CarData) as
(
select ID,
cast(substring(CarData+',', 1, charindex(',', CarData+',')-1) as varchar(100)),
stuff(CarData, 1, charindex(',', CarData), '')+','
from @T
where len(CarData) > 0
union all
select ID,
cast(substring(CarData, 1, charindex(',', CarData)-1) as varchar(100)),
stuff(CarData, 1, charindex(',', CarData), '')
from cte
where len(CarData) > 0
)
-- Use parsename to split the car data
select ID,
parsename(replace(Car, ':', '.'), 3) as Manufacture,
parsename(replace(Car, ':', '.'), 2) as Color,
parsename(replace(Car, ':', '.'), 1) as Cost
from cte
order by ID
结果:
ID Manufacture Color Cost
-- ----------- ------ -----
1 Nissan blue 20000
1 Ford green 10000
2 Nissan steel 20001
2 Ford blue 10001
2 Chevy blue 10000
2 Ford olive 10000
修改1
如果颜色,费用或制造商名称包含parsename
,您将无法使用.
。如果是这种情况,你应该尝试这样做。
-- Sample data
declare @T table(ID int, CarData varchar(100))
insert into @T values
(1, 'Nissan:blue:20000,Ford:green:10000'),
(2, 'Nissan:steel:20001,Ford:blue:10001,Chevy:blue:10000,Ford:olive:10000')
-- Recursice CTE to get one row for each car
;with cte(ID, Car, CarData) as
(
select ID,
cast(substring(CarData+',', 1, charindex(',', CarData+',')-1) as varchar(100)),
stuff(CarData, 1, charindex(',', CarData), '')+','
from @T
where len(CarData) > 0
union all
select ID,
cast(substring(CarData, 1, charindex(',', CarData)-1) as varchar(100)),
stuff(CarData, 1, charindex(',', CarData), '')
from cte
where len(CarData) > 0
)
-- Split the car data with substring
select ID,
substring(Car, 1, P1.Pos-1) as Manufacture,
substring(Car, P1.Pos+1, P2.Pos-P1.Pos-1) as Color,
substring(Car, P2.Pos+1, len(Car)-P2.Pos) as Cost
from cte
cross apply (select charindex(':', Car)) as P1(Pos)
cross apply (select charindex(':', Car, P1.Pos+1)) as P2(Pos)
order by ID
答案 1 :(得分:1)
使用this string splitting function生成结果表。
我首先使用dbo.split()
作为分隔符调用,
。然后你会有一个项目列表,如:
Nissan:blue:20000
Ford:green:10000
Nissan:steel:20001
Ford:blue:10001
Chevy:blue:10000
Ford:olive:10000
您可以使用dbo.split()
作为分隔符再次致电:
。每次调用都会产生三条记录(假设您的设计至少是“正常”)。
正如@JNK在评论中提到的那样,希望这不是你想要定期运行的东西。
修改强>
一些示例代码可帮助您入门:
SELECT *
INTO #YuckyCar
FROM (
SELECT 1 ID, 'Nissan:blue:20000,Ford:green:10000' CarData
UNION
SELECT 2, 'Nissan:steel:20001,Ford:blue:10001,Chevy:blue:10000,Ford:olive:10000'
) T;
-- Shows logical step #1
SELECT ID, X.items MoreCarData
FROM #YuckyCar CROSS APPLY dbo.Split(CarData, ',') X;
-- Shows logical step #2
SELECT Q.ID, Y.items
FROM (
SELECT ID, X.items MoreCarData
FROM #YuckyCar CROSS APPLY dbo.Split(CarData, ',') X) Q CROSS APPLY dbo.Split(Q.MoreCarData, ':') Y
DROP TABLE #YuckyCar;
最后一部分的问题是你不能保证第1行=制造商,第2行=颜色,第3行=成本。
答案 2 :(得分:1)
这可以解决您的问题:
[编辑] 您的ID是varchar(100)并且您没有指定它是否是主键,所以我做了一些更改... ID不一定是主键在这种情况下。
declare @T table(ID varchar(100), CarData varchar(1000))
declare @OUT table(pk INT IDENTITY(1,1), ID varchar(100), Manufacture varchar(100), Color VARCHAR(100), Cost INT)
insert into @T (ID, CarData) values
('1', 'Nissan:blue:20000,Ford:green:10000'),
('2', 'Nissan:steel:20001,Ford:blue:10001,Chevy:blue:10000,Ford:olive:10000')
DECLARE @x XML, @i INT, @ID VARCHAR(100), @maxi INT;
;WITH list AS (SELECT pk = ROW_NUMBER() OVER(ORDER BY ID), * FROM @T)
SELECT @i=1, @maxi=MAX(pk) FROM list;
WHILE @i <= @maxi
BEGIN
;WITH list AS (SELECT pk = ROW_NUMBER() OVER(ORDER BY ID), * FROM @T)
SELECT
@x = CAST( '<root><car><prop>' +
REPLACE(
REPLACE(
CarData
,':'
,'</prop><prop>'
)
,','
,'</prop></car><car><prop>'
) +
'</prop></car></root>'
AS XML)
, @ID = ID
FROM list
WHERE pk = @i
INSERT INTO @OUT
SELECT
ID = @ID
,Manufacture = x.value('./prop[1]','VARCHAR(100)')
,Color = x.value('./prop[2]','VARCHAR(100)')
,Cost = x.value('./prop[3]','INT')
FROM @x.nodes('/root/car') AS T(x)
SET @i = @i + 1;
END
SELECT * FROM @OUT
/* -- OUTPUT
ID Manufacture Color Cost
--------------------------------
1 Nissan blue 20000
1 Ford green 10000
2 Nissan steel 20001
2 Ford blue 10001
2 Chevy blue 10000
2 Ford olive 10000
*/