mysql获取分割字段字符串根据原数据分割后修改成新数据

查询sys_menu表，字段perms，把字段分割前22个字符和23-100的字符

SELECT menu_id,substring(perms,23,100),substring(perms,1,22) FROM  `sys_menu`  WHERE perms like "tiancheng:ErpStoreDrug%"

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根据一个表修改另一个表内容

UPDATE student1 a,student2 b SET a.score=b.score WHERE a.name = b.name 

SELECT   CONCAT("tiancheng:ErpMgStore","Drug:",substring(a.perms,23,100)) AS perms  FROM sys_menu a  where  a.perms like "tiancheng:ErpStoreDrug%"

UPDATE sys_menu a set a.perms = CONCAT("tiancheng:ErpMgStore","Drug:",substring(a.perms,23,100))  where  a.perms like "tiancheng:ErpStoreDrug%"

one:

declare @s varchar(20) 
declare @i varchar(20) 
set @i='' 
set @s='新会员必须购买350元产品' 
while PATINDEX ('%[0-9]%', @s)>0 
begin 
set @i=@i+substring(@s,PATINDEX ('%[0-9]%', @s),1) 
set @s=stuff(@s,1,PATINDEX ('%[0-9]%', @s),'') 
end 

select @i

--
300
two:

declare @a table(id int identity(1,1),a varchar(100)) 
insert @a select '新会员必须购买350元产品' 
union all select '新店首次定货必须满20000元' 

select left(right(a,len(a)-patindex('%[0-9]%',a)+1),len(right(a,len(a)-patindex('%[0-9]%',a)+1))-1) from @a 

上在的

select substring(所查询字符串,patindex('%[^0-9][0-9]%',所查询字符串)+1,patindex('%[0-9][^0-9]%',所查询字符串)-patindex('%[^0-9][0-9]%',所查询字符串)) 
 #这个只能查询第一次在字符串出现的数字串 

那么如果出现字符串什么样子的呢 sss8989sss //www.jb51.net ss8989ss8989ss8989 7879aafds789 432432432543534 应该怎么取呢

实例
复制代码 代码如下:

create function fn_GetNum(@s varchar(8000)) 
returns varchar(8000) 
as 
begin 
select @s = stuff(stuff(@s, 1, patindex('%[0-9, .]%', @s) - 1, ''), 
patindex('%[^0-9, .]%', stuff(@s, 1, patindex('%[0-9, .]%', @s) - 1, '')), 
len(@s), '') 
return @s 
end 

declare @t table(s varchar(8000)) 
insert @t select 'aaa11112bbb' 
union all select 'ccc212sss' 
union all select 'sss21a' 
select dbo.fn_GetNum(s) as result from @t 

select substring(s,patindex('%[^0-9][0-9]%',s)+1,patindex('%[0-9][^0-9]%',s)-patindex('%[^0-9][0-9]%',s)) from @t 

/*功能:获取字符串中的字母*/ 
CREATE FUNCTION dbo.F_Get_STR (@S VARCHAR(100)) 
RETURNS VARCHAR(100) 
AS 
BEGIN 
WHILE PATINDEX('%[^a-z]%',@S)>0 
BEGIN 
set @s=stuff(@s,patindex('%[^a-z]%',@s),1,'') 
END 
RETURN @S 
END 
GO 

--测试

select dbo.F_Get_STR('测试ABC123ABC')

GO 

/* 
功能:获取字符串中的数字 
*/ 
create function dbo.F_Get_Number (@S varchar(100)) 
returns int 
AS 
begin 
while PATINDEX('%[^0-9]%',@S)>0 
begin 
set @s=stuff(@s,patindex('%[^0-9]%',@s),1,'') 
end 
return cast(@S as int) 
end 

--测试 
---select dbo.F_Get_Number('测试AB3C123AB5C') 

GO