InStr and Mid crap [resolved]

ok i was told how to get just part of a line from a text file using InStr and Mid, and ive looked all over for how to do that, because i have no idea how to use InStr and Mid at all. i see things like

Code:

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InStr(strCheck, Mid(sInvalidChars, i, 1)

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but i have no idea why its setup like that, or what it means (ill post the entire script of the program dealy that had it inside). thats the closest thing to instructions on what InStr and Mid are that ive seen

Code:

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Dim bCK As Boolean
Dim strDomainType As String
Dim strDomainName As String
Const sInvalidChars As String = "!#$%^&*()=+{}[]|\;:'/?>,< "
Dim i As Integer

bCK = Not InStr(1, strCheck, Chr(34)) > 0 'Check to see if there is a double quote
If Not bCK Then GoTo ExitFunction

bCK = Not InStr(1, strCheck, "..") > 0 'Check to see if there are consecutive dots
If Not bCK Then GoTo ExitFunction

' Check for invalid characters.
If Len(strCheck) > Len(sInvalidChars) Then
    For i = 1 To Len(sInvalidChars)
        If InStr(strCheck, Mid(sInvalidChars, i, 1)) > 0 Then
            bCK = False
            GoTo ExitFunction
        End If
    Next
Else
    For i = 1 To Len(strCheck)
        If InStr(sInvalidChars, Mid(strCheck, i, 1)) > 0 Then
            bCK = False
            GoTo ExitFunction
        End If
    Next
End If

If InStr(1, strCheck, "@") > 1 Then 'Check for an @ symbol
    bCK = Len(Left(strCheck, InStr(1, strCheck, "@") - 1)) > 0
Else
    bCK = False
End If
If Not bCK Then GoTo ExitFunction

strCheck = Right(strCheck, Len(strCheck) - InStr(1, strCheck, "@"))
bCK = Not InStr(1, strCheck, "@") > 0 'Check to see if there are too many @'s
If Not bCK Then GoTo ExitFunction

strDomainType = Right(strCheck, Len(strCheck) - InStr(1, strCheck, "."))
bCK = Len(strDomainType) > 0 And InStr(1, strCheck, ".") < Len(strCheck)
If Not bCK Then GoTo ExitFunction

strCheck = Left(strCheck, Len(strCheck) - Len(strDomainType) - 1)
Do Until InStr(1, strCheck, ".") <= 1
    If Len(strCheck) >= InStr(1, strCheck, ".") Then
        strCheck = Left(strCheck, Len(strCheck) - (InStr(1, strCheck, ".") - 1))
    Else
        bCK = False
        GoTo ExitFunction
    End If
Loop
If strCheck = "." Or Len(strCheck) = 0 Then bCK = False

ExitFunction:
ValidEmail = bCK
End Function

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PS: i dont have any help files oO
go figure...

PS: thats not my code...

Have you looked in Help?

Syntex of InStr is
(Start, Original String, String to be found)
This function will tell you where the String to be found is located in the Original String.
For example:

VB Code:

Msgbox InStr(1,"Hey You","You")

This code will show you 5, cuz the string "You" is starting in the 5th place of the original string.

This appears to be your problem:

VB Code:

Const sInvalidChars As String = "!#$%^&*()=+{}[]|\;:'/?>,< "

If you define your constant like this, the program will only do what you want when it encounters all those characters in that specific sequence. I'm not sure how to declare a multiple-setting constant, but I do know that a delimiter is used.

Quote:

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PS: i dont have any help files oO
go figure...

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oooooo ok so thats how it works.
and now 1 more question:
how would i save "Hey" to a variable?

You could do something like this:

VB Code:

VariableName = Left$("Hey You", 3)

That will return the first three characters from the left end of the string. In this case, it would return "Hey."

Now if I did something like this:

VB Code:

VariableName = Right$("Hey You", 5)

That would return the first 5 characters from the right end of the string. This would return "y You." Mid$ has nothing to do with what you're trying to accomplish, although this would achieve the same thing as my Right$ example:

VB Code:

VariableName = Mid$("Hey You", 3)

It would skip to the 3rd position in the string, and return everything from that point. You would need to do a reverse Mid$ to use that function in this case, and I don't think there is a reverse Mid$.

I see you pretty much have InStr down pat though.

Get it now?

BTW Stiletto, you can also use InStr to check for specific characters. I do it all the time. Something like this would check for ('s and return a message box if one was found:

VB Code:

If InStr(1, strCheck, "(" Then
    MsgBox "There is a ( in the string."
End If

nice....thx 4 the help, i finally understand WOOTTTTTTTT

It's not a problem. These functions are quite intimidating at first, but once you get the hang of them, you'll find they're actually quite easy.

Hothead, and all of you out there who think something "is easy". After all these year, I've finally come up with "THE ABSOLUTE TRUTH". It is:

BEFORE YOU KNOW HOW TO DO SOMETHING, "IT'S IMPOSSIBLE"; ONCE YOU LEARN HOW TO DO IT, "IT'S SO EASY"

That's my point. Re-read my post. ;)