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I'm outputting a couple of objects from a data table to a .csv file and wanted to change the object names.

This works fine:

$dataset.tables[0] | select-object System.ItemName, System.ItemPathDisplay | Export-Csv -Path D:\SEARCH_RESULT.csv -NoTypeInformation

However I'd like to change System.ItemName to SKU and have tried the following:

$dataset.tables[0] | select-object @{N='SKU';E={$_.System.ItemName}}, System.ItemPathDisplay | Export-Csv -Path D:\SEARCH_RESULT.csv -NoTypeInformation

This gives the right column heading but blank rows. So tried this which give rows but they all say SYSTEM.ITEMNAME:

$dataset.tables[0] | select-object @{N='SKU';E={$dataset.tables[0].Columns['SYSTEM.ITEMNAME']}}, System.ItemPathDisplay | Export-Csv -Path D:\SEARCH_RESULT.csv -NoTypeInformation

Clearly I'm not referencing the object correctly. Any help greatly appreciated.

ColinA
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1 Answers1

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Change $_.System.ItemName to $_.'System.ItemName'

Your property name is System.ItemName, and if a property name itself contains ., it must be quoted - otherwise, PowerShell interprets it as a nested property access

That is, $_.System.ItemName looks for a property on object $_ named System first, and then for an ItemName property on the first property's value; PowerShell defaults to $null when accessing non-existent properties.

Contrast this with the use of System.ItemName as an argument passed to the (positionally implied) -Property parameter of the Select-Object cmdlet, where the argument is always interpreted as a single property name (whether you quote it or not).
In other words: Select-Object doesn't directly support nested property access, but you can do it via a calculated property, as in your approach.

mklement0
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