How To Split column In SQL

SET NOCOUNT ON
 
DECLARE @Table table (Col001 varchar(100) NOT NULL)
 
INSERT INTO @Table VALUES('AABY TAMARA J')
INSERT INTO @Table VALUES('AADNESEN MARJORIE')
INSERT INTO @Table VALUES('Paola Godoy Calcagno Jensen')
INSERT INTO @Table VALUES('AARANT MAUREEN R')
INSERT INTO @Table VALUES('AARON MICHELLEE')
INSERT INTO @Table VALUES('AARON PATRICIA J')
INSERT INTO @Table VALUES('ABBA DAVID W')
INSERT INTO @Table VALUES('ABBATE ANTHONY P')
 
SELECT LEFT(Col001,CHARINDEX(' ',Col001)) AS FName
,CASE
WHEN LEN(Col001)-(CHARINDEX(' ',Col001)+CHARINDEX(' ',REVERSE(Col001))) > 0 THEN
SUBSTRING(Col001,CHARINDEX(' ',Col001)+1,LEN(Col001)-(CHARINDEX(' ',Col001)+CHARINDEX(' ',REVERSE(Col001))))
ELSE
''
END AS MName
,RIGHT(Col001,CHARINDEX(' ',REVERSE(Col001))-1) AS LName
FROM @Table
 
 
 
 
CREATE TABLE Inventory (ItemDescription VARCHAR(99))
INSERT Inventory VALUES ('245 HELIUM, COMPRESSED 2.2 UN1046' )
INSERT Inventory VALUES ('24adada5 HELIsadasdadUM, sdsdsd 6.6 UN99' )
INSERT Inventory VALUES ('24adada5 HELIsadasdadUM, sdsdsd 446.6777 UN9988888' )
INSERT Inventory VALUES ('24adada5 HEdUM, sdsdsd 446.0 UN9988' )
 
 
SELECT RIGHT(ItemDescription,PATINDEX('% %',
REVERSE(ItemDescription))-1) AS COL1,
LTRIM(REVERSE(LEFT(REVERSE(LEFT(ItemDescription,(LEN(ItemDescription)-PATINDEX('% %', REVERSE(ItemDescription))))),
PATINDEX('% %',REVERSE(LEFT(ItemDescription,(LEN(ItemDescription)-PATINDEX('% %', REVERSE(ItemDescription)))))))))
AS COL2
FROM Inventory
 
Result:
 
 
 
CREATE TABLE Names
    (
        FirstName VARCHAR(20),
        LastName VARCHAR(20) NULL
    )

INSERT Names SELECT 'Anthony Smith', NULL
INSERT Names SELECT 'Peter Ward', NULL
INSERT Names SELECT 'John Brown', NULL
INSERT Names SELECT 'Prince', NULL
INSERT Names SELECT 'Mary Jane Smith', NULL

UPDATE    Names
SET    FirstName =
        CASE
            WHEN CHARINDEX(' ', FirstName) > 0 THEN SUBSTRING(FirstName, 1,
CHARINDEX(' ', FirstName) - 1)
            ELSE FirstName
        END,
    LastName =
        CASE WHEN CHARINDEX(' ', FirstName) > 0 THEN SUBSTRING(FirstName,
CHARINDEX(' ', FirstName) + 1, LEN(FirstName) - CHARINDEX(FirstName, ' '))
        END

SELECT * FROM Names

Returns:

FirstName            LastName
-------------------- --------------------
Anthony              Smith
Peter                Ward
John                 Brown
Prince               NULL
Mary                 Jane Smith
 
 
 
 
create table #t ( [Name] varchar(40))
 
insert into #t ([Name]) values ('Smith,John E')
insert into #t ([Name]) values ('Smith,Bill')
insert into #t ([Name])  values ('Smith,Adam F')
insert into #t ([Name])  values ('St,Smith deWaal III')
 
---go
 
 
select LastName, FirstName, MiddleName
from (
  select
    Name,
    substring(Name,1,Comma-1) LastName,
    substring(Name,Comma+1,Spce-Comma-1) FirstName,
    nullif(substring(Name,Spce+1,40),'') MiddleName
  from (
    select
      Name,
      charindex(',',Name) Comma,
      charindex(' ',Name+space(1),charindex(',',Name)) Spce
    from #t
  ) D
) SplitNames
 
drop table #t
 
Result:
 
 
 
 
 
CREATE FUNCTION [dbo].[udf_Split2]
    (
      @List VARCHAR(8000),
      @Delimitter VARCHAR(5),
      @SplitOn VARCHAR(5)
    )
RETURNS @RtnValue TABLE
    (
      ID INT IDENTITY(1, 1),
      Value1 VARCHAR(100),
      Value2 VARCHAR(100)
    )
AS BEGIN
    DECLARE @SubString AS VARCHAR(50)
    DECLARE @Index1 AS INT
    DECLARE @Index2 AS INT
    WHILE( CHARINDEX(@SplitOn, @List) > 0 )
        BEGIN
            SELECT  @Index1 = CHARINDEX(@Delimitter, @List)
            IF @Index1 = 0
                SET @Index1 = 200
            SELECT  @SubString = LTRIM(RTRIM(SUBSTRING(@List, 1, @Index1 - 1)))
            SELECT  @Index2 = CHARINDEX(@SplitOn, @SubString)
 
