单独的逗号分隔值并存储在 sql server 的表中 [英] separate comma separated values and store in table in sql server
问题描述
我有一个存储过程,它获取逗号分隔值作为输入.我需要将它分开并需要将它作为单独的行存储在表中.
I am having a stored procedure which gets the comma separated value as an input. I need to separate it and needs to store it in a table as individual rows.
让 SP 的输入为:
Rule_ID ListType_ID Values
1 2 319,400,521,8465,2013
我需要将它存储在一个名为 DistributionRule_x_ListType
的表中,格式如下:
I need to store it in a table called DistributionRule_x_ListType
in the below format:
Rule_ID ListType_ID Value
1 2 319
1 2 400
1 2 521
1 2 8465
1 2 2013
我的 SP 如下所示:
My SP looks like below:
ALTER PROCEDURE [dbo].[spInsertDistributionRuleListType]
(@Rule_ID int,
@ListType_ID int,
@Values VARCHAR(MAX)=NULL
)
AS
BEGIN
INSERT INTO DistributionRule_x_ListType (Rule_ID,ListType_ID,Value)
VALUES (@Rule_ID,@ListType_ID,@Values)
END
推荐答案
您将需要创建一个类似于此的拆分函数:
You will need to create a split function similar to this:
create FUNCTION [dbo].[Split](@String varchar(MAX), @Delimiter char(1))
returns @temptable TABLE (items varchar(MAX))
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;
然后在您的存储过程中,您将调用该函数来拆分您的字符串:
Then in your stored procedure, you will call the function to split your string:
ALTER PROCEDURE [dbo].[spInsertDistributionRuleListType]
(
@Rule_ID int,
@ListType_ID int,
@Values VARCHAR(MAX)=NULL
)
AS
BEGIN
INSERT INTO DistributionRule_x_ListType (Rule_ID, ListType_ID, Value)
SELECT @Rule_ID, @ListType_ID, items
FROM [dbo].[Split] (@Values, ',') -- call the split function
END
当您执行存储过程时,它将拆分值并将多行插入到您的表中:
When you execute the stored procedure, it will split the values and insert the multiple rows into your table:
exec spInsertDistributionRuleListType 1, 2, '319,400,521,8465,2013';
请参阅SQL Fiddle with Demo.这将插入以下结果:
See SQL Fiddle with Demo. This will insert the following result:
| RULE_ID | LISTTYPE_ID | VALUE |
---------------------------------
| 1 | 1 | 10 |
| 1 | 2 | 319 |
| 1 | 2 | 400 |
| 1 | 2 | 521 |
| 1 | 2 | 8465 |
| 1 | 2 | 2013 |
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