nth character (1 Viewer)

[dgkindy]

I have recently seen a function that will find the position of the nth character in a string but cannot remember what it was?

 

[georgedwilkinson]

You mean:

x = mid(mystring,n,1)

?

 

[dgkindy]

x = mid(mystring,n,1)
n represents the starting point of the function, while the 1 represents the lenght of the trimming

I will provide an example

C:mydocuments\floder1\folder2\folder3

I want to locate the second \ in the string

 

[LPurvis]

George offered what he did because of the phrase "find the position of the nth character in a string". It doesn't make much sense as is.
Technically - the answer to that question is N.

But you're wanting the position of the nth occurance of a specified character.
Why do you want that character's position?

Here's a non-rigourously tested different take on the task than normal.

Function fGetCharPos(strText As String, strChar As String, Optional intPos As Integer = 1)
On Error Resume Next
 
    Dim strArr() As String
 
    strArr = Split(strText, strChar)
    ReDim Preserve strArr(intPos - 1)
    fGetCharPos = Len(Join(strArr)) + (Len(strChar) - 1) * (intPos - 1) + 1
 
End Function

 

[dgkindy]

Why I want to be able to do is so that I can easily trim the directory string at the nth \ and then add in a new directory location

C:mydocuments\floder1\folder2\folder3
C:mydocuments\floder1\folder2\folder4
C:mydocuments\floder1\folder5

 

[LPurvis]

Fair enough - the function as suggested should be enough for you.
Or you could amend it to return the string to the left, rather than just the position.
e.g.

Function fGetChars(strText As String, strChar As String, Optional intPos As Integer = 1)
On Error Resume Next
 
    Dim strArr() As String
 
    strArr = Split(strText, strChar)
    ReDim Preserve strArr(intPos - 1)
    fGetChars = Join(strArr, strChar) & strChar
 
End Function

Giving:
?fGetChars("C:\mydocuments\folder1\folder2\folder4","\",3)
C:\mydocuments\folder1\

 

[ions]

what about two nested inStr()? find the first postion using inStr (use 1 as the start parameter), then if it finds it alter your start parameter to be the result of the first inStr.

If instr(1,"a","abcabc")  > 0 then
       intposition = instr(instr(1,"a","abcabc")+ 1, "a","abcabc")
end if

 

[LPurvis]

Yeah. Using InStr would a more traditional method - though, obviously, you'd have to write a function to call it iteratively on the string provided for the required number of occurances (hard coding it to two levels - or more - would be much too limited as a solution).

I offered the function solution as presented mainly because Split has become the common trend for more abbreviated code (i.e. it's cooler ;-).