比较第一与第二个字符串,是否有连续的5个字符相同.sql
生活随笔
收集整理的這篇文章主要介紹了
比较第一与第二个字符串,是否有连续的5个字符相同.sql
小編覺(jué)得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.
--比較第一與第二個(gè)字符串,是否有連續(xù)的5個(gè)字符相同,如果有,返回1,否則返回0
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_compstr]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[f_compstr]
GO
if exists (select * from dbo.sysobjects where id = object_id(N'[序數(shù)表]') and OBJECTPROPERTY(id, N'IsUserTable') = 1)
drop table [序數(shù)表]
GO
--為提高效率,最好創(chuàng)建一個(gè)固定的序數(shù)表
select top 1600 id=identity(int,1,1) into 序數(shù)表 from sysobjects a,sysobjects b
go
create function f_compstr(
@str1 varchar(8000),
@str2 varchar(8000)
) returns bit
as
begin
?? ?declare @re bit
?? ?/*--如果使用動(dòng)態(tài)序數(shù),效率極低.
?? ?declare @tb table(id int identity(1,1),a int)
?? ?insert into @tb(a) select top 1600 null from sysobjects a,sysobjects b
?? ?--*/
?? ?
?? ?if len(@str1)>len(@str2) set @re=case when exists(
?? ??? ?select 1
?? ??? ?from (select name=@str2)a,序數(shù)表 b? --使用動(dòng)態(tài)序數(shù),改為: ,@tb b
?? ??? ?where b.id<=len(@str2)-4 and @str1 like '%'+substring(name,b.id,5)+'%'
?? ??? ?) then 1 else 0 end
?? ?else set @re=case when exists(
?? ??? ?select 1
?? ??? ?from (select name=@str1)a,序數(shù)表 b? --使用動(dòng)態(tài)序數(shù),改為: ,@tb b
?? ??? ?where b.id<=len(@str1)-4 and @str2 like '%'+substring(name,b.id,5)+'%'
?? ??? ?) then 1 else 0 end
?? ?return(@re)
end
go
--調(diào)用示例
select dbo.f_compstr('浦東:田園小區(qū)148號(hào)405室','浦:田園小區(qū)148幢405室')
go
--使用例子,查詢(xún)不重復(fù)的記錄,重復(fù)的,只顯示第一條:
declare @t table(id int identity(1,1),address varchar(1000))
insert into @t
select '浦東:田園小區(qū)148號(hào)405室'
union all select '浦:田園小區(qū)148幢405室'
union all select '浦:田園小區(qū)148幢405室'
union all select '浦:田園小區(qū)148幢405室'
union all select 'abcdefghijklmn'
union all select 'abcdefg'
union all select 'abcdefghijklmn'
union all select 'abcdefg'
union all select '123456768898'
union all select '123345'
select * from @t a
where id=(select min(id) from @t where dbo.f_compstr(a.address,address)=1)
/*--測(cè)試結(jié)果
id????????? address??????????????????? ?
----------- ----------------------------
1?????????? 浦東:田園小區(qū)148號(hào)405室
5?????????? abcdefghijklmn
9?????????? 123456768898
10????????? 123345
--*/
if exists (select * from dbo.sysobjects where id = object_id(N'[dbo].[f_compstr]') and xtype in (N'FN', N'IF', N'TF'))
drop function [dbo].[f_compstr]
GO
if exists (select * from dbo.sysobjects where id = object_id(N'[序數(shù)表]') and OBJECTPROPERTY(id, N'IsUserTable') = 1)
drop table [序數(shù)表]
GO
--為提高效率,最好創(chuàng)建一個(gè)固定的序數(shù)表
select top 1600 id=identity(int,1,1) into 序數(shù)表 from sysobjects a,sysobjects b
go
create function f_compstr(
@str1 varchar(8000),
@str2 varchar(8000)
) returns bit
as
begin
?? ?declare @re bit
?? ?/*--如果使用動(dòng)態(tài)序數(shù),效率極低.
?? ?declare @tb table(id int identity(1,1),a int)
?? ?insert into @tb(a) select top 1600 null from sysobjects a,sysobjects b
?? ?--*/
?? ?
?? ?if len(@str1)>len(@str2) set @re=case when exists(
?? ??? ?select 1
?? ??? ?from (select name=@str2)a,序數(shù)表 b? --使用動(dòng)態(tài)序數(shù),改為: ,@tb b
?? ??? ?where b.id<=len(@str2)-4 and @str1 like '%'+substring(name,b.id,5)+'%'
?? ??? ?) then 1 else 0 end
?? ?else set @re=case when exists(
?? ??? ?select 1
?? ??? ?from (select name=@str1)a,序數(shù)表 b? --使用動(dòng)態(tài)序數(shù),改為: ,@tb b
?? ??? ?where b.id<=len(@str1)-4 and @str2 like '%'+substring(name,b.id,5)+'%'
?? ??? ?) then 1 else 0 end
?? ?return(@re)
end
go
--調(diào)用示例
select dbo.f_compstr('浦東:田園小區(qū)148號(hào)405室','浦:田園小區(qū)148幢405室')
go
--使用例子,查詢(xún)不重復(fù)的記錄,重復(fù)的,只顯示第一條:
declare @t table(id int identity(1,1),address varchar(1000))
insert into @t
select '浦東:田園小區(qū)148號(hào)405室'
union all select '浦:田園小區(qū)148幢405室'
union all select '浦:田園小區(qū)148幢405室'
union all select '浦:田園小區(qū)148幢405室'
union all select 'abcdefghijklmn'
union all select 'abcdefg'
union all select 'abcdefghijklmn'
union all select 'abcdefg'
union all select '123456768898'
union all select '123345'
select * from @t a
where id=(select min(id) from @t where dbo.f_compstr(a.address,address)=1)
/*--測(cè)試結(jié)果
id????????? address??????????????????? ?
----------- ----------------------------
1?????????? 浦東:田園小區(qū)148號(hào)405室
5?????????? abcdefghijklmn
9?????????? 123456768898
10????????? 123345
--*/
總結(jié)
以上是生活随笔為你收集整理的比较第一与第二个字符串,是否有连续的5个字符相同.sql的全部?jī)?nèi)容,希望文章能夠幫你解決所遇到的問(wèn)題。
- 上一篇: 什么时候考虑使用神经网络
- 下一篇: forbiden django1.4 t