I am stuck with an issue. I've done my research and found out that I can use InSTR function to search for a specific character in a string.
What I am trying to do is get a file name extracted from a file path.
Currently I have
InStr(StrFrom(pName), "\")
The issue here is, it returns the first occurrence of slash, where as I want to get the last occurrence of the slash (so that I can use a 'right' function wrapped around the above code to capture the file name)
Any help is appreciated on how to get the last slash in a string!
Thanks!
Instr looks from the start of the text string, InstrRev starts looking from the other end.
Public Function FileNameOnly(ByVal FileNameAndPath As String) As String
FileNameOnly = Mid(FileNameAndPath, InStrRev(FileNameAndPath, "\") + 1, Len(FileNameAndPath))
End Function
Consider:
Sub marine()
Dim s As String, ary
s = "C:\whatever\sub1\sub2\reallydeep\x.xlsm"
ary = Split(s, "\")
MsgBox ary(UBound(ary))
End Sub
Use InStrRev() to find the first occurrence of the slash from the right side of the string.
https://msdn.microsoft.com/en-us/library/t2ekk41a(v=vs.90).aspx
Assuming StrFrom is some user defined function, the following will do what you want:
Dim filename as String
Dim path as String
path = StrFrom(pName)
filename = Mid$(path, InstrRev(path, "\") + 1)
Note that its easier to use Mid$ than Right$, as InstrRev returns the character position from the left of the string. Omitting the final parameter of Mid$, returns the rest of the string from that position.
Related
I am using regex.ismatch to check a string doesn't contain any one of a list of characters such as £&+(/?!;:* And also a quotation mark " not sure how to place that...
But can't get to to work...
If Regex.ismatch(Line, "^[^##£&+()*']"). Then
Msgbox("error")
End If
But doesn't work for me?
Any suggestions
You could do this pretty easily without Regex by simply doing something like this:
Public Shared Function HasSpecialChars(ByVal str As String) As Boolean
Const specialChars As String = "!##$%^&*()"
Dim indexOf As Integer = str.IndexOfAny(specialChars.ToCharArray())
Return indexOf <> -1
End Function
I need to pull the code from the following string: 72381 Test 4Dx for Worms. The code is 72381 and the function that I'm using does a wonderful job of pulling ALL the numbers from a string and gives me back 723814, which pulls the 4 from the description of the code. The actual code is only the 72381. The codes are of varying length and are always followed by a space before the description begins; however there are spaces in the descriptions as well. This is the function I am using that I found from a previous search:
Function OnlyNums(sWord As String)
Dim sChar As String
Dim x As Integer
Dim sTemp As String
sTemp = ""
For x = 1 To Len(sWord)
sChar = Mid(sWord, x, 1)
If Asc(sChar) >= 48 And _
Asc(sChar) <= 57 Then
sTemp = sTemp & sChar
End If
Next
OnlyNums = Val(sTemp)
End Function
If the first character in the description part of your string is never numeric, you could use the VBA Val(string) function to return all of the numeric characters before the first non-numeric character.
Function GetNum(sWord As String)
GetNum = Val(sWord)
End Function
See the syntax of the Val(string) function for full details of it's usage.
You're looking for the find function.. Example:
or in VBA instr() and left()
Since you know the pattern is always code followed by space just use left of the string for the number of characters to the first space found using instr. Sample in immediate window above. Loop is going to be slow, and while it may validate they are numeric why bother if you know pattern is code then space?
In similar situations in C# code, I leave the loop early after finding the first instance of a space character (32). In VBA, you'd use Exit For.
You can get rid of the function altogether and use this:
split("72381 Test 4Dx for Worms"," ")(0)
This will split the string into an array using " " as the split char. Then it shows us address 0 in the array (the first element)
In the context of your function if you are dead set on using one it is this:
Function OnlyNums(sWord As String)
OnlyNums = Split(sWord, " ")(0)
End Function
While I like the simplicity of Mark's solution, you could use an efficient parser below to improve your character by character search (to cope with strings that don't start with numbers).
test
Sub test()
MsgBox StrOut("72381 Test 4Dx")
End Sub
code
Function StrOut(strIn As String)
Dim objRegex As Object
Set objRegex = CreateObject("vbscript.regexp")
With objRegex
.Pattern = "^(\d+)(\s.+)$"
If .test(strIn) Then
StrOut = .Replace(strIn, "$1")
Else
StrOut = "no match"
End If
End With
End Function
I want to filter out other characters from a string as well as split the remaining numbers with periods.
