Question

我在表中有一个列（full_location_id），其中包含一个由' - '分隔的字符串，我需要在视图中将其拆分为4列（Test_SplitColumn）。并非full_location_id中的每条记录都包含相同长度的ID。有些可能有像A1-BF-35-B1这样的ids，而有些可能只有AR-B3。我不确定这样做的最佳方式。我能够检索第一列，但到目前为止，不是其他列。

CREATE VIEW [dbo].[Test_SplitColumn]
AS
select p.name, location.aisle_id, location.full_location_id, SUBSTRING(location.full_location_id, 0,charindex('-', location.full_location_id )) as Aisle,
SUBSTRING(location.full_location_id, charindex('-', location.full_location_id ) + 1, charindex('-', location.full_location_id, LEN(SUBSTRING(location.full_location_id, 0,charindex('-', location.full_location_id ))) )) as ShelvingUnit
from location 
inner join product p on p.id = location.product_id
GO

任何帮助或指导都将不胜感激。

Answer 1

这是一种快速，简单的方法：

DECLARE @T TABLE(full_location_id varchar(100));

INSERT INTO @T 
VALUES  ('A1-BF-35-B1'),
        ('AR-B3');

WITH CTE AS
(
    SELECT  full_location_id,
            LEN(full_location_id)-LEN(REPLACE(full_location_id,'-','')) N
    FROM @T
)
SELECT  full_location_id,
        PARSENAME(REPLACE(full_location_id,'-','.'),N+1),
        PARSENAME(REPLACE(full_location_id,'-','.'),N),
        PARSENAME(REPLACE(full_location_id,'-','.'),N-1),
        PARSENAME(REPLACE(full_location_id,'-','.'),N-2)
FROM CTE

结果：

╔══════════════════╦══════╦══════╦══════╦══════╗
║ full_location_id ║ Col1 ║ Col2 ║ Col3 ║ Col4 ║
╠══════════════════╬══════╬══════╬══════╬══════╣
║ A1-BF-35-B1      ║ A1   ║ BF   ║ 35   ║ B1   ║
║ AR-B3            ║ AR   ║ B3   ║ NULL ║ NULL ║
╚══════════════════╩══════╩══════╩══════╩══════╝

和here is an sqlfiddle一起演示。

Answer 2

这是您模型的失败。不是将位置存储为分隔字符串，而是建立1-n表来存储位置可能是个好主意。事实上，对你的问题的正确“回答”可能是“重新设计数据库的这一部分！”

然而，要做你想做的事，你可以做这样的事情：

USE tempdb
GO

/* udfSplit (A Fast String Splitter) **************************************************************
 *
 * Uses a number table to *very* quickly split the text (@text). Splits on the delimiter (@d)
 * Returns Table of ( [RowID], [SplitText] ). Inlineable for CROSS APPLY etc.
 *
 * Charlie
 *
 *************************************************************************************************/
CREATE FUNCTION [dbo].[udfSplit] (@text NVARCHAR(4000), @d NVARCHAR(50))
RETURNS TABLE AS RETURN (
WITH numbers(n) AS (
SELECT ROW_NUMBER() OVER (ORDER BY a.[n])
  FROM
             ( VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9) ) AS  a ([n])
  CROSS JOIN ( VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9) ) AS  b ([n])
  CROSS JOIN ( VALUES (0), (1), (2), (3), (4), (5), (6), (7), (8), (9) ) AS  c ([n])
  CROSS JOIN ( VALUES (0), (1), (2), (3), (4)) AS  d ([n])
  )
SELECT
      [RowID] = ROW_NUMBER() OVER ( ORDER BY [n] ASC )
    , [SplitText] = SUBSTRING(
        @d + @text + @d
        , [n] + LEN(@d)
        , CHARINDEX(@d, @d + @text + @d, [n] + LEN(@d)) - [n] - LEN(@d)
        )
FROM numbers AS n
WHERE [n] <= LEN(@d + @text + @d) - LEN(@d)
  AND SUBSTRING(@d + @text + @d, [n], LEN(@d)) = @d
)
GO

IF OBJECT_ID('tempdb..#sample') IS NOT NULL DROP TABLE #sample
GO

CREATE TABLE #sample (
      name VARCHAR(255)
    , locations VARCHAR(MAX)
    )

INSERT #sample (name, locations)    
VALUES ('a', 'ab-cd')
     , ('b', 'ab-cd-ef')
     , ('c', 'gh')
     , ('d', NULL)

; WITH SPLIT AS (
    SELECT [name], l.*
    FROM #sample AS s
    OUTER APPLY dbo.[udfSplit](s.locations,'-') AS l
    )
SELECT
      s.name
    , MAX(CASE WHEN s.rowId = 1 THEN s.SplitText ELSE '' END) AS a
    , MAX(CASE WHEN s.rowId = 2 THEN s.SplitText ELSE '' END) AS b
    , MAX(CASE WHEN s.rowId = 3 THEN s.SplitText ELSE '' END) AS c
    , MAX(CASE WHEN s.rowId = 4 THEN s.SplitText ELSE '' END) AS d
FROM
    SPLIT AS s
GROUP BY
    s.name

这可能看起来非常复杂。函数udfSplit是一个非常快速的字符串拆分器 - 它将分隔的字符串转换为表格，返回位置（1-4）和拆分字符串。除非你真的想进入它，否则不要担心它是如何工作的。如果您做想要了解在数据库中拆分字符串（以及通常为什么这是一个糟糕的计划） - 请在此处阅读：

http://www.sqlservercentral.com/articles/Tally+Table/72993/

其余代码组成一个示例表，然后对其进行选择以获得所需的输出：

(4 row(s) affected)
name                 a     b     c     d
-------------------- ----- ----- ----- -----
a                    ab    cd          
b                    ab    cd    ef    
c                    gh                
d

MAX（CASE ....）表达式是sql server 2000中的一个转发技巧。我从来没有得到PIVOT运营商的支持。

SQL小提琴： http://sqlfiddle.com/#!3/80f74/1

Answer 3

这是一种更通用的方法。这假设您知道您将拥有的最大列数。

CREATE TABLE #tmp (
    full_location_id varchar(255) );

INSERT INTO dbo.#tmp VALUES ( 'A1-BF-35-B1' );
INSERT INTO dbo.#tmp VALUES ( 'AR-B3' );
INSERT INTO dbo.#tmp VALUES ( 'A1-BF-35' );
INSERT INTO dbo.#tmp VALUES ( 'A1' );


with tmp( full_location_id, c, position, single ) as (
select #tmp.full_location_id
     , STUFF( #tmp.full_location_id, 1, CHARINDEX('-', #tmp.full_location_id + ' -'), '') AS c
     , 1 AS position
     , convert(nvarchar(max),left(#tmp.full_location_id, CHARINDEX('-', #tmp.full_location_id + ' -') -1)) AS single
  from #tmp
 union all
select full_location_id
     , STUFF(c, 1, CHARINDEX('-', c + ' -'), '')
     , position + 1
     , convert(nvarchar(max),left(c, CHARINDEX('-', c + ' -') -1))
  from tmp
 where c > ''
)

SELECT pvt.full_location_id
     , [1]
     , [2]
     , [3]
     , [4]
 FROM 
( SELECT full_location_id
       , single
       , position
    FROM tmp ) AS src
PIVOT
(
    MAX( single )
    FOR position IN ( [1], [2], [3], [4] )

) AS pvt;

如何将字符串拆分为视图的列

3 个答案: