If a substring is found that matches my regular expression
pattern, I would like each character to be replaced with a
"*" character. I thought the following code would work,
but I'm missing something? :

'--------------------------------------------------------
Sub myregex()
Dim regEx As New VBScript_RegExp_55.RegExp
Dim str As String

str = "Call me at (982)555-1234!"

regEx.Pattern = "(...)...-...."
regEx.Global = True
regEx.IgnoreCase = True

MsgBox regEx.Replace(str, "*")

End Sub

"Robert Crandal" wrote:
'--------------------------------------------------------
Sub myregex()
Dim regEx As New VBScript_RegExp_55.RegExp
Dim str As String

str = "Call me at (982)555-1234!"

regEx.Pattern = "(...)...-...."
regEx.Global = True
regEx.IgnoreCase = True

MsgBox regEx.Replace(str, "*")

End Sub


If I wasn't clear, the output message should be:

*************

But, I would also be happy if the output was:
(***)***-****

Either way is fine, but the first is better.

Hi Robert,

Am Thu, 2 Apr 2015 03:08:31 -0700 schrieb Robert Crandal:

If I wasn't clear, the output message should be:

*************

But, I would also be happy if the output was:
(***)***-****

try:

Sub ReplaceGlobal()
Dim str As String
Dim ptrn As String, re As Object

str = "Call me at (982)555-1234!"
ptrn = "[0-9]"


Set re = New RegExp
re.Pattern = ptrn
re.IgnoreCase = False
re.Global = True

MsgBox re.Replace(str, "*")
End Sub


Regards
Claus B.
--
Vista Ultimate / Windows7
Office 2007 Ultimate / 2010 Professional

Hi again,

Am Thu, 2 Apr 2015 03:08:31 -0700 schrieb Robert Crandal:

If I wasn't clear, the output message should be:

*************

Sub ReplaceGlobal()
Dim str As String
Dim ptrn As String, re As Object

str = "Call me at (982)555-1234!"
ptrn = "\(|\)|-|!|\d"


Set re = New RegExp
re.Pattern = ptrn
re.IgnoreCase = False
re.Global = True

MsgBox re.Replace(str, "*")
End Sub


Regards
Claus B.
--
Vista Ultimate / Windows7
Office 2007 Ultimate / 2010 Professional

"Claus Busch" wrote:
Hi again,

str = "Call me at (982)555-1234!"
ptrn = "\(|\)|-|!|\d"


Wow, I never woulda guessed that expression.
Can you explain how it works? All I see are
a couple of alternation "|" symbols and one
digit "\d" metacharater.

Also, what pattern works for Social Security
numbers?

Hi Robert,

Am Thu, 2 Apr 2015 09:47:12 -0700 schrieb Robert Crandal:

Can you explain how it works? All I see are
a couple of alternation "|" symbols and one
digit "\d" metacharater.

\d is equivalent to [0-9]
To find parentheses you have to put a back slash in front
\( and \)
But I am not an expert with Regulkar Expressions. I always play around
with try and error.


Regards
Claus B.
--
Vista Ultimate / Windows7
Office 2007 Ultimate / 2010 Professional

"Claus Busch" wrote:

\d is equivalent to [0-9]
To find parentheses you have to put a back slash in front
\( and \)
But I am not an expert with Regulkar Expressions. I always play around
with try and error.


Yup, I'm not a regular expression expert either. For the pattern,
my guess was: "\(\d\d\d\)\d\d\d-\d\d\d\d"

The only problem was that the entire string got replaced with
a single "*", instead of filling in all 13 characters with a "*".

Hi again,

Am Thu, 2 Apr 2015 09:47:12 -0700 schrieb Robert Crandal:

Can you explain how it works? All I see are
a couple of alternation "|" symbols and one
digit "\d" metacharater.

the alternation sign splits all characters in your string and the
characters are searched one by one.
If you want to do it with a pattern try:

ptrn = "[\(\d\)\d-\d!]"

can you post an example of Social Security numbers?
I don't know how they look like in your country.


Regards
Claus B.
--
Vista Ultimate / Windows7
Office 2007 Ultimate / 2010 Professional

"Claus Busch" wrote:

can you post an example of Social Security numbers?
I don't know how they look like in your country.


In my country, they are typically 123-45-6789.

So, my guess for the pattern would be:

regEx.Pattern = "\d\d\d-\d\d-\d\d\d\d"