This is my string: major.number=9minor.number=10revision.number=0build.number=804
and this is the expected output: 9.10.0.804
Any suggestions?
As to my comment, if your text is going to be constant you can use String.Split to remove the text and String.Join to add your deliminators. Quick example using your string.
Sub Main()
Dim value As String = "major.number=9minor.number=10revision.number=0build.number=804"
Dim seperator() As String = {"major.number=", "minor.number=", "revision.number=", "build.number="}
Console.WriteLine(String.Join(".", value.Split(seperator, StringSplitOptions.RemoveEmptyEntries)))
Console.ReadLine()
End Sub
If your string does not always follow a specific pattern, you could use Regex.Replace:
Sub Main()
Dim value as String = "major.number=9minor.number=10revision.number=0build.number=804"
Dim version as String = Regex.Replace(value, "\D*(\d+)\D*", "$1.") ' Run the regex
version = version.Substring(0, version.Length - 1) ' Trim the last dot
End
Note you should Imports System.Text.RegularExpressions.
BEFORE:
Johnson0, Yvonne
AFTER:
Johnson, Yvonne
String functions for Access can be found at http://www.techonthenet.com/access/functions/string/replace.php
In your example, code like
Replace("Johnson0", "0", "")
will do the trick for the particular string Johnson0. If you need to only remove the zero if it is the last character, play with the additional start and count parameters described in the link above.
You can try executing following query..
UPDATE table set
columnName = REPLACE(columnName,'0','')
WHERE columnName LIKE "%0%";
This will replace all occurrence of "0" with "".
The answer you submitted clarifies your requirement. Based on that, you don't need to create a user-defined function if your Access version is 2000 or later. You can get the same result with the Replace() function.
MsgBox Replace("Jonson0, Yvonne", "0,", ",")
One approach is to create a custom function
See http://www.techonthenet.com/access/functions/misc/alphanumeric.php for an example. You could do something similar, but in the loop you would only keep the alpha characters.
Public Sub xxx()
MsgBox RemoveStr0("Jonson0, Yvonne")
End Sub
Public Function RemoveStr0(sString As String) As String
Dim ipos As Long, sTemp As String
ipos = InStr(1, sString, "0,")
sTemp = Mid$(sString, 1, ipos - 1)
sTemp = sTemp & Mid$(sString, ipos + 1)
RemoveStr0 = sTemp
End Function
if you can pull it out to java or another OO lang you can just do a matching using regexes.
This question already has answers here:
How to extract file name from path?
(16 answers)
Closed 1 year ago.
I need to extract the filename from a path (a string):
e.g.,
"C:\folder\folder\folder\file.txt" = "file" (or even "file.txt" to get me started)
Essentially everything before and including the last \
I've heard of using wildcards in place of Regex (as it's an odd implementation in VBA?) but can't find anything solid.
Cheers in advance.
