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How to add a column to default output of export-csv

Posted on 2012-04-10
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Last Modified: 2012-04-12
Is there a way to do add a second column to the CSV in the following code snip?  I'm putting this in a loop that goes through multiple files.  I need another column so I know what file I need to look at when I get a get diff, and I can make it into a table in Excel for easy viewing.

I need the output to be one file, and am going to add Dmitri's funtion from http://dmitrysotnikov.wordpress.com/2010/01/19/export-csv-append/.  

I'm comparing one master set of source files against files from different servers, which is why I need to path as a column.

compare-object -referenceobject $(get-content $sourceFile) -differenceobject $(get-content $serverFile) | export-csv -Path d:\some.csv -notype -append
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Question by:1974Widget
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3 Comments
 
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Accepted Solution

by:
Qlemo earned 2000 total points
ID: 37827331
Well, I would add two columns, to have both reference and target file names:
compare-object (get-content $sourceFile) (get-content $serverFile) |
  Add-Member NoteProperty SourceFile $sourceFile -PassThru|
  Add-Member NoteProperty TargetFile $serverFile -PassThru |
  export-csv -Path d:\some.csv -notype -append

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Expert Comment

by:slidingfox
ID: 37827380
Give the code below a try.


$comparison = compare-object -referenceobject $(get-content $sourceFile) -differenceobject $(get-content $serverFile)

$comparison | % {

    $obj = New-Object -type PSObject
    $obj | Add-Member -membertype NoteProperty -name InputObject -Value $_.InputObject
    $obj | Add-Member -membertype NoteProperty -name SideIndicator -Value $_.SideIndicator
    $obj | Add-Member -membertype NoteProperty -name Path -Value $source

    return $obj
   
}

$obj | export-csv -Path d:\some.csv -notype -append
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Author Closing Comment

by:1974Widget
ID: 37840407
Just what I asked for!  Thanks!
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