Regular expression for converting file string in PowerShell

Question

I have some zip files that all ends with:

Some nameA 1.0.0 rev. 110706.zip
Some nameB 1.0.0 rev. 110806.zip
Some name moreC 1.0 rev. 120904.zip
name 1.1 rev. 130804.zip

In PowerShell I would like to read those file names, create a new text file that contains only the versions but converted into the following format:

1.0.0.110706
1.0.0.110806
1.0.120904
1.1.130804

For now I do:

$files = Get-ChildItem "."  -Filter *.zip
for ($i=0; $i -lt $files.Count; $i++) {
    $FileName = $files[$i].Name
    $strReplace = [regex]::replace($FileName, " rev. ", ".")
    $start = $strReplace.LastIndexOf(" ")
    $end = $strReplace.LastIndexOf(".")
    $length = $end-$start

    $temp = $strReplace.Substring($start+1,$length-1)
    $temp
}

I have looked at: Use Powershell to replace subsection of regex result

To see if I can get a more compact version of this. Any suggestions given the above pattern?

Mathias R. Jessen · Accepted Answer · 2015-10-18 22:49:03Z

3

You could do a single -replace operation:

$FileName -replace '^\D+([\.\d]+)\srev\.\s(\d+)','$1.$2'

Breakdown:

^\D+       # 1 or more non-digits - matches ie. "Some nameA "  
([\.\d]+)  # 1 or more dots or digits, capture group - matches the version  
\srev\.\s  # 1 whitespace, the characters r, e and v, a dot and another whitespace  
(\d+)      # 1 or more digits, capture group - matches the revision number

In the second argument, we refer to the two capture groups with $1 and $2

You can pipe Get-ChildItem to ForEach-Object instead of using a for loop and indexing into $files:

Get-ChildItem . *.zip |ForEach-Object {
    $_.BaseName -replace '^\D+([\.\d]+)\srev\.\s(\d+)','$1.$2'
} | Out-File output.txt

edited Oct 18, 2015 at 22:49

answered Oct 18, 2015 at 20:57

Mathias R. Jessen

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4 Comments

u123 Over a year ago

hm cool but I still get the zip extension in result if I use the above, e.g. '1.0.0.150806.zip'

Mathias R. Jessen Over a year ago

@u123 sorry, thought you were trying to rename the files, just updated the answer to output the versions to a file instead

Keith Hill Over a year ago

You don't need to escape the . in [\.\d].

Mathias R. Jessen Over a year ago

@KeithHill true, but it makes it more obvious what he's trying to do

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