I believe this works, using VBA:
Dim strPath As String
strPath = "C:\folder\folder\folder\file.txt"
Dim strFile As String
strFile = Right(strPath, Len(strPath) - InStrRev(strPath, "\"))
InStrRev looks for the first instance of "\" from the end, and returns the position. Right makes a substring starting from the right of given length, so you calculate the needed length using Len - InStrRev
Thanks to kaveman for the help. Here is the full code I used to remove both the path and the extension (it is not full proof, does not take into consideration files that contain more than 2 decimals eg. *.tar.gz)
sFullPath = "C:\dir\dir\dir\file.txt"
sFullFilename = Right(sFullPath, Len(sFullPath) - InStrRev(sFullPath, "\"))
sFilename = Left(sFullFilename, (InStr(sFullFilename, ".") - 1))
sFilename = "file"
I was looking for a solution without code. This VBA works in the Excel Formula Bar:
To extract the file name:
=RIGHT(A1,LEN(A1)-FIND("~",SUBSTITUTE(A1,"\","~",LEN(A1)-LEN(SUBSTITUTE(A1,"\","")))))
To extract the file path:
=MID(A1,1,LEN(A1)-LEN(MID(A1,FIND(CHAR(1),SUBSTITUTE(A1,"\",CHAR(1),LEN(A1)-LEN(SUBSTITUTE(A1,"\",""))))+1,LEN(A1))))
`You can also try:
Sub filen()
Dim parts() As String
Dim Inputfolder As String, a As String
'Takes input as any file on disk
Inputfolder = Application.GetOpenFilename("Folder, *")
parts = Split(Inputfolder, "\")
a = parts(UBound(parts()))
MsgBox ("File is: " & a)
End Sub
This sub can display Folder name of any file
Here's simpler method: a one-line function to extract only the name — without the file extension — as you specified in your example:
Function getName(pf):getName=Split(Mid(pf,InStrRev(pf,"\")+1),".")(0):End Function
...so, using your example, this:
MsgBox getName("C:\folder\folder\folder\file.txt")
returns:
For cases where you want to extract the filename while retaining the file extension, or if you want to extract the only the path, here are two more single-line functions:
Extract Filename from x:\path\filename:
Function getFName(pf)As String:getFName=Mid(pf,InStrRev(pf,"\")+1):End Function
Extract Path from x:\path\filename:
Function getPath(pf)As String: getPath=Left(pf,InStrRev(pf,"\")): End Function
Examples:
(Source)
Using Java:
String myPath="C:\folder\folder\folder\file.txt";
System.out.println("filename " + myPath.lastIndexOf('\\'));
I used kaveman's suggestion successfully as well to get the Full File name but sometimes when i have lots of Dots in my Full File Name, I used the following to get rid of the .txt bit:
FileName = Left(FullFileName, (InStrRev(FullFileName, ".") - 1))
You can use a FileSystemObject for that.
First, include a reference for de Microsoft Scripting Runtime (VB Editor Menu Bar > Tools > References).
After that, you can use a function such as this one:
Function Get_FileName_fromPath(myPath as string) as string
Dim FSO as New Scripting.FileSystemObject
'Check if File Exists before getting the name
iF FSO.FileExists(myPath) then
Get_FileName_fromPath = FSO.GetFileName(myPath)
Else
Get_FileName_fromPath = "File not found!"
End if
End Function
File System Objects are very useful for file manipulation, especially when checking for their existence and moving them around. I like to use them early bound (Dim statement), but you can use them late bound if you prefer (CreateObject statement).
'[/vba]
' Keep It Simple
' .. why use FileSystemObject or Split when Left and Mid will do it
' the FSO has some 33 Subs or Functions
that have to be loaded each time it is created.
' and needs the file to exist ... yet is only a bit slower
... under twice time.. some good code in FSO
' conservation is good .. spare a few electrons. ????... save a few millionths of a sec
'Also
' .. why the format of a function that we all seem to use like
'
' .. Function GetAStr(x) as string
' dim extraStr as string
' a lot of work with extraStr..
' that could have been done with the string variable GetAStr
already created by the function
' then .. GetAStr=extraStr to put it in its right place
' .. End Function
Function GetNameL1$(FilePath$, Optional NLess& = 1)
' default Nless=1 => name only
' NLess =2 => xcopya.xls xcopyb.xls xcopy7.xlsm all as xcopy to get find latest version
' Nless = - 4 or less => name with name.ext worka.xlsm
GetNameL1 = Mid(FilePath, InStrRev(FilePath, "") + 1)
GetNameL1 = Left(GetNameL1, InStrRev(GetNameL1, ".") - NLess)
End Function
Function LastFold$(FilePath$)
LastFold = Left(FilePath, InStrRev(FilePath, "") - 1)
LastFold = Mid(LastFold, InStrRev(LastFold, "") + 1)
End Function
Function LastFoldSA$(FilePath$)
Dim SA$(): SA = Split(FilePath, "")
LastFoldSA = SA(UBound(SA) - 1)
End Function
[<vba]