Past only select fields to out-gridview

Question

How can I past only a selection of fields from an object to show in out-gridview, but still keep the whole object as a result.

For example, I retrieve an object that gives me:

$listcreds = get-listofcredentials

Id                    : 03c0e0c0-0951-4698-9ba9-a70508b5467f
IsLocalProtect        : True
Name                  : vsphere.local\Administrator
CurrentUser           : False
UserName              : vsphere.local\Administrator
UserNameAtNotation    : Administrator@vsphere.local
UserNameSlashNotation : vsphere.local\Administrator
UserNameOnly          : Administrator
DomainName            : vsphere.local
EncryptedPassword     : Veeam.Backup.Common.CCodedPassword

In the gridview I want to only see Name and ID. After the user selects the row desired, I would like to have the result as the same type of object again.

When I use select-object,

$selectedcred = $listofcredentials | select-object Name, Id | out-gridview -passthru

I get the result back but I would now have to search the original object again to get the other properties of that row.

What better way to do this?

Sometimes there no easy shortcut. Sometimes you have to go long way. ;-) — Olaf, Sep 23 '21 at 10:15
Normally, `Out-GridView` does not return any objects. When using the `PassThru` parameter, the objects representing the **selected rows** are returned to the pipeline. — JosefZ, Sep 23 '21 at 12:46
I understand, but would like to only show certain columns of the object presented to out-gridview — Gabrie, Sep 23 '21 at 13:01
`$cred = $listofcredentials | Where-Object { $_.Id -eq $selectedcred.Id }` — Theo, Sep 23 '21 at 14:59
Yes, that is how I do it now and where I have the luck that ID indeed is unique, but this isn't always the case. But I learned from the other replies, there is no other way. — Gabrie, Sep 24 '21 at 06:12

spicy.dll · Accepted Answer · 2021-09-23T17:07:29.203

1

You need to find the full object again in the list with Where-Object

This answer assumes the Id property is unique for every item in $listcreds

$listcreds = get-listofcredentials
$selectedcred = $listofcredentials | select-object Name, Id | out-gridview -passthru
$selectedcred = $listcreds | Where-Object {$_.Id -eq $selectedcred.Id}

I don't think there's a better solution here. If performance is a concern, you can convert the Where-Object into a foreach as below:

$listcreds = get-listofcredentials
$selectedcred = $listofcredentials | select-object Name, Id | out-gridview -passthru

foreach ($cred in $listcreds) {
    if ($cred.Id -eq $selectedcred.Id) {
        $selectedcred = $cred
        break
    }
}

However, the performance difference may be negligible or even negative.

edited Sep 23 '21 at 17:07

answered Sep 23 '21 at 16:45

spicy.dll

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OP did already state this in their post :) "I get the result back but I would now have to search the original object again to get the other properties of that row." – Daniel Sep 23 '21 at 16:53
1

@Daniel I didn't see that, whoops. However, I don't think there's a better way to do this. – spicy.dll Sep 23 '21 at 17:00
Thank you for all the replies. Too bad there is no other way. The example using ID is indeed easy since the ID is unique, but I don't always have the option to make it completely unique without exposing some fields that I'd rather not show. But that is just the way it is :-) – Gabrie Sep 24 '21 at 06:11

Past only select fields to out-gridview

1 Answers1