下面是一个简单的测试sql语句，可以直接在查询分析器中运行。
DECLARE @intLength AS INTEGER
DECLARE @vchRegularExpression AS VARCHAR(50)
DECLARE @vchSourceString as VARCHAR(50)
DECLARE @vchSourceString2 as VARCHAR(50)
DECLARE @bitHasNoSpecialCharacters as BIT
-- 初始化变量
SET @vchSourceString = 'Test one This is a test!!'
SET @vchSourceString2 = 'Test two This is a test'
-- 我们的正则表达式应该类似于
-- [a-zA-Z ]{}
-- 如： [a-zA-Z ]{10} 一个十字符的字符串
-- 获得字符串长度
SET @intLength = LEN(@vchSourceString)
-- 设置完整的正则表达式
SET @vchRegularExpression = '[a-zA-Z ]{' + CAST(@intLength as varchar) + '}'
-- 是否有任何特殊字符
SET @bitHasNoSpecialCharacters = dbo.find_regular_expression(@vchSourceString, @vchRegularExpression,0)
PRINT @vchSourceString
IF @bitHasNoSpecialCharacters = 1 BEGIN
PRINT 'No special characters.'
END ELSE BEGIN
PRINT 'Special characters found.'
END
PRINT '**************'
-- 获得字符串长度
SET @intLength = LEN(@vchSourceString2)
-- 设置完整的正则表达式
SET @vchRegularExpression = '[a-zA-Z ]{' + CAST(@intLength as varchar) + '}'
-- 是否有任何特殊字符
SET @bitHasNoSpecialCharacters = dbo.find_regular_expression(@vchSourceString2, @vchRegularExpression,0)
PRINT @vchSourceString2
IF @bitHasNoSpecialCharacters = 1 BEGIN
PRINT 'No special characters.'
END ELSE BEGIN
PRINT 'Special characters found.'
END
GO
最后得出的查询语句如下：
select * FROM TaskDeploy where 1=1 And dbo.find_regular_expression(Participants,'\bsamlin\b',0) = 1
DECLARE @vchRegularExpression AS VARCHAR(50)
DECLARE @vchSourceString as VARCHAR(50)
DECLARE @vchSourceString2 as VARCHAR(50)
DECLARE @bitHasNoSpecialCharacters as BIT
-- 初始化变量
SET @vchSourceString = 'Test one This is a test!!'
SET @vchSourceString2 = 'Test two This is a test'
-- 我们的正则表达式应该类似于
-- [a-zA-Z ]{}
-- 如： [a-zA-Z ]{10} 一个十字符的字符串
-- 获得字符串长度
SET @intLength = LEN(@vchSourceString)
-- 设置完整的正则表达式
SET @vchRegularExpression = '[a-zA-Z ]{' + CAST(@intLength as varchar) + '}'
-- 是否有任何特殊字符
SET @bitHasNoSpecialCharacters = dbo.find_regular_expression(@vchSourceString, @vchRegularExpression,0)
PRINT @vchSourceString
IF @bitHasNoSpecialCharacters = 1 BEGIN
PRINT 'No special characters.'
END ELSE BEGIN
PRINT 'Special characters found.'
END
PRINT '**************'
-- 获得字符串长度
SET @intLength = LEN(@vchSourceString2)
-- 设置完整的正则表达式
SET @vchRegularExpression = '[a-zA-Z ]{' + CAST(@intLength as varchar) + '}'
-- 是否有任何特殊字符
SET @bitHasNoSpecialCharacters = dbo.find_regular_expression(@vchSourceString2, @vchRegularExpression,0)
PRINT @vchSourceString2
IF @bitHasNoSpecialCharacters = 1 BEGIN
PRINT 'No special characters.'
END ELSE BEGIN
PRINT 'Special characters found.'
END
GO
最后得出的查询语句如下：
select * FROM TaskDeploy where 1=1 And dbo.find_regular_expression(Participants,'\bsamlin\b',0) = 1
