Visual Basic, Finding if certain characters exist in a string

Alex
#1 / 7

Finding if certain characters exist in a string

I have a string and i would like to be able to find out if it contains 4
points "." is this possible in vb 5? Also would it be the same for words?

Thanks

Alex

Wed, 18 Jun 1902 08:00:00 GMT

Philipp Lensse
#2 / 7

Finding if certain characters exist in a string

I don't think there is a VB method to do this, I wrote my own

Public Function countString(ByVal stringToSearchThrough As String, _
ByVal stringToCount As String, Optional ByVal _
compareMethod As VbCompareMethod = vbTextCompare) As Integer
' returns how often a string appears in another one
Dim positionFound As Long, quantityFound As Long
Do
positionFound = InStr(positionFound + 1, _
stringToSearchThrough, stringToCount, _
compareMethod)
If positionFound >= 1 Then
quantityFound = quantityFound + 1
Else
Exit Do
End If
Loop
countString = quantityFound
End Function

Now you would write "if countString(myString, ".") = 4 then..."

--
http://www.hitnet-ev.de/~lenssen

Quote:

> I have a string and i would like to be able to find out if it contains 4
> points "." is this possible in vb 5? Also would it be the same for words?

> Thanks

> Alex

Wed, 18 Jun 1902 08:00:00 GMT

dsavits
#3 / 7

Finding if certain characters exist in a string

i am not sure if you want to look for
"XXXX.XXXX.XXXX.XXXX.XXXX" - four "."
or
"XXXXXXXXXXX....XXXXXXXXX" - four "."'s

to do the first
strX = "XXXX.XXXX.XXXX.XXXX.XXXX"
Dim r as Variant
r = split(strX, ".")
(r is now a 1 dimensional string array, so count through its items (with a
for each loop) to see if there are 5.)

to do the second
strX = "XXXXXXXXXXX....XXXXXXXXX"
if InStr(strX, "....") <> 0 then
'Do something here
end if

Quote:

> Also would it be the same for words?

yes

Quote:

> I don't think there is a VB method to do this, I wrote my own

> Public Function countString(ByVal stringToSearchThrough As String, _
> ByVal stringToCount As String, Optional ByVal _
> compareMethod As VbCompareMethod = vbTextCompare) As Integer
> ' returns how often a string appears in another one
> Dim positionFound As Long, quantityFound As Long
> Do
> positionFound = InStr(positionFound + 1, _
> stringToSearchThrough, stringToCount, _
> compareMethod)
> If positionFound >= 1 Then
> quantityFound = quantityFound + 1
> Else
> Exit Do
> End If
> Loop
> countString = quantityFound
> End Function

> Now you would write "if countString(myString, ".") = 4 then..."

> --
> http://www.hitnet-ev.de/~lenssen

> > I have a string and i would like to be able to find out if it contains 4
> > points "." is this possible in vb 5? Also would it be the same for
words?

> > Thanks

> > Alex

Wed, 18 Jun 1902 08:00:00 GMT

Philipp Lensse
#4 / 7

Finding if certain characters exist in a string

If split works in VB5 already, then the function could be a oneliner...
actually, one could just remember the approach

if UBound(Split(myString, ".")) = 4 then...

--
http://www.hitnet-ev.de/~lenssen

Quote:

> i am not sure if you want to look for
> "XXXX.XXXX.XXXX.XXXX.XXXX" - four "."
> or
> "XXXXXXXXXXX....XXXXXXXXX" - four "."'s

> to do the first
> strX = "XXXX.XXXX.XXXX.XXXX.XXXX"
> Dim r as Variant
> r = split(strX, ".")
> (r is now a 1 dimensional string array, so count through its items (with a
> for each loop) to see if there are 5.)

> to do the second
> strX = "XXXXXXXXXXX....XXXXXXXXX"
> if InStr(strX, "....") <> 0 then
> 'Do something here
> end if
> > Also would it be the same for words?
> yes

> > I don't think there is a VB method to do this, I wrote my own

> > Public Function countString(ByVal stringToSearchThrough As String, _
> > ByVal stringToCount As String, Optional ByVal _
> > compareMethod As VbCompareMethod = vbTextCompare) As Integer
> > ' returns how often a string appears in another one
> > Dim positionFound As Long, quantityFound As Long
> > Do
> > positionFound = InStr(positionFound + 1, _
> > stringToSearchThrough, stringToCount, _
> > compareMethod)
> > If positionFound >= 1 Then
> > quantityFound = quantityFound + 1
> > Else
> > Exit Do
> > End If
> > Loop
> > countString = quantityFound
> > End Function

> > Now you would write "if countString(myString, ".") = 4 then..."

