I am automatically obtaining directories from an application but I can't seem to get the actual directories with the correct case of letters.
For example I get $a='C:\test\dir\log\wqerst' but the actual directory is C:\test\dir\log\WQERST.
What I want is to uppercase only wqerst so it would show C:\test\dir\log\WQERST
I've already tried using substring but I don't know how I would be able to connect it to the whole directory once it is uppercase.
Windows system is not case sensitive, but if you want really this result, you can do it :
$a='C:\test\dir\log\wqerst'
$parentpath=Split-Path -Path $a
$file=(Split-Path -Path $a -Leaf).ToUpper()
$result=Join-Path $parentpath $file
$result
As James C. and vonPryz already wrote, there is not much point to get the case sensitive folder path. However you can use this helper method:
function Get-CaseSensitiveFilePath
{
Param
(
[string]$FilePath
)
$parent = Split-Path $FilePath
$leaf = Split-Path -Leaf $FilePath
$result = Get-ChildItem $parent | where { $_ -like $leaf }
$result.FullName
}
usage:
Get-CaseSensitiveFilePath -FilePath 'C:\test\dir\log\WQERST'
This will give you the case sensitive folder name but the directory must exist on the computer you execute the script...
Related
I am having a strange problem in Powershell (Version 2021.8.0) while creating folders and naming them. I start with a number of individual ebook files in a folder that I set using Set-Location. I use the file name minus the extension to create a new folder with the same name as the e-book file. The code works fine the majority of the time with various file extensions I have stored in an array beginning of the code.
What's happening is that the code creates the proper folder name the majority of the time and moves the source file into the folder after it's created.
The problem is, if the last letter of the source file name, on files with the extension ".epub" end in an "e", then the "e" is missing from the end of the created folder name. I thought that I saw it also drop "r" and "p" but I have been unable to replicate that error recently.
Below is my code. It is set up to run against file extensions for e-books and audiobooks. Please ignore the error messages that are being generated when files of a specific type don't exist in the working folder. I am just using the array for testing and it will be filled automatically later by reading the folder contents.
This Code Creates a Folder for Each File and moves the file into that Folder:
Clear-Host
$SourceFileFolder = 'N:\- Books\- - BMS\- Books Needing Folders'
Set-Location $SourceFileFolder
$MyArray = ( "*.azw3", "*.cbz", "*.doc", "*.docx", "*.djvu", "*.epub", "*.mobi", "*.mp3", "*.pdf", "*.txt" )
Foreach ($FileExtension in $MyArray) {
Get-ChildItem -Include $FileExtension -Name -Recurse | Sort-Object | ForEach-Object { $SourceFileName = $_
$NewDirectoryName = $SourceFileName.TrimEnd($FileExtension)
New-Item -Name $NewDirectoryName -ItemType "directory"
$OriginalFileName = Join-Path -Path $SourceFileFolder -ChildPath $SourceFileName
$DestinationFilename = Join-Path -Path $NewDirectoryName -ChildPath $SourceFileName
$DestinationFilename = Join-Path -Path $SourceFileFolder -ChildPath $DestinationFilename
Move-Item $OriginalFileName -Destination $DestinationFilename
}
}
Thanks for any help you can give. Driving me nuts and I am pretty sure it's something that I am doing wrong, like always.
String.TrimEnd()
Removes all the trailing occurrences of a set of characters specified in an array from the current string.
TrimEnd method will remove all characters that matches in the character array you provided. It does not look for whether or not .epub is at the end of the string, but rather it trims out any of the characters in the argument supplied from the end of the string. In your case, all dots,e,p,u,b will be removed from the end until no more of these characters are within the string. Now, you will eventually (and you do) remove more than what you intended for.
I'd suggest using EndsWith to match your extensions and performing a substring selection instead, as below. If you deal only with single extension (eg: not with .tar.gz or other double extensions type), you can also use the .net [System.IO.Path]::GetFileNameWithoutExtension($MyFileName) method.