            INSERT  INTO @RtnValue ( Value1, Value2 )
                    SELECT  Value1 = LTRIM(RTRIM(SUBSTRING(@SubString, 1,
                                                           @Index2 - 1))),
                            Value2 = LTRIM(RTRIM(SUBSTRING(@SubString,
                                                           @Index2 + 1,
                                                           LEN(@SubString)
                                                           - @Index2)))
 
            SET @List = SUBSTRING(@List, @Index1 + LEN(@SplitOn), LEN(@List))
        END
    RETURN
   END
 
 
TEST:
SELECT * FROM dbo.udf_Split2('This is,aTest|of the, Split2 function','|',',')
 
Result:
This is     aTest
of the      Split2 function
 
 
函数

CREATE FUNCTION [dbo].[Fun_Split] 
(
 @field varchar(1000),
 @ContainValue varchar(8000)
)
RETURNS varchar(4000)
AS
BEGIN
 declare @strRet varchar(10)
 declare @strTemp varchar(1000)
 declare @StrSeprate varchar(10)
 DECLARE @i int
 set @StrSeprate=','
 set @strRet='1'
 SET @ContainValue=rtrim(ltrim(@ContainValue))
 SET @i=charindex(@StrSeprate,@ContainValue)

 WHILE (@i>=1)
 BEGIN
  set @strTemp=(left(@ContainValue,@i-1))
  if(charindex(','+@strTemp+ ',',',' + @field + ',')=0)
  begin
    set @strRet='0'
  end
  SET @ContainValue=substring(@ContainValue,@i+1,len(@ContainValue)-@i)
  if(@strRet='0')
    begin
        SET @i=0
    end
  else
    begin
        SET @i=charindex(@StrSeprate,@ContainValue)
    end
 END

  if(@strRet='1')
  begin
   if(charindex(','+@ContainValue+',',','+@field+',')=0)
      begin
        set @strRet='0'
      end
  end
  
  return @strRet
 end

 以上函数有两个输入参数，@field 是需要判断的参数，@ContainValue是否被包含的值。

当 @field 中包含@ContainValue 的所有值时，返回1，否者返回0.

以下是实际中的使用：

tablename: test

id	name	value
1	A	1,2,3,4
2	B	2,3,4
3	C	3,4,5

 

SQL->select * from test where dbo.Fun_Split(value,'2,4')=1

查询结果：

id	name	value
1	A	1,2,3,4
2	B	2,3,4

 

以上是我最近工作中编写的一个函数，是为了解决一个特殊的查询的问题。其实，产生这样的问题是源于自己在设计表结构时不合理的缘故。

关于这样的问题，还有一个更简单的方法（这里不谈数据库结构的设计），就是在Visual studio IDE 中创建一个SqlServerProject ，编写一个如下的函数：

 

代码

[Microsoft.SqlServer.Server.SqlFunction]
    public static string StrContain(string field,string value)
    {
        bool isContain = true;
        if (!string.IsNullOrEmpty(value))
        {
            string[] values = value.Split(',');
            foreach (var item in values)
            {
                if ((',' + field + ",").IndexOf(',' + item + ',') < 0)
                {
                    isContain = false;
                    break;
                }
            }
        }
       return isContain?"1":"0";
    }

 

 

启用sqlserver2005 的 Clr(我用的sqlserver2005 和 sqlserver2008).

 

代码

--启动clr
exec sp_configure 'show advanced options', '1'
go
reconfigure
go
exec sp_configure 'clr enabled', '1'
go
reconfigure
go
exec sp_configure 'show advanced options', '1'
go

--关闭clr
exec sp_configure 'clr enabled', '0'
go
reconfigure
go

 

select * from test where dbo.strContain(value,@input)=1;

查询的结果与上边的sql function一样的。

由于我考虑到客户服务器权限的问题，所以还是使用了 sql function的方式。 

 
 
 
CREATE FUNCTION dbo.Split
(
    @RowData nvarchar(2000),
    @SplitOn nvarchar(5)
) 
RETURNS @RtnValue table
(
    Id int identity(1,1),
    Data nvarchar(100)
)
AS 
BEGIN
    Declare @Cnt int
    Set @Cnt = 1
 
    While (Charindex(@SplitOn,@RowData)>0)
    Begin
       Insert Into @RtnValue (data)
       Select
           Data = ltrim(rtrim(Substring(@RowData,1,Charindex(@SplitOn,@RowData)-1)))
 
       Set @RowData = Substring(@RowData,Charindex(@SplitOn,@RowData)+1,len(@RowData))
       Set @Cnt = @Cnt + 1
    End
   
    Insert Into @RtnValue (data)
    Select Data = ltrim(rtrim(@RowData))
 
    Return
END
 
Test:
SELECT  *
FROM    dbo.Split('This is,aTest,of the, Split2 function',  ',')
Result:
1     This is
2     aTest
3     of the
4     Split2 function