在SQL Server中，我想评估类似1.2.30.4.50的字符串，并使用0102300450输出。基本上在周期之前评估每个数字，如果它有一个数字并输出两位数，则用零填充它。如果数字已有两位数，则应保持不变。
2.5.99.10.1之类的字符串应评估为0205991001。请注意，99和10之前没有零。
我知道以下查询将在第一个句点出现之前返回第一个数字。但我无法进一步采取这一点。任何帮助表示赞赏！
如果col包含一个字符串1.2.30.4.50：
SELECT left(col, charindex('.', col) - 1) FROM table
将返回以下内容：
column1
----------
1
答案 0 :(得分：2)
您可以如下
CREATE FUNCTION PrepareString
(
@val varchar(max),
@delimiter varchar(10)
)
RETURNS VARCHAR(MAX)
AS
BEGIN
DECLARE @xml as xml
SET @xml = cast(('<X>'+replace(@val,@delimiter ,'</X><X>')+'</X>') as xml)
RETURN
(
SELECT
(
SELECT
CASE WHEN LEN(N.value('.', 'varchar(MAX)')) = 1 THEN '0' + N.value('.', 'varchar(MAX)') ELSE N.value('.', 'varchar(MAX)') END
FROM @xml.nodes('X') as T(N)
FOR XML PATH ('')
) A
)
END
GO
示例：
SELECT dbo.PrepareString('2.5.99.10.1', '.') -- 0205991001
SELECT dbo.PrepareString('1.2.30.4.50', '.') -- 0102300450
SELECT dbo.PrepareString('1.2.30.4.50.123.3.443.2', '.') -- 01023004501230344302
答案 1 :(得分：1)
我的第一个想法是使用ParseName（），但观察次数太多
Declare @String varchar(50) = '2.5.99.10.1'
Select right('0'+Pos1,2)+right('0'+Pos2,2)+right('0'+Pos3,2)+right('0'+Pos4,2)+right('0'+Pos5,2)
From [dbo].[udf-Str-Parse-Row](@String,'.')
返回
0205991001
行解析UDF
CREATE FUNCTION [dbo].[udf-Str-Parse-Row] (@String varchar(max),@Delimiter varchar(10))
Returns Table
As
Return (
Select Pos1 = xDim.value('/x[1]','varchar(max)')
,Pos2 = xDim.value('/x[2]','varchar(max)')
,Pos3 = xDim.value('/x[3]','varchar(max)')
,Pos4 = xDim.value('/x[4]','varchar(max)')
,Pos5 = xDim.value('/x[5]','varchar(max)')
,Pos6 = xDim.value('/x[6]','varchar(max)')
,Pos7 = xDim.value('/x[7]','varchar(max)')
,Pos8 = xDim.value('/x[8]','varchar(max)')
,Pos9 = xDim.value('/x[9]','varchar(max)')
From (Select Cast('<x>' + Replace(@String,@Delimiter,'</x><x>')+'</x>' as XML) as xDim) A
)
--Select * from [dbo].[udf-Str-Parse-Row]('Dog,Cat,House,Car',',')
--Select * from [dbo].[udf-Str-Parse-Row]('John Cappelletti',' ')
答案 2 :(得分：1)
假设您不在SQL Server 2016上且无法使用string_split函数，则此类问题非常适合Jeff Moden's Tally Table based string split table valued function。
创建功能：
CREATE FUNCTION [dbo].[DelimitedSplit8K]
--===== Define I/O parameters
(@pString VARCHAR(8000), @pDelimiter CHAR(1))
--WARNING!!! DO NOT USE MAX DATA-TYPES HERE! IT WILL KILL PERFORMANCE!
RETURNS TABLE WITH SCHEMABINDING AS
RETURN
--===== "Inline" CTE Driven "Tally Table" produces values from 1 up to 10,000...
