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
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