$MyFileName = "Teste.epub"
$FileExt = '.epub'
# Wrong approach
$output = $MyFileName.TrimEnd($FileExt)
write-host $output -ForegroundColor Yellow
#Output returns Test
# Proper method
if ($MyFileName.EndsWith($FileExt)) {
$output = $MyFileName.Substring(0,$MyFileName.Length - $FileExt.Length)
Write-Host $output -ForegroundColor Cyan
}
# Returns Tested
#Alternative method. Won't work if you want to trim out double extensions (eg. tar.gz)
if ($MyFileName.EndsWith($FileExt)) {
$Output = [System.IO.Path]::GetFileNameWithoutExtension($MyFileName)
Write-Host $output -ForegroundColor Cyan
}
You're making this too hard on yourself. Use the .BaseName to get the filename without extension.
Your code simplified:
$SourceFileFolder = 'N:\- Books\- - BMS\- Books Needing Folders'
$MyArray = "*.azw3", "*.cbz", "*.doc", "*.docx", "*.djvu", "*.epub", "*.mobi", "*.mp3", "*.pdf", "*.txt"
(Get-ChildItem -Path $SourceFileFolder -Include $MyArray -File -Recurse) | Sort-Object Name | ForEach-Object {
# BaseName is the filename without extension
$NewDirectory = Join-Path -Path $SourceFileFolder -ChildPath $_.BaseName
$null = New-Item -Path $NewDirectory -ItemType Directory -Force
$_ | Move-Item -Destination $NewDirectory
}
How to get the 4th folder name and store it in a variable while looping through the files stored in a parent folder. For example, if the path is
C:\ParentFolder\Subfolder1\subfolder2\subfolder3\file.extension
C:\ParentFolder\Subfolder1\subfolder2\subfolder4\file.extension
C:\ParentFolder\Subfolder1\subfolder2\subfolder5\file.extension
then subfolder2 name should be stored in a variable. Can any one please help me on this?
get-childitem $DirectorNane -Recurse | Where-Object{!($_.PSIsContainer)} | % {
$filePath = $_.FullName
#Get file name
$fileName = Split-Path -Path $filePath -Leaf
} $FileI = Split-Path -Path $filePath -Leaf
Thanks in advance!
You can use the -split operator on the $filePath variable to get what you want.
$split = $filePath -split '\\'
$myvar = $split[3]
We use the double backslash to split with, because the first slash escapes the slash character to split the path by. Then, we can reference the part of the path we want in the array that gets generated in the "split" variable.
Additionally, you can solve this with a one liner using the following code:
$myvar = $filepath.split('\')[3]
This would ensure that you're always getting the fourth element in the array, but is a little less flexible since you can't specify what exactly you want based on conditions with additional scripting.
If you are asking how to get the parent directory of the directory containing a file, you can call Split-Path twice. Example:
$filePath = "C:\ParentFolder\Subfolder1\subfolder2\subfolder3\file.extension"
$parentOfParent = Split-Path (Split-Path $filePath)
# $parentOfParent now contains "C:\ParentFolder\Subfolder1\subfolder2"
I want to extract filename from below path:
D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv
Now I wrote this code to get filename. This working fine as long as the folder level didn't change. But in case the folder level has been changed, this code need to rewrite. I looking a way to make it more flexible such as the code can always extract filename regardless of the folder level.
($outputFile).split('\')[9].substring(0)
If you are ok with including the extension this should do what you want.
$outputPath = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
$outputFile = Split-Path $outputPath -leaf
Use .net:
[System.IO.Path]::GetFileName("c:\foo.txt") returns foo.txt.
[System.IO.Path]::GetFileNameWithoutExtension("c:\foo.txt") returns foo
Using the BaseName in Get-ChildItem displays the name of the file and and using Name displays the file name with the extension.
$filepath = Get-ChildItem "E:\Test\Basic-English-Grammar-1.pdf"
$filepath.BaseName
Basic-English-Grammar-1
$filepath.Name
Basic-English-Grammar-1.pdf
Find a file using a wildcard and get the filename:
Resolve-Path "Package.1.0.191.*.zip" | Split-Path -leaf
$(Split-Path "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv" -leaf)
Get-ChildItem "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
|Select-Object -ExpandProperty Name
You could get the result you want like this.
$file = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
$a = $file.Split("\")
$index = $a.count - 1
$a.GetValue($index)
If you use "Get-ChildItem" to get the "fullname", you could also use "name" to just get the name of the file.
Just to complete the answer above that use .Net.
In this code the path is stored in the %1 argument (which is written in the registry under quote that are escaped: \"%1\" ). To retrieve it, we need the $arg (inbuilt arg). Don't forget the quote around $FilePath.
