Export Log file to new CSV file in Powershell

Question

I feel like I'm going about this the wrong way, but I've gotten myself half way there... I have a log file called DataSet.log that is formatted as:

First Dataset;24226382;2020-10-01 00:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;2;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
First Dataset;24421469;2020-10-01 01:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;4;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
First Dataset;24667838;2020-10-01 02:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;6;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0
First Dataset;24667839;2020-10-01 02:00;Second.Data.Set 1.0;0;Third.DataSet 1.0;1;Fourth.DataSet 1.0;0;Fifth.Dataset 1.0;0

I'm trying to convert this to a new CSV file and get it to display only the third and seventh columns. I've gotten it to display in PowerShell properly with:

Import-Csv .\DataSet.log -Header A,B,C,D,E,F,G,H,I,J,K -Delimiter ';' | Format-Table C,G

I've tried to export it as:

$testPush = Import-Csv .\DataSet.log -Header A,B,C,D,E,F,G,H,I,J,K -Delimiter ';' | Format-Table C,G

$testPush | Out-File .\test.csv

And it does create a new csv file, but it's only displaying the third column with a header of "C G." Also, that header is in A2, where A1 is blank, and A3 is populated with "- -" before the datetime from the third column populates the remaining rows... What the heck am I doing wrong?

Answer 1

Try this:

 $testPush = Import-Csv .\DataSet.log -Header A,B,C,D,E,F,G,H,I,J,K -Delimiter ';' | Select-Object C,G
 $testPush | Export-CSv -Path .\test.csv -NoTypeInformation