Powershell 如何将除新创建的文件之外的所有内容移动到另一个目录

Question

I have a directory that is getting populated with text files every morning and it generates one single output file. I have also created a Dump directory that will get populated with the files that were used to create the output file.

My question is, how can I move everything except the output file to the dump folder to be deleted? I am not quite sure how to do that.

My code so far:

# If the path does not exist, force it to create it
if(!(Test-Path $Path)) {
    New-Item -ItemType Directory -Force -Path $Path 
    # New-Item -ItemType Directory -Force -Path $SOF
    # New-Item -ItemType Directory -Force -Path $EOF
}

if ($Path) {
    # Move-Item $PathWithFiles -Destination $Path # move (not copy) files into new directory to concat
    Get-ChildItem $PathWithFiles | ForEach-Object {    # Output all except first and last line of current file 
        Get-Content $_ | Select-Object -Skip 1 | Select-Object -SkipLast 1
    
        ''  # Output an empty line
    
    } | Add-Content $OutPutFile
}

Answer 1

In reference to my answer in your earlier question , you can change that code to

$Path     = 'C:\RemoveFirst\*.txt'
$PathDump = 'C:\RemoveFirst\DumpARoo'
$Output   = 'C:\RemoveFirst\TestingFile.txt'

if(!(Test-Path -Path $PathDump)) {
    # create the folder if it does not yet exist
    $null = New-Item -ItemType Directory $PathDump
}
# move all *.txt items from 'C:\RemoveFirst\txt' to 'C:\RemoveFirst\DumpARoo'
# EXCEPT the output file itself
$moveThese =(Get-ChildItem -Path $Path -Filter '*.txt' -File).FullName | Where-Object { $_ -ne $Output }
Move-Item -Path $moveThese -Destination $PathDump # move (not copy) files into new directory to concat
Get-ChildItem -Path $PathDump -Filter '*.txt' -File | ForEach-Object {
    $_ | Get-Content | 
    Select-Object -Skip 1 | 
    Select-Object -SkipLast 1 |
    Add-Content -Path $OutPut
}

in order to move all txt files, except the one used for output