# Get the File path:
$FilePath = $args
Write-Host "FilePath: " $FilePath
# Get the complete file name:
$file_name_complete = [System.IO.Path]::GetFileName("$FilePath")
Write-Host "fileNameFull :" $file_name_complete
# Get File Name Without Extension:
$fileNameOnly = [System.IO.Path]::GetFileNameWithoutExtension("$FilePath")
Write-Host "fileNameOnly :" $fileNameOnly
# Get the Extension:
$fileExtensionOnly = [System.IO.Path]::GetExtension("$FilePath")
Write-Host "fileExtensionOnly :" $fileExtensionOnly
You can try this:
[System.IO.FileInfo]$path = "D:\Server\User\CUST\MEA\Data\In\Files\CORRECTED\CUST_MEAFile.csv"
# Returns name and extension
$path.Name
# Returns just name
$path.BaseName
$file = Get-Item -Path "c:/foo/foobar.txt"
$file.Name
Works with both relative and absolute paths
I'm trying to process a list of files that may or may not be up to date and may or may not yet exist. In doing so, I need to resolve the full path of an item, even though the item may be specified with relative paths. However, Resolve-Path prints an error when used with a non-existant file.
For example, What's the simplest, cleanest way to resolve ".\newdir\newfile.txt" to "C:\Current\Working\Directory\newdir\newfile.txt" in Powershell?
Note that System.IO.Path's static method use with the process's working directory - which isn't the powershell current location.
You want:
c:\path\exists\> $ExecutionContext.SessionState.Path.GetUnresolvedProviderPathFromPSPath(".\nonexist\foo.txt")
returns:
c:\path\exists\nonexists\foo.txt
This has the advantage of working with PSPaths, not native filesystem paths. A PSPath may not map 1-1 to a filesystem path, for example if you mount a psdrive with a multi-letter drive name.
What's a pspath?
ps c:\> new-psdrive temp filesystem c:\temp
...
ps c:\> cd temp:
ps temp:\>
temp:\ is a drive-qualified pspath that maps to a win32 (native) path of c:\temp.
-Oisin
When Resolve-Path fails due to the file not existing, the fully resolved path is accessible from the thrown error object.
You can use a function like the following to fix Resolve-Path and make it work like you expect.
function Force-Resolve-Path {
<#
.SYNOPSIS
Calls Resolve-Path but works for files that don't exist.
.REMARKS
From http://devhawk.net/blog/2010/1/22/fixing-powershells-busted-resolve-path-cmdlet
#>
param (
[string] $FileName
)
$FileName = Resolve-Path $FileName -ErrorAction SilentlyContinue `
-ErrorVariable _frperror
if (-not($FileName)) {
$FileName = $_frperror[0].TargetObject
}
return $FileName
}
I think you're on the right path. Just use [Environment]::CurrentDirectory to set .NET's notion of the process's current dir e.g.:
[Environment]::CurrentDirectory = $pwd
[IO.Path]::GetFullPath(".\xyz")
Join-Path (Resolve-Path .) newdir\newfile.txt
This has the advantage of not having to set the CLR Environment's current directory:
[IO.Path]::Combine($pwd,"non\existing\path")
NOTE
This is not functionally equivalent to x0n's answer. System.IO.Path.Combine only combines string path segments. Its main utility is keeping the developer from having to worry about slashes. GetUnresolvedProviderPathFromPSPath will traverse the input path relative to the present working directory, according to the .'s and ..'s.
I've found that the following works well enough.
$workingDirectory = Convert-Path (Resolve-Path -path ".")
$newFile = "newDir\newFile.txt"
Do-Something-With "$workingDirectory\$newFile"
Convert-Path can be used to get the path as a string, although this is not always the case. See this entry on COnvert-Path for more details.
function Get-FullName()
{
[CmdletBinding()]
Param(
[Parameter(ValueFromPipeline = $True)] [object[]] $Path
)
Begin{
$Path = #($Path);
}
Process{
foreach($p in $Path)
{
if($p -eq $null -or $p -match '^\s*$'){$p = [IO.Path]::GetFullPath(".");}
elseif($p -is [System.IO.FileInfo]){$p = $p.FullName;}
else{$p = [IO.Path]::GetFullPath($p);}
$p;
}
}
}
I ended up with this code in my case. I needed to create a file later in the the script, so this code presumes you have write access to the target folder.
