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I think we are in standby mode for the next regex question today
Any suggestions, ideas, or 'constructive criticism' are always welcome.
"There's no such thing as a stupid question, only stupid people." - Mr. Garrison
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Hi,
Thanks a lot friends. I got it. It is completed.
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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Yes, but is it tested completely?? Did you check all boundry conditions and possible inputs??
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Hi,
Can anybody please check and clear any problem exists in the below code. My actual requirement is I get values as "900.09" and "90009", but I have return "90009" as numeric and "900.09" as non-numeric.
I am not getting the pattern or may be my testing senario is wrong please clear me this as I am new to VB 6.0.
With objRegExp
.Pattern = "\d" '"[0-9]"
.IgnoreCase = True
.Global = True
End With
If iLen > 0 Then
If (objRegExp.Test(sTemp) = True) Then
IsNumericOnly = True
End If
End If
Thanks,
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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Quit spamming the board. You've already asked this question below.
If this vb6 code or .Net?
Any suggestions, ideas, or 'constructive criticism' are always welcome.
"There's no such thing as a stupid question, only stupid people." - Mr. Garrison
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Hi,
It is VB 6.0 code not .Net.
Thanks,
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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kissdznuts wrote: this vb6 code or .Net?
He says he is new to vb6. I don't know whether to laugh or feel sorry for him.
"The clue train passed his station without stopping." - John Simmons / outlaw programmer
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
"Not only do you continue to babble nonsense, you can't even correctly remember the nonsense you babbled just minutes ago." - Rob Graham
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Judging from your previous posts, why do you even need a RegEx?? Try to convert the string to a Double, or some other appropriate numeric type. If the conversion works, the string was valid, if not, it'll throw up an error, so use an "On Error" to catch this and react appropriately.
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Hi,
Will this work, may be not because I have to separate only the integers not even floats or doubles. Thats why choosen Regular expression.
Thanks,
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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You really have a problem with specifying your requirements. As I counted, this is the third time you've changed them.
In that case, since you're need the number to be only integers, it appears that you've already got the RegEx correct, or maybe you need to specify that you're looking for 'x' number of digits in a row. Maybe this field needs to be a 5 digit number at all times?? In that case, the expression would be something like \d{5} . Or, it's the rest of the code that's bad, but since you haven't posted the entire function, nor given a complete description of your requirements and input samples, it's very difficult to say.
Also, you say you're NEW to VB6?? Did you know that VB6 is now a dead technology?? Microsoft stopped supporting it a while ago and there will be no future support for running VB6 apps in Windows 7 and beyond.
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Hi.
I have an issue with VB 6.0 Binary Compatibility. I have one dll (Let's say a.dll which is in VC++) and another VB Activex Dll (Let's say b.dll) is using a.dll. b.dll is binary compatible.
I have made small changes to the a.dll and i have referred to the new version of a.dll in b.dll project. Now when I try to compile I get an error "Unexpected Error (32810)".
I have spent lot of time on this and unable to find out the root cause. Can anybody please explain the issue and solution for this issue.
Best regards
Shankar
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i think u have to re-register the a.dll
use regsvr32 <dll name="" with="" path="">
before that ensure that your a.dll is successfully compiled in th vc++.
Rajesh B --> A Poor Workman Blames His Tools <--
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This[^] should clear things up a bit.
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Nothing to do with VB.net either - stop cross posting
Bob
Ashfield Consultants Ltd
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srisant wrote: stop demotivating people,
Didn't know I was - you posted the same question 4 times in 3 different forums. Multiple posts of the same question are considered bad manners.
What else have you tried? A quick google revealed that ONENOTE from Microsoft does what you want.
Bob
Ashfield Consultants Ltd
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Ditto
Any suggestions, ideas, or 'constructive criticism' are always welcome.
"There's no such thing as a stupid question, only stupid people." - Mr. Garrison
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Hi All,
Can anybody please help me whats "RegularExpression" and how to use it. If any links can be provided will be great help.
Thanks,
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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Hi,
Can u please help me in resolving it, I have a some 100 strings of numerical values. some will like "900.09" and some will be like "90009".
Then I should take values like "90009" as numerics and return "True" and if any char other than that and like this "900.09" as non-numeric return 'False".
I have written code as below, but they are saying that it is performance issue. Can u help me in writing it by using Regular expressions.
sTemp = TestString
iLen = Len(sTemp)
If iLen > 0 Then
For iCtr = 1 To iLen
sChar = Mid(sTemp, iCtr, 1)
If Not sChar Like "[0-9]" Then Exit Function
Next
'If IsNumeric(sTemp) = True Then
'If Not sTemp Like "[0-9]" Then MsgBox "Hi"
'End If
IsNumericOnly = True
End If
Thanks a Lot,
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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The keyword "Like" in VB actually is a very weak form of regular expression. The performance problem here is that you are iterating through every single character. With a proper regular expression you can match against the whole string in one go.
You need to create a new Regex object, and set it up with the required pattern string.
then you can call Match on the regex object and check the Success flag.
To work out the pattern string you need to use, look at the MSDN link I provided in my last reply.
You can find a reference to the regular expression pattern language here[^]
Also, try working through this tutorial[^] to start you off.
Simon
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try this
Private Function validatedata(data As String) As Boolean
Dim innlop As Integer
For innlop = 1 To Len(data)
If IsNumeric(Mid(data, innlop, 1)) = False Then
validatedata = False
Exit Function
End If
validatedata = True
Next
End Function
Rajesh B --> A Poor Workman Blames His Tools <--
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Hi,
Thanks good help, but can u please help me in doing the same by using the Regular Expression object.
Thanks,
Aleem Mohammad.
Thanks & Regards,
Md. Abdul Aleem
NIIT technologies
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try to implement your self
Rajesh B --> A Poor Workman Blames His Tools <--
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