There are two strings a
and b
有两个字符串a和b
The a
string contains comma. I would like to split the a string by comma, then go through every element .
一个字符串包含逗号。我想用逗号分割一个字符串,然后遍历每个元素。
IF the b
string contains any element which split by comma will return 0
如果b字符串包含任何以逗号分割的元素将返回0
(e.g: a = "4,6,8"
; b = "R3799514"
because the b
string contains 4 so return 0)
(例如:a =“4,6,8”; b =“R3799514”因为b字符串包含4所以返回0)
How to achieve this with a stored procedure? Thanks in advance!
如何使用存储过程实现此目的?提前致谢!
I have seen a split function:
我见过一个split函数:
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
select top 10 * from dbo.split('Chennai,Bangalore,Mumbai',',')
1 个解决方案
#1
6
Following will work -
以下将工作 -
DECLARE @A VARCHAR (100)= '4,5,6'
DECLARE @B VARCHAR (100)= 'RXXXXXX'
DECLARE @RETURN_VALUE BIT = 1 --DEFAULT 1
SELECT items
INTO #STRINGS
FROM dbo.split(@A,',')
IF EXISTS(SELECT 1 FROM #STRINGS S WHERE CHARINDEX(items, @B) > 0)
SET @RETURN_VALUE = 0
PRINT @RETURN_VALUE
DROP TABLE #STRINGS
You can also use CONTAINS instead of CHARINDEX -
您也可以使用CONTAINS而不是CHARINDEX -
IF EXISTS(SELECT 1 FROM #STRINGS S WHERE CONTAINS(items, @B))
SET @RETURN_VALUE = 0
#1
6
Following will work -
以下将工作 -
DECLARE @A VARCHAR (100)= '4,5,6'
DECLARE @B VARCHAR (100)= 'RXXXXXX'
DECLARE @RETURN_VALUE BIT = 1 --DEFAULT 1
SELECT items
INTO #STRINGS
FROM dbo.split(@A,',')
IF EXISTS(SELECT 1 FROM #STRINGS S WHERE CHARINDEX(items, @B) > 0)
SET @RETURN_VALUE = 0
PRINT @RETURN_VALUE
DROP TABLE #STRINGS
You can also use CONTAINS instead of CHARINDEX -
您也可以使用CONTAINS而不是CHARINDEX -
IF EXISTS(SELECT 1 FROM #STRINGS S WHERE CONTAINS(items, @B))
SET @RETURN_VALUE = 0