Allows Caps letter in password form

neideb

Registered User.
Local time
Today, 00:32
Joined
Oct 9, 2001
Messages
42
Hi all,
Here is a piece of code I have to check if PW is OK. I do not know why it says that PW is not OK when I use Capital Letters. Can anyone help? I believe my problem is here but not sure.:confused:
...
For K = 1 To Len(Me!PW)
strPW1 = strPW1 & Asc(Mid$(Me!PW, K, 1))
strPW2 = strsPW2 & Asc(Mid$(.Column(2), K, 1))
Next K

If strPW1 = strPW2 Then
...
Tks a lot
 
furthermore, check out the function STRCOMP(). That can compare apples to apples with regard to cases.
Funny how that is what the article that JBB was referring to was talking about. :confused: You didn't read it before posting, did you?
 
Funny how that is what the article that JBB was referring to was talking about. :confused: You didn't read it before posting, did you?

of course I didn't. You know me, Bob. I never contribute anything useful. Isn't that what your last negative reputation to me said? :rolleyes:

Actually though, since you've invited yet more arguing AGAIN, I might as well say that if you theoretically consider my post from a time-saving standpoint, it IS useful. Hence, the OP can now look at one sentence of mine and see a function that him/her can look up in Access alone, instead of reading through a mundane MS article and hope that it's useful and/or comprehendable.

Has my likeability gone up in Oregon yet? :p
 
Hi Bob, happy birthday to you. Tks for helping.
Tks to John Big Booty and the Net 2.0. I will read article and let´s see I can fix my code.
Latter I will let you know.
Tks again
 
Hello, I solved the problem to let user type PW in both caps and small caracter as follows:
...
For K = 1 To Len(Me!PW)
strPW1 = strPW1 & Asc(Mid$(Me!PW, K, 0)) 'where there was 1 put 0
strPW2 = strsPW2 & Asc(Mid$(.Column(2), K, 0)) 'where there was 1 put 0

Next K

If strPW1 = strPW2 Then
...

Up to now is working.
Tks again
 

Users who are viewing this thread

Back
Top Bottom