Visual Basic

Regular Expression in VB 6.0

12-Nov-08 1:52

Ditto

Any suggestions, ideas, or 'constructive criticism' are always welcome.

"There's no such thing as a stupid question, only stupid people." - Mr. Garrison

indian14311-Nov-08 23:18

11-Nov-08 23:18

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

Simon P Stevens11-Nov-08 23:51

11-Nov-08 23:51

http://www.google.co.uk/search?hl=en&q=regular%20expression&meta=[^]

Regular expressions are ways of matching patterns in text, validating formats, and sometimes extracting specifically formatted information.

MSDN Guide[^]

Simon

indian14312-Nov-08 0:05

12-Nov-08 0:05

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

Simon P Stevens12-Nov-08 0:18

12-Nov-08 0:18

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

Rajesh Anuhya12-Nov-08 0:21

Rajesh Anuhya

12-Nov-08 0:21

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 <--

indian14312-Nov-08 0:44

12-Nov-08 0:44

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

Rajesh Anuhya12-Nov-08 0:54

Rajesh Anuhya

12-Nov-08 0:54

try to implement your self

Rajesh B --> A Poor Workman Blames His Tools <--

indian14312-Nov-08 1:03

12-Nov-08 1:03

Hi,

Can anybody please tell me the pattern to take only the integer values as numerics and decimal values not. Like "90009" should be taken as numeric and "900.09" should be taken as non numeric.
Just give me one pattern.

Thanks,
Aleem Mohammad.

Thanks & Regards,

Md. Abdul Aleem
NIIT technologies

Simon P Stevens12-Nov-08 1:15

12-Nov-08 1:15

No one here is going to do your job for you for free.

I will be happy to write the code for you, but you will have to pay me my standard hourly rate of £1000 per hour.

If you don't want to pay, you will have to figure it out yourself. I've guided you as much as I can, I've given you links to all the information you need. You now need to go away and read those links and understand what needs doing.

Good luck.

Simon

indian14312-Nov-08 1:33

12-Nov-08 1:33

Hi,

Can u please help whether I have any method or preperty in RegExp class to check if our expression is matching with the pattern, am asking u this as am new to VB 6.0. And intellisense is also not working for this object in my editor.

Thanks,
Aleem Mohammad.

Thanks & Regards,

Md. Abdul Aleem
NIIT technologies

Simon P Stevens12-Nov-08 2:54

12-Nov-08 2:54

No, sorry. I've given you enough links and suggestions. If you can't work it out yourself then you need to go and spend some time learning properly.

These links explain regular expressions in VB6. Read them.

http://www.regular-expressions.info/vb.html[^]

http://support.microsoft.com/kb/818802[^]

Simon

Jon_Boy12-Nov-08 3:22

12-Nov-08 3:22

And if that isn't enough, download something like Expresso to help you test the expressions.

Cheers!

Any suggestions, ideas, or 'constructive criticism' are always welcome.

"There's no such thing as a stupid question, only stupid people." - Mr. Garrison

Paul Conrad12-Nov-08 3:42

Paul Conrad

12-Nov-08 3:42

kissdznuts wrote:
download something like Expresso

I second that. It is a great tool, I have found it helpful many of times when dealing with regular expressions.

"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

indian14312-Nov-08 4:02

12-Nov-08 4:02

Hi,

Is it free download or costs some thing and another one is, can I replace some of the characters not matching with my regular expression, with null string.
If it is then it will be of great help for me.

Thanks,
Aleem Mohammad.

Thanks & Regards,

Md. Abdul Aleem
NIIT technologies

Paul Conrad12-Nov-08 4:07

Paul Conrad

12-Nov-08 4:07

indian143 wrote:
it free download

Ummm, yes it is. Familiarize yourself with the search textbox at the top of this forum.

Here is your assignment: type in the word Expresso and then hit the Go! button. Hint: there are two articles done by Jim Hollenhorst that are both excellent.

Jon_Boy12-Nov-08 6:09

12-Nov-08 6:09

Paul 'FTW'!

Any suggestions, ideas, or 'constructive criticism' are always welcome.

"There's no such thing as a stupid question, only stupid people." - Mr. Garrison

Paul Conrad12-Nov-08 6:38

Paul Conrad

12-Nov-08 6:38

My assignment for him can't be that hard. Fine quotation from Mr. Garrison. Smile | :)

Jon_Boy12-Nov-08 4:09

12-Nov-08 4:09

Free registration to unlock Expresso.

Enjoy!

Any suggestions, ideas, or 'constructive criticism' are always welcome.

"There's no such thing as a stupid question, only stupid people." - Mr. Garrison

indian14312-Nov-08 3:37

12-Nov-08 3:37

Ok thanks for the help I got it but only one thing in last I want to ask you in this series is : please give me a pattern to mis the special characters like "@", "#" etc.

I have done it as below its working fine for charates, float all, but giving true when I add "@" or "#" to it. Can u please help in this.

Set objRegExp = New VBScript_RegExp_10.RegExp

   With objRegExp
       .Pattern = "\d{" & iLen & "}"
       .IgnoreCase = True
       .Global = True
   End With

   If iLen > 0 Then
        If (objRegExp.Test(sTemp) = True) Then
           IsNumericOnly = True
        End If
   End If

Thanks a Lot,
Aleem Mohammad.

Thanks & Regards,

Md. Abdul Aleem
NIIT technologies

SQL query problem

sunil goyalG11-Nov-08 22:53

sunil goyalG

11-Nov-08 22:53

hi all,
i run a query which update a field , is failed when it contains single or double quote.Otherwise it is running successfully. sample query is following
update table1 set name='abcd'efg'

please suggest how can i solve this problem ?

nishkarsh_k11-Nov-08 23:13

nishkarsh_k

11-Nov-08 23:13

sunil goyalG wrote:
update table1 set name='abcd'efg'

try this
update table1 set name='abcd\'efg'

' and \ are excape characters so one has to put "\" before them to send as proper query

sunil goyalG12-Nov-08 0:17

sunil goyalG

12-Nov-08 0:17

This solution is not works for me. query is failed. I am using SQL Server Express 2005

Nanda_MR12-Nov-08 20:00

Nanda_MR

12-Nov-08 20:00

use the instant of ' Asc char. It will work i also tried in my system it is working properly .

Ananda