-- enough to cover VARCHAR(8000)
WITH E1(N) AS (
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL
SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1
), --10E+1 or 10 rows
E2(N) AS (SELECT 1 FROM E1 a, E1 b), --10E+2 or 100 rows
E4(N) AS (SELECT 1 FROM E2 a, E2 b), --10E+4 or 10,000 rows max
cteTally(N) AS (--==== This provides the "base" CTE and limits the number of rows right up front
-- for both a performance gain and prevention of accidental "overruns"
SELECT TOP (ISNULL(DATALENGTH(@pString),0)) ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) FROM E4
),
cteStart(N1) AS (--==== This returns N+1 (starting position of each "element" just once for each delimiter)
SELECT 1 UNION ALL
SELECT t.N+1 FROM cteTally t WHERE SUBSTRING(@pString,t.N,1) = @pDelimiter
),
cteLen(N1,L1) AS(--==== Return start and length (for use in substring)
SELECT s.N1,
ISNULL(NULLIF(CHARINDEX(@pDelimiter,@pString,s.N1),0)-s.N1,8000)
FROM cteStart s
)
--===== Do the actual split. The ISNULL/NULLIF combo handles the length for the final element when no delimiter is found.
SELECT ItemNumber = ROW_NUMBER() OVER(ORDER BY l.N1),
Item = SUBSTRING(@pString, l.N1, l.L1)
FROM cteLen l
GO
然后与for xml结合使用将行连接到列以获得结果。
-- Create some data.
declare @a table(String nvarchar(100));
insert into @a values
('1.2.30.4.50')
,('06.7.80.90.10')
,('01.20.3.40.5');
-- This is how the function returns the data.
select Item
,right('0' + cast(Item as varchar(10)),2) as PadItem
from dbo.DelimitedSplit8K('1.2.30.4.50','.');
-- This is how you use FOR XML to concatenate the rows back together.
select (select '' + right('0' + cast(Item as varchar(10)),2)
from dbo.DelimitedSplit8K(a.String,'.')
for xml path('')
) as Result
from @a a;
给你的结果如下：
Result
----------
0102300450
0607809010
0120034005
答案 3 :(得分：0)
请尝试以下方法：
DECLARE @Str VARCHAR(100) = '2.5.99.10.1' DECLARE @EvlStr VARCHAR(100) = '',@OutPut VARCHAR(100) = ''
WHILE LEN（@Str）＆gt; 0 BEGIN
IF CHARINDEX('.',@Str) = 0
BEGIN
SET @EvlStr = @Str
SET @Str = ''
END
ELSE
BEGIN
SELECT @EvlStr = SUBSTRING(@Str,0,CHARINDEX('.',@Str))
SELECT @Str = SUBSTRING(@Str,CHARINDEX('.',@Str)+1,LEN(@Str))
END
IF LEN(@EvlStr) = 1
SET @OutPut = @OutPut + '0' + @EvlStr
ELSE
SET @OutPut = @OutPut + @EvlStr
END
SELECT @OutPut
答案 4 :(得分：0)
此解决方案基于以下内容：
source 'https://rubygems.org'
# Bundle edge Rails instead: gem 'rails', github: 'rails/rails'
gem 'rails', '~> 5.0.0', '>= 5.0.0.1'
# Use Puma as the app server
gem 'puma', '~> 3.0'
# Use SCSS for stylesheets
gem 'sass-rails', '~> 5.0'
# Use Uglifier as compressor for JavaScript assets
gem 'uglifier', '>= 1.3.0'
# Use CoffeeScript for .coffee assets and views
gem 'coffee-rails', '~> 4.2'
# See https://github.com/rails/execjs#readme for more supported runtimes
# gem 'therubyracer', platforms: :ruby
# Use jquery as the JavaScript library
gem 'jquery-rails'
# Turbolinks makes navigating your web application faster. Read more: https://github.com/turbolinks/turbolinks
gem 'turbolinks', '~> 5'
# Build JSON APIs with ease. Read more: https://github.com/rails/jbuilder
gem 'jbuilder', '~> 2.5'
gem 'mongoid', '~> 6.0'
gem 'bson_ext'
gem 'moped'
gem 'devise', '~> 4.0'
group :development, :test do
gem 'byebug', platform: :mri
gem 'rspec-rails', '~> 3.0'
gem 'factory_girl_rails', '~> 4.0'
gem 'database_cleaner', '~> 1.0'
end
group :development do
gem 'web-console'
gem 'listen', '~> 3.0.5'
gem 'spring'
gem 'spring-watcher-listen', '~> 2.0.0'
end
gem 'tzinfo-data', platforms: [:mingw, :mswin, :x64_mingw, :jruby]