[AccessD] Find characters in one string that are inanother:w asRE: Treat string as array

Well Heenan, that was well beyond the time when I had already just caved and
implemented the mid() solution.  I will be going back and rewriting using
Gustav's solution, I just haven't had the time yet.

The nice thing about this list is that so many people are willing to
respond, and I REALLY DO appreciate all responses, just want to get that
out.  I may grouse a bit but people are willing to take time and make
suggestions.  And of course we have so many keen minds that someone will
have the answer if there is one. 

Don't you agree that

Dim byArray() As Byte
Dim c As Variant
    byArray = StrConv(str, vbFromUnicode)
    For Each c In byArray()
        Debug.Print c; " "; Chr(c);
    Next c

Is just an elegant solution?  It may not be any faster than the mid() but is
sooooo much more readable.  Not to mention the fact that once it is in an
array you can do all kinds of comparisons on each element.


John W. Colby
Colby Consulting
www.ColbyConsulting.com 
-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Heenan, Lambert
Sent: Tuesday, March 25, 2008 11:15 AM
To: 'Access Developers discussion and problem solving'
Subject: Re: [AccessD] Find characters in one string that are inanother:w
asRE: Treat string as array

But John, didn't you see Gustav's response very early on in the thread
suggesting you use StrConv? Armed with that clue your dream of using a
simple for next loop is easily achieved...

Function StringArray(str As String) As Variant Dim byArray() As Byte
    byArray = StrConv(str, vbFromUnicode)
    StringArray = byArray
End Function

Sub PlayWithStringArray(str As String)
Dim byArray() As Byte
Dim c As Variant
    byArray = StringArray(str)
    For Each c In byArray()
        Debug.Print c; " "; Chr(c);
    Next c
    Debug.Print vbCrLf; str
End Sub

You're on the money though about the number of answers that seemed not to be
connected with the question! :-)

Lambert

-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com] On Behalf Of jwcolby
Sent: Tuesday, March 25, 2008 11:01 AM
To: 'Access Developers discussion and problem solving'
Subject: Re: [AccessD] Find characters in one string that are in
another:wasRE: Treat string as array


LOL.  

Turn a string into an array.  Turn a string into an ARRAY.  TURN A STRING
INTO AN ARRAY.  

That was my question.  Why I want to do it is irrelevant.

And yea, Regex probably would work, but I did not have time to spend a week
learning a useful but EXTREMELY arcane syntax to solve a 5 minute problem.
In fact I didn't have time to answer 25 suggestions having nothing to do
with TURNING A STRING INTO AN ARRAY. 

Sigh.

;-)

John W. Colby
Colby Consulting
www.ColbyConsulting.com
-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Max Wanadoo
Sent: Tuesday, March 25, 2008 10:43 AM
To: 'Access Developers discussion and problem solving'
Subject: Re: [AccessD] Find characters in one string that are in
another:wasRE: Treat string as array

Aha!  Now we get the question...

>I was trying to search for characters in a password that are also in a
"must contain" string.
>"The password must contain one 'special character' ~!@#$%^&*()_+ and 
>the
password must contain one 'number' 1234567890.
>Given a password "The thread that went south", how do I determine if it
contains any special characters or numbers.

Whew!  Took a while to drag it out of him folks but we got there in the end.
Moral: Give JC rubbish answers, he gets frustrated, QED he posts the correct
question.  Rolls Eyes!

JC: Don't know if RegExp will solve it elegantly ;-) but worth a sly glance.

Max

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