> > --
> > http://www.hitnet-ev.de/~lenssen

> > > I have a string and i would like to be able to find out if it contains
4
> > > points "." is this possible in vb 5? Also would it be the same for
> words?

> > > Thanks

> > > Alex

Wed, 18 Jun 1902 08:00:00 GMT

Mr.
#5 / 7

Finding if certain characters exist in a string

Quote:

> to do the first
> strX = "XXXX.XXXX.XXXX.XXXX.XXXX"
> Dim r as Variant
> r = split(strX, ".")
> (r is now a 1 dimensional string array, so count through its items (with a
> for each loop) to see if there are 5.)

why not dimension r as a dynamic string array? and instead of looping
through the array, just check the upper bound and see if it is 4 (0 to 4 =
5). the last will either be blank or whatever comes after the last "." which
gives the extra 5th.

strX = "XXXX.XXXX.XXXX.XXXX.XXXX"
Dim r() as String
r = Split(strX, ".")
If UBound(r) = 4 Then 'there are 4 points "."

or make it a function:

strX = "XXXX.XXXX.XXXX.XXXX.XXXX"

Function FindNumOfStrings(ByVal SearchString As String, ByVal SearchFor As
String, HowMany As Long) As Boolean
Dim r() As String
r = Split(SearchString, SearchFor)
If UBound(r) = HowMany Then SomeFunc = True
End Function

Private Sub Command1_Click()
If FindNumOfStrings(strX, ".", 4) Then Debug.Print "There are 4 points
""."""
End Sub

Boolean values are False by default and the function will not retain its
last value, so if it is not set to true within itself it will be False so
there is no need to add and Else statement to set the function to False.

Mr. X

Wed, 18 Jun 1902 08:00:00 GMT

KOOM
#6 / 7

Finding if certain characters exist in a string

Quote:

> I have a string and i would like to be able to find out if it contains 4
> points "." is this possible in vb 5? Also would it be the same for words?

in VB6 works InStr

Wed, 18 Jun 1902 08:00:00 GMT

Howard Henry Schlunde
#7 / 7

Finding if certain characters exist in a string

Split() is only supported in VB6.

--
Howard Henry Schlunder
AutoSigAmp Plug-In by Howard:
Winamp is currently playing:
chumbawumba--the big issue wma

Quote:

> If split works in VB5 already, then the function could be a oneliner...
> actually, one could just remember the approach

> if UBound(Split(myString, ".")) = 4 then...

> --
> http://www.hitnet-ev.de/~lenssen

> > i am not sure if you want to look for
> > "XXXX.XXXX.XXXX.XXXX.XXXX" - four "."
> > or
> > "XXXXXXXXXXX....XXXXXXXXX" - four "."'s

> > to do the first
> > strX = "XXXX.XXXX.XXXX.XXXX.XXXX"
> > Dim r as Variant
> > r = split(strX, ".")
> > (r is now a 1 dimensional string array, so count through its items (with
a
> > for each loop) to see if there are 5.)

> > to do the second
> > strX = "XXXXXXXXXXX....XXXXXXXXX"
> > if InStr(strX, "....") <> 0 then
> > 'Do something here
> > end if
> > > Also would it be the same for words?
> > yes

> > > I don't think there is a VB method to do this, I wrote my own

> > > Public Function countString(ByVal stringToSearchThrough As String, _
> > > ByVal stringToCount As String, Optional ByVal _
> > > compareMethod As VbCompareMethod = vbTextCompare) As Integer
> > > ' returns how often a string appears in another one
> > > Dim positionFound As Long, quantityFound As Long
> > > Do
> > > positionFound = InStr(positionFound + 1, _
> > > stringToSearchThrough, stringToCount, _
> > > compareMethod)
> > > If positionFound >= 1 Then
> > > quantityFound = quantityFound + 1
> > > Else
> > > Exit Do
> > > End If
> > > Loop
> > > countString = quantityFound
> > > End Function

> > > Now you would write "if countString(myString, ".") = 4 then..."

> > > --
> > > http://www.hitnet-ev.de/~lenssen

> > > > I have a string and i would like to be able to find out if it
contains
> 4
> > > > points "." is this possible in vb 5? Also would it be the same for
> > words?

> > > > Thanks

> > > > Alex

Wed, 18 Jun 1902 08:00:00 GMT