Export-CSV with the path as a variable
export-to-csv, powershell, variables
Solution
Sorry, but it won't work for any path. Reason is simple - you read/write to the same file. It would work, if you would read into variable, and work on data in memory.
IMO having a path (elements) stored in variable have nothing to do with it. BTW: you can join paths with Join-Path cmdlet... :)
Problem
I have a script that basically writes down each file and folder on a remote share and dumps it to a csv. Everything works wonderfully (there is still a lot of work to be done on the script itself tho) except for the last part. ``` Import-Csv -Path ($savelocation + '\' + $filename) -Delimiter ',' | ForEach-Object { if ($audioarray -contains $_.Extension) { $_.MediaType = 'Audio' $_ } elseif ($videoarray -contains $_.Extension) { $_.MediaType = 'Video' $_ } elseif ($otherarray -contains $_.Extension) { $_.MediaType = 'Other' $_ } else { $_ } } | Export-Csv -Path ($savelocation + '\' + $filename) -Force -Delimiter ',' -NoTypeInformation ``` If I change the path to a static location something like `\\servername\share\testfolder` it works fine. But if I use the above or even if I join the `($savelocation + '\' + $filename)` into something like `$filename` it still just writes an empty file.