拆分值对并使用 UDF 创建表

问题描述 投票:0回答:2

我一直在尝试编写一个表值函数,它将值对作为参数并返回一个包含两列的表。

下面是我正在尝试做的函数签名。

FUNCTION [dbo].[ValuePairParser]( @DelimitedValuePairs VARCHAR(MAX),
                                  @Delimiter CHAR(1), 
                                  @ValuePairDelimiter CHAR(1) ) 
  RETURNS @ValuePairTable
  TABLE ( Id INT, Code INT ) 

我想调用如下方法

@ValuePairs VARCHAR(MAX) = '1:1, 1:2, 1:4, 2:3, 1000:230, 130:120,'

ValuePairParser (@ValuePairs, ',', ':')

你能看到任何好的方法来分割上面的 ValuePairs 字符串并创建一个包含两列的表吗?

sql sql-server sql-server-2005 t-sql user-defined-functions
2个回答
6
投票

十年后回顾,今天肯定有更好的方法来做到这一点。

SQL Server 2022、Azure SQL 数据库、托管实例

示例

CREATE OR ALTER FUNCTION dbo.SplitWithPairs
(
    @List           nvarchar(max),
    @MajorDelimiter varchar(3) = ',',
    @MinorDelimiter varchar(3) = ':'
)
RETURNS TABLE WITH SCHEMABINDING
AS
  RETURN 
  (
    SELECT LeftItem = [1], RightItem = [2], Position
    FROM
    (
      SELECT Position = o.ordinal, 
             value    = TRIM(i.value),
             i.ordinal
        FROM        STRING_SPLIT(@List,   @MajorDelimiter, 1) AS o
        CROSS APPLY STRING_SPLIT(o.value, @MinorDelimiter, 1) AS i
      WHERE o.value > ''
    ) AS s PIVOT (MAX(value) FOR ordinal IN ([1],[2])) AS p
  );

SQL Server 2016 - 2019

示例

在 SQL Server 2016 到 2019 中,进行了一些更改:

  • STRING_SPLIT
    在 SQL Server 2022 之前不支持序数,因此我们可以使用
    OPENJSON
    代替
  • 我们需要使用
    LTRIM/RTRIM
    ,因为SQL Server 2017中添加了
    TRIM
  • key
    是从 0 开始的,所以我们增加 1 以获得等效的从 1 开始的位置

关于此函数的一个注意事项是,如果不更改函数,则不能使用

,
作为次分隔符,因为
,
OPENJSON
的分隔方式。您可能希望保留其他字符作为两个分隔符。

CREATE OR ALTER FUNCTION dbo.SplitWithPairs
(
    @List           nvarchar(max),
    @MajorDelimiter varchar(3) = ',',
    @MinorDelimiter varchar(3) = ':' -- can't be ,
)
RETURNS TABLE WITH SCHEMABINDING
AS
  RETURN 
  (
    SELECT LeftItem = [1], RightItem = [2], Position
    FROM
    (  
      SELECT Position = o.[key] + 1, 
             value    = LTRIM(RTRIM(i.value)),
             ordinal  = i.[key] + 1
      FROM OPENJSON
      (
        CONCAT('["', REPLACE(STRING_ESCAPE(@List, 'JSON'), 
          @MajorDelimiter, '","'), '"]')
      ) AS o
      CROSS APPLY OPENJSON
      (
        REPLACE(CONCAT('[', QUOTENAME(o.value, '"'), ']'), 
          @MinorDelimiter, '","')
      ) AS i
      WHERE o.value > ''
    ) AS s PIVOT (MAX(value) FOR ordinal IN ([1],[2])) AS p
  );

不再支持的版本

最后,原始答案适用于 SQL Server 2016 之前的版本,我将保持原样,但有一个免责声明,它可以改进(不乏阅读有关那里的内容here,并且一般来说,多语句表值函数 - 特别是循环 - 是一个很大的危险信号):

CREATE FUNCTION [dbo].[SplitWithPairs]
(
    @List NVARCHAR(MAX),
    @MajorDelimiter VARCHAR(3) = ',',
    @MinorDelimiter VARCHAR(3) = ':'
)
RETURNS @Items TABLE
(
    Position  INT IDENTITY(1,1) NOT NULL,
    LeftItem  INT NOT NULL,
    RightItem INT NOT NULL
)
AS
BEGIN
    DECLARE
        @Item      NVARCHAR(MAX),
        @LeftItem  NVARCHAR(MAX),
        @RightItem NVARCHAR(MAX),
        @Pos       INT;

    SELECT
        @List = @List + ' ',
        @MajorDelimiter = LTRIM(RTRIM(@MajorDelimiter)),
        @MinorDelimiter = LTRIM(RTRIM(@MinorDelimiter));

    WHILE LEN(@List) > 0
    BEGIN
        SET @Pos = CHARINDEX(@MajorDelimiter, @List);

        IF @Pos = 0 
            SET @Pos = LEN(@List) + LEN(@MajorDelimiter);

        SELECT
            @Item = LTRIM(RTRIM(LEFT(@List, @Pos - 1))),
            @LeftItem = LTRIM(RTRIM(LEFT(@Item,
            CHARINDEX(@MinorDelimiter, @Item) - 1))),
            @RightItem = LTRIM(RTRIM(SUBSTRING(@Item,
            CHARINDEX(@MinorDelimiter, @Item)
            + LEN(@MinorDelimiter), LEN(@Item))));

        INSERT @Items(LeftItem, RightItem)
            SELECT @LeftItem, @RightItem;

        SET @List = SUBSTRING(@List,
            @Pos + LEN(@MajorDelimiter), DATALENGTH(@List));
    END
    RETURN;
END
GO

DECLARE @ValuePairs VARCHAR(MAX) = '1:1, 1:2, 1:4, 2:3,1000:230, 130:120,';

SELECT LeftItem, RightItem
  FROM dbo.SplitWithPairs(@ValuePairs, ',', ':')
  ORDER BY Position;
GO

1
投票
create function ValuePairParser(@DelimitedValuePairs varchar(MAX),
                                @Delimiter char(1), 
                                @ValuePairDelimiter char(1)) 
returns @ValuePairTable table(Id int, Code int) as
begin 
  with Split(ValuePair, Rest) as
  (
    select left(@DelimitedValuePairs, charindex(@Delimiter, @DelimitedValuePairs)-1),
           stuff(@DelimitedValuePairs, 1, charindex(@Delimiter, @DelimitedValuePairs), '')
    where charindex(@Delimiter, @DelimitedValuePairs) > 0
    union all
    select left(Rest, charindex(@Delimiter, Rest)-1),
           stuff(Rest, 1, charindex(@Delimiter, Rest), '')
    from Split         
    where charindex(@Delimiter, Rest) > 0
  )               
  insert into @ValuePairTable
  select left(ValuePair, charindex(@ValuePairDelimiter, ValuePair)-1),
         stuff(ValuePair, 1, charindex(@ValuePairDelimiter, ValuePair), '')
  from Split                 
  option (maxrecursion 0)

  return
end
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