merge several .csv files into one array

Question

I'm trying to load several .csv files (with the same columns) into the same array from different directories.

$csv1 = Import-Csv "PATH1"
$csv1 = Import-Csv "PATH2"
$csv1 | Export-Csv C:\test.csv

This just outputs the last .csv loaded, what would be the best way to do this?

Hey this should to the trick if i can stop importing the headers of the 2nd .csv file? — Bunion
– Bunion, Commented Apr 27, 2017 at 19:21
found out i could use the += on the 2nd csv file to add it to the array. thanks — Bunion
– Bunion, Commented Apr 27, 2017 at 19:30
@TessellatingHeckler - Get-Content ...| Set-Content ... might "work", except that if it's a CSV in standard format (first line is headers) he'll end up with a record in the middle that duplicates his headers, e.g., Name="Name" Address="Address" etc. His solution using += is better. — Jeff Zeitlin
– Jeff Zeitlin, Commented Apr 27, 2017 at 19:43
hey @JeffZeitlin if i wanted to add an additional colum per line stating what path the file came from how would i do this? — Bunion
– Bunion, Commented Apr 27, 2017 at 19:45
@JeffZeitlin No need to make a new object when you could use -PassThru on Add-Member or Select-Object to make a calculated property. @Bunion Here is a one liner that adds that property: ($(Import-Csv "PATH1" | Add-Member -Name "SourcePath" -Value "PATH1" -MemberType NoteProperty -PassThru) + $(Import-Csv "PATH2" | Add-Member -Name "SourcePath" -Value "PATH2" -MemberType NoteProperty -PassThru)) | Export-CSV C:\test.csv — BenH
– BenH, Commented Apr 27, 2017 at 21:42

Ansgar Wiechers · Accepted Answer · 2017-04-28 12:29:57Z

2

When in doubt, read the documentation. The Import-Csv cmdlet accepts an array of path strings as input, so all you need to do (assuming that all your CSVs have the same fields) is something like this:

$src = 'C:\path\to\input1.csv', 'C:\path\to\input2.csv', ...
$dst = 'C:\path\to\output.csv'

Import-Csv $src | Export-Csv $dst -NoType

If you want an additional column with the path of the source file you need some additional steps, though:

$src | ForEach-Object {
  $path = $_
  Import-Csv $path | Select-Object *,@{n='Path';e={$path}}
} | Export-Csv $dst -NoType

edited Apr 28, 2017 at 12:29

answered Apr 28, 2017 at 12:24

Ansgar Wiechers

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