【www.bbyears.com--php函数】
在sql中我判断包含字符串我们可使用很多方法,如like,replace,charindex函数都可实现我们要的功能,下面我来给各位介绍判断字符串包含字符串sql语句。
通过2个函数CHARINDEX和PATINDEX以及通配符的灵活使用
函数:CHARINDEX和PATINDEX
CHARINDEX:查某字符(串)是否包含在其他字符串中,返回字符串中指定表达式的起始位置。
PATINDEX:查某字符(串)是否包含在其他字符串中,返回指定表达式中某模式第一次出现的起始位置;如果在全部有效的文本和字符数据类型中没有找到该模式,则返回零。特殊:可以使用通配符!
例子:
1. 查询字符串中是否包含非数字字符
SELECT PATINDEX("%[^0-9]%", "1235X461") SELECT PATINDEX("%[^0-9]%", "12350461")
2. 查询字符串中是否包含数字字符
SELECT PATINDEX("%[0-9]%", "SUYLLGoO") SELECT PATINDEX("%[0-9]%", "SUYLLG0O")
3.函数判断字符串只包含数字
CREATE FUNCTION [dbo].fn_IsNumeric ( @pString VARCHAR(8000) ) RETURNS bit WITH ENCRYPTION AS BEGIN DECLARE @vJudge int SET @vJudge = 0 SELECT @vJudge = CASE WHEN PATINDEX("%[0-9]%", LOWER(@pString)) > 0 THEN 0 WHEN PATINDEX("%[0-9]%", LOWER(@pString)) = 0 THEN 1 END RETURN @vJudge END
4.函数判断字符串只包含字母(忽略大小写)
CREATE FUNCTION [dbo].fn_IsAlpha ( @pString VARCHAR(8000) ) RETURNS bit WITH ENCRYPTION AS BEGIN DECLARE @vJudge int SET @vJudge = 0 SELECT @vJudge = CASE WHEN PATINDEX("%[a-z]%", LOWER(@pString)) > 0 THEN 0 WHEN PATINDEX("%[a-z]%", LOWER(@pString)) = 0 THEN 1 END RETURN @vJudge END
5. 函数判断字符串不包含任何符号(包括空格)
CREATE FUNCTION [dbo].fn_IsAlphanumeric ( @pString VARCHAR(8000) ) RETURNS bit WITH ENCRYPTION AS BEGIN DECLARE @vJudge int SET @vJudge = 0 SELECT @vJudge = CASE WHEN PATINDEX("%[^a-z0-9]%", LOWER(@pString)) > 0 THEN 0 WHEN PATINDEX("%[^a-z0-9]%", LOWER(@pString)) = 0 THEN 1 END RETURN @vJudge END
6. 函数判断字符串不包含任何符号(除空格外)
CREATE FUNCTION [dbo].fn_IsAlphanumericBlank ( @pString VARCHAR(8000) ) RETURNS bit WITH ENCRYPTION AS BEGIN DECLARE @vJudge int SET @vJudge = 0 SELECT @vJudge = CASE WHEN PATINDEX("%[^a-z0-9 ]%", LOWER(@pString)) > 0 THEN 0 WHEN PATINDEX("%[^a-z0-9 ]%", LOWER(@pString)) = 0 THEN 1 END RETURN @vJudge END -- 注意:[^a-z0-9 ]模式中最后有一个空格。
用charindex()——charindex(字符,字符串)>0 –>包含 查看一段话、一篇文章里面包含什么词
select ID,title,author from Article where CHARINDEX(title,@item)>0
7.用like——
select * from tablename where field1 like like ‘%key%"
8.使用replace()函数
declare @item nvarchar(100) set @item="英语好难"; select ID,title,author from Article where LEN(REPLACE(@item,title,""))2、