Adding columns to multiple .csv files with powershell

Question

I have a number of .csv files that I need to modify to have 2 extra columns. 1 column is labelled 'site' and will be the same for all files in the same directory (0 in the example). The other is simply the original filename so that I can later merge these files but still preserve their original relationships.

I tried the following, but it just blanks out all the .csv files:

$files = Get-ChildItem ".\"

for ($i=0; $i -lt $files.Count; $i++) {
    $outfile = $files[$i].FullName + "out" 
    Import-Csv $files[$i].FullName | Select-Object *,@{Name='site';Expression={'0'}},@{Name='trace';Expression={$files[$i].FullName}} | Export-Csv $files[$i].FullName -NoTypeInformation
}

In your title, you say you need to add rows. In the post, you say you need to add columns. Please clarify - which is it? — alroc
– alroc, Commented May 6, 2014 at 17:48

Hunter Eidson · Accepted Answer · 2014-05-06 18:50:14Z

1

How about something along these lines:

$files = Get-ChildItem ".\" -filter "*.csv"

for ($i=0; $i -lt $files.Count; $i++) {
    $outfile = $files[$i].FullName + "out" 
    $csv = Import-Csv $files[$i].FullName 
    $newcsv = @()
    foreach ( $row in $csv ) {
        $row | Add-Member -MemberType NoteProperty -Name 'site' -Value '0'
        $row | Add-Member -MemberType NoteProperty -Name 'trace' -Value $files[$i].FullName
        $newcsv += $row
    }
    $newcsv | Export-Csv $files[$i].FullName -NoTypeInformation
}

answered May 6, 2014 at 18:50

Hunter Eidson

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4 Comments

Scott Over a year ago

Thanks! Is there a way to just append the filename and not the full path?

Hunter Eidson Over a year ago

Instead of .FullName you could use .Name in the Add-Member line

Scott Over a year ago

That makes sense. Thanks, I'm not very familiar with powershell.

Hunter Eidson Over a year ago

Using Get-Member and Format-List can be helpful in digging some of this stuff up... For instance, Get-ChildItem foo.csv | Get-Member will list off of the methods and properties of a directory listing. Get-ChildItem foo.csv | Format-List * will display the values of the properties of that directory listing. (All of that assumes you have a file named foo.csv in your current directory, of course) :)

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