Remove path from output

Question

Using the following Select-String command in PowerShell:

Select-String -Path E:\Documents\combined0.txt -Pattern "GET /ccsetup\.exe" -AllMatches > E:\Documents\combined3.txt

creates an output file with each line starting with the path and file name followed by a colon. For example:

E:\Documents\combined0.txt:255:255.255.255 - - [31/Dec/2014:04:15:16 -0800] "GET /ccsetup.exe HTTP/1.1" 301 451 "-" "Mozilla/5.0 (compatible; SearchmetricsBot; http://www.xxxx.com/en/xxxx-bot/)"

How do I get rid of the output file path name, output file name and colon in the results?

Answer 1

Select-String outputs an object from which you can pick off properties that you want. The Get-Member command will show you these object members if you pipe into it e.g.:

Select-String -Path E:\Documents\combined0.txt -Pattern "GET /ccsetup\.exe" -AllMatches  | 
    Get-Member

One of those properties is Line. So try it this way:

Select-String -Path E:\Documents\combined0.txt -Pattern "GET /ccsetup\.exe" -AllMatches | 
    Foreach {$_.Line} > E:\Documents\combined3.txt

Answer 2

As usual powershell returns things as objects, by default select-string returns several properties including LineNumber, Filename, etc; the one you want with the data in is just called "Line". So no need for anything fancy, just pipe it to "select line".

Eg:

Select-String "bla" filename.txt | select-object -ExpandProperty Line | out-file E:\bla_text.txt

Answer 3

If you're looking for (sub)strings rather than patterns, using the -like operator might be a better approach, performance-wise and with respect to ease-of-use.

$searchString = 'GET /ccsetup.exe'

Get-Content 'E:\Documents\combined0.txt' |
  ? { $_ -like "*$searchString*" } |
  Set-Content 'E:\Documents\combined3.txt'

If you do need pattern matches, you can easily replace the -like operator with the -match operator:

$pattern = 'GET /ccsetup\.exe'

Get-Content 'E:\Documents\combined0.txt' |
  ? { $_ -match $pattern } |
  Set-Content 'E:\Documents\combined3.txt'