$File = ".\newdir\newfile.txt"
If (Test-Path $File) {
$Resolved = (Resolve-Path $File).Path
} else {
New-Item $File -ItemType File | Out-Null
$Resolved = (Resolve-Path $File).Path
Remove-Item $File
}
I also enclosed New-Item in try..catch block, but that goes out of this question.
I had a similar issue where I needed to find the folder 3 levels up from a folder that does not exist yet to determine the name for a new folder I wanted to create... It's complicated. Anyway, this is what I ended up doing:
($path -split "\\" | select -SkipLast 3) -join "\\"
You can just set the -errorAction to "SilentlyContinue" and use Resolve-Path
5 > (Resolve-Path .\AllFilerData.xml -ea 0).Path
C:\Users\Andy.Schneider\Documents\WindowsPowerShell\Scripts\AllFilerData.xml
6 > (Resolve-Path .\DoesNotExist -ea 0).Path
7 >
There is an accepted answer here, but it is quite lengthy and there is a simpler alternative available.
In any recent version of Powershell, you can use Test-Path -IsValid -Path 'C:\Probably Fake\Path.txt'
This simply verifies that there are no illegal characters in the path and that the path could be used to store a file. If the target doesn't exist, Test-Path won't care in this instance -- it's only being asked to test if the provided path is potentially valid.
Both most popular answers don't work correctly on paths on not existing drives.
function NormalizePath($filename)
{
$filename += '\'
$filename = $filename -replace '\\(\.?\\)+','\'
while ($filename -match '\\([^\\.]|\.[^\\.]|\.\.[^\\])[^\\]*\\\.\.\\') {
$filename = $filename -replace '\\([^\\.]|\.[^\\.]|\.\.[^\\])[^\\]*\\\.\.\\','\'
}
return $filename.TrimEnd('\')
}
Check if the file exists before resolving:
if(Test-Path .\newdir\newfile.txt) { (Resolve-Path .\newdir\newfile.txt).Path }
I am still pretty new to scripting and "programming" at all. if you miss any information here let me know.
This is my working zip function:
$folder = "C:\zipthis\"
$destinationFilePath = "C:\_archive\zipped"
function create-7zip{
param([string] $folder,
[String] $destinationFilePath)
write-host $folder $destinationFilePath
[string]$pathToZipExe = "C:\Program Files (x86)\7-Zip\7zG.exe";
[Array]$arguments = "a", "-tzip", "$destinationFilePath", "$folder";
& $pathToZipExe $arguments;
}
Get-ChildItem $folder | ? { $_.PSIsContainer} | % {
write-host $_.BaseName $_.Name;
$dest= [System.String]::Concat($destPath,$_.Name,".zip");
(create-7zip $_.FullName $dest)
}
create-7zip $folder $destinationFilePath
now I want him to zip special folders which I already sorted out :
get-childitem "C:\zipme\" | where-Object {$_.name -eq "www" -or $_.name -eq "sql" -or $_.name -eq "services"}
This small function finds the 3 folders I need called www, sql and services. But I didn't manage to insert this into my zip function, so that exactly this folders are zipped and put into C:\_archive\zipped
Because a string is used instead of an array, he tried always to look for a folder called wwwsqlservice which is not there. I tried to put an array using #(www,sql,services) but i had no success, so whats the right way, if there is one?
It should compatible with powershell 2.0, no ps3.0 cmdlets or functions please.
thanks in advance!
Here's a really simple example of what you want to do, removed from the context of your function. It assumes that your destination folders already exist (You can just use Test-Path and New-Item to create them if they don't), and that you're using 7z.exe.
$directories = #("www","sql","services")
$archiveType = "-tzip"
foreach($dir in $directories)
{
# Use $dir to update the destination each loop to prevent overwrites!
$sourceFilePath = "mySourcePath\$dir"
$destinationFilePath = "myTargetPath\$dir"
cmd /c "$pathToZipExe a $archiveType $destinationFilePath $sourceFilePath"
}
Overall it looks like you got pretty close to a solution, with some minor changes needed to support the foreach loop. If you're confident that create-7zip works fine for a single folder, you can substitute that for the cmd /c line above. Here's a listing of some handy example usages for 7zip on the command line.