验证非浮点整数值 [英] validating a non float integer value
问题描述
代码如下:
函数isInt(输入)
isInt = true
for i = 1到len(输入)
d =中(输入,i,1)
如果Asc(d)< 48或Asc(d)> 57然后
isInt = false
退出
结束如果
next
结束功能
我尝试更换:
功能isInt(输入)
设置re =新的RegExp
re.Pattern =" [0-9]"
isInt = re.test(输入)
结束功能
但似乎没有用。
如何解决问题?
非常感谢您的帮助。
问候
Eugene Anthony
***发送来自开发人员指南 http://www.developersdex.com ***
更正:
函数isInt(输入)
dim re,match
set re = new RegExp
re.Pattern =" \D"
re.Global = True
set match = re.Execute(输入)
如果match.count> 0然后
isInt = false
else
isInt = true
结束如果
set match = nothing
set re = nothing
结束功能
-
罗兰大厅
/ *这些信息的分发是希望它有用,但
没有任何保证;甚至没有适销性的暗示保证
或特定用途的适用性。 * /
Technet脚本中心 - http:// www .microsoft.com / technet / scriptcenter /
WSH 5.6文档 - http://msdn.microsoft.com/downloads/list/webdev.asp
MSDN Library - http://msdn.microsoft.com/library/default.asp
" Eugene Anthony"在消息中写道
news:eK ************** @ tk2msftngp13.phx.gbl ...
:function isInt(输入)
:isInt = true
:for i = 1 to len(输入)
:d = Mid(输入,i,1)
:如果Asc(d)< 48或Asc(d)> 57然后
:isInt = false
:退出
:结束如果
:next
:结束功能
:
:我尝试更换:
:
:function isInt (输入)
:设置re = new RegExp
:re.Pattern =" [0-9]"
:isInt = re.test(输入)
:结束功能
功能isInt(输入)
dim re,匹配
set re = new RegExp
re.Pattern =" \D"
re.Global = True
set match = re.Execute(输入)
如果match.count> 0然后
isInt = false
else
isInt = true
结束如果
set re = nothing
结束功能
-
Roland Hall
/ *这个信息的分发是希望它有用,但
没有任何保证;甚至没有适销性的暗示保证
或特定用途的适用性。 * /
Technet脚本中心 - http:// www .microsoft.com / technet / scriptcenter /
WSH 5.6文档 - http://msdn.microsoft.com/downloads/list/webdev.asp
MSDN Library - http://msdn.microsoft.com/library/default.asp
Eugene Anthony写道:我试过替换为:
函数isInt(输入)
设置re = new RegExp
re.Pattern =" [0-9]"
isInt = re.test(输入)
结束功能
但似乎没有工作。
如何解决问题?。
单程:
函数IsInt(str)
IsInt = CDbl(str)= Int(str)
结束功能
-
戴夫安德森
未经请求的商业电子邮件将以
The code bellow:
function isInt(input)
isInt = true
for i=1 to len(input)
d = Mid(input,i,1)
if Asc(d) < 48 OR Asc(d) > 57 then
isInt = false
exit for
end if
next
end function
I tried replacing with:
function isInt(input)
Set re = new RegExp
re.Pattern = "[0-9]"
isInt = re.test(input)
end function
But dont seem to work.
How do I solve the problem?.
Your help is kindly appreciated.
Regards
Eugene Anthony
*** Sent via Developersdex http://www.developersdex.com ***
Correction:
function isInt(input)
dim re, match
set re = new RegExp
re.Pattern = "\D"
re.Global = True
set match = re.Execute(input)
if match.count > 0 then
isInt = false
else
isInt = true
end if
set match = nothing
set re = nothing
end function
--
Roland Hall
/* This information is distributed in the hope that it will be useful, but
without any warranty; without even the implied warranty of merchantability
or fitness for a particular purpose. */
Technet Script Center - http://www.microsoft.com/technet/scriptcenter/
WSH 5.6 Documentation - http://msdn.microsoft.com/downloads/list/webdev.asp
MSDN Library - http://msdn.microsoft.com/library/default.asp
"Eugene Anthony" wrote in message
news:eK**************@tk2msftngp13.phx.gbl...
: function isInt(input)
: isInt = true
: for i=1 to len(input)
: d = Mid(input,i,1)
: if Asc(d) < 48 OR Asc(d) > 57 then
: isInt = false
: exit for
: end if
: next
: end function
:
: I tried replacing with:
:
: function isInt(input)
: Set re = new RegExp
: re.Pattern = "[0-9]"
: isInt = re.test(input)
: end function
function isInt(input)
dim re, match
set re = new RegExp
re.Pattern = "\D"
re.Global = True
set match = re.Execute(input)
if match.count > 0 then
isInt = false
else
isInt = true
end if
set re = nothing
end function
--
Roland Hall
/* This information is distributed in the hope that it will be useful, but
without any warranty; without even the implied warranty of merchantability
or fitness for a particular purpose. */
Technet Script Center - http://www.microsoft.com/technet/scriptcenter/
WSH 5.6 Documentation - http://msdn.microsoft.com/downloads/list/webdev.asp
MSDN Library - http://msdn.microsoft.com/library/default.asp
Eugene Anthony wrote:I tried replacing with:
function isInt(input)
Set re = new RegExp
re.Pattern = "[0-9]"
isInt = re.test(input)
end function
But dont seem to work.
How do I solve the problem?.
One way:
Function IsInt(str)
IsInt = CDbl(str) = Int(str)
End Function
--
Dave Anderson
Unsolicited commercial email will be read at a cost of
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