使用SQL Server,我如何分割一个字符串,以便我可以访问项目x?

拿一根“你好,约翰·史密斯”的绳子。我如何通过空格分割字符串并访问索引1应该返回“约翰”的项目?


当前回答

试试这个:

CREATE function [SplitWordList]
(
 @list varchar(8000)
)
returns @t table 
(
 Word varchar(50) not null,
 Position int identity(1,1) not null
)
as begin
  declare 
    @pos int,
    @lpos int,
    @item varchar(100),
    @ignore varchar(100),
    @dl int,
    @a1 int,
    @a2 int,
    @z1 int,
    @z2 int,
    @n1 int,
    @n2 int,
    @c varchar(1),
    @a smallint
  select 
    @a1 = ascii('a'),
    @a2 = ascii('A'),
    @z1 = ascii('z'),
    @z2 = ascii('Z'),
    @n1 = ascii('0'),
    @n2 = ascii('9')
  set @ignore = '''"'
  set @pos = 1
  set @dl = datalength(@list)
  set @lpos = 1
  set @item = ''
  while (@pos <= @dl) begin
    set @c = substring(@list, @pos, 1)
    if (@ignore not like '%' + @c + '%') begin
      set @a = ascii(@c)
      if ((@a >= @a1) and (@a <= @z1))  
        or ((@a >= @a2) and (@a <= @z2))
        or ((@a >= @n1) and (@a <= @n2))
      begin
        set @item = @item + @c
      end else if (@item > '') begin
        insert into @t values (@item)
        set @item = ''
      end
    end 
    set @pos = @pos + 1
  end
  if (@item > '') begin
    insert into @t values (@item)
  end
  return
end

像这样测试它:

select * from SplitWordList('Hello John Smith')

其他回答

试试这个:

CREATE function [SplitWordList]
(
 @list varchar(8000)
)
returns @t table 
(
 Word varchar(50) not null,
 Position int identity(1,1) not null
)
as begin
  declare 
    @pos int,
    @lpos int,
    @item varchar(100),
    @ignore varchar(100),
    @dl int,
    @a1 int,
    @a2 int,
    @z1 int,
    @z2 int,
    @n1 int,
    @n2 int,
    @c varchar(1),
    @a smallint
  select 
    @a1 = ascii('a'),
    @a2 = ascii('A'),
    @z1 = ascii('z'),
    @z2 = ascii('Z'),
    @n1 = ascii('0'),
    @n2 = ascii('9')
  set @ignore = '''"'
  set @pos = 1
  set @dl = datalength(@list)
  set @lpos = 1
  set @item = ''
  while (@pos <= @dl) begin
    set @c = substring(@list, @pos, 1)
    if (@ignore not like '%' + @c + '%') begin
      set @a = ascii(@c)
      if ((@a >= @a1) and (@a <= @z1))  
        or ((@a >= @a2) and (@a <= @z2))
        or ((@a >= @n1) and (@a <= @n2))
      begin
        set @item = @item + @c
      end else if (@item > '') begin
        insert into @t values (@item)
        set @item = ''
      end
    end 
    set @pos = @pos + 1
  end
  if (@item > '') begin
    insert into @t values (@item)
  end
  return
end

像这样测试它:

select * from SplitWordList('Hello John Smith')

我在网上寻找解决方案,下面的工作对我来说。 Ref。

然后像这样调用函数:

SELECT * FROM dbo.split('ram shyam hari gopal',' ')

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO

CREATE FUNCTION [dbo].[Split](@String VARCHAR(8000), @Delimiter CHAR(1))       
RETURNS @temptable TABLE (items VARCHAR(8000))       
AS       
BEGIN       
    DECLARE @idx INT       
    DECLARE @slice VARCHAR(8000)        
    SELECT @idx = 1       
    IF len(@String)<1 OR @String IS NULL  RETURN       
    WHILE @idx!= 0       
    BEGIN       
        SET @idx = charindex(@Delimiter,@String)       
        IF @idx!=0       
            SET @slice = LEFT(@String,@idx - 1)       
        ELSE       
            SET @slice = @String       
        IF(len(@slice)>0)  
            INSERT INTO @temptable(Items) VALUES(@slice)       
        SET @String = RIGHT(@String,len(@String) - @idx)       
        IF len(@String) = 0 break       
    END   
    RETURN       
END

几乎所有其他答案都是替换正在分割的字符串,这浪费了CPU周期并执行不必要的内存分配。

我在这里介绍了一种更好的进行字符串拆分的方法:http://www.digitalruby.com/split-string-sql-server/

代码如下:

SET NOCOUNT ON

-- You will want to change nvarchar(MAX) to nvarchar(50), varchar(50) or whatever matches exactly with the string column you will be searching against
DECLARE @SplitStringTable TABLE (Value nvarchar(MAX) NOT NULL)
DECLARE @StringToSplit nvarchar(MAX) = 'your|string|to|split|here'
DECLARE @SplitEndPos int
DECLARE @SplitValue nvarchar(MAX)
DECLARE @SplitDelim nvarchar(1) = '|'
DECLARE @SplitStartPos int = 1

SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)

WHILE @SplitEndPos > 0
BEGIN
    SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, (@SplitEndPos - @SplitStartPos))
    INSERT @SplitStringTable (Value) VALUES (@SplitValue)
    SET @SplitStartPos = @SplitEndPos + 1
    SET @SplitEndPos = CHARINDEX(@SplitDelim, @StringToSplit, @SplitStartPos)
END

SET @SplitValue = SUBSTRING(@StringToSplit, @SplitStartPos, 2147483647)
INSERT @SplitStringTable (Value) VALUES(@SplitValue)

SET NOCOUNT OFF

-- You can select or join with the values in @SplitStringTable at this point.

基于纯集的解决方案,使用TVF和递归CTE。您可以将此函数JOIN和APPLY到任何数据集。

create function [dbo].[SplitStringToResultSet] (@value varchar(max), @separator char(1))
returns table
as return
with r as (
    select value, cast(null as varchar(max)) [x], -1 [no] from (select rtrim(cast(@value as varchar(max))) [value]) as j
    union all
    select right(value, len(value)-case charindex(@separator, value) when 0 then len(value) else charindex(@separator, value) end) [value]
    , left(r.[value], case charindex(@separator, r.value) when 0 then len(r.value) else abs(charindex(@separator, r.[value])-1) end ) [x]
    , [no] + 1 [no]
    from r where value > '')

select ltrim(x) [value], [no] [index] from r where x is not null;
go

用法:

select *
from [dbo].[SplitStringToResultSet]('Hello John Smith', ' ')
where [index] = 1;

结果:

value   index
-------------
John    1

在我看来,你们把事情搞得太复杂了。只需要创建一个CLR UDF就可以了。

using System;
using System.Data;
using System.Data.SqlClient;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
using System.Collections.Generic;

public partial class UserDefinedFunctions {
  [SqlFunction]
  public static SqlString SearchString(string Search) {
    List<string> SearchWords = new List<string>();
    foreach (string s in Search.Split(new char[] { ' ' })) {
      if (!s.ToLower().Equals("or") && !s.ToLower().Equals("and")) {
        SearchWords.Add(s);
      }
    }

    return new SqlString(string.Join(" OR ", SearchWords.ToArray()));
  }
};