How to find hash key from an array, and update the array based on the key of the hash using powershell?

Mike 246 Reputation points

I really have been research and fell short on where to start on this.

My objective is to find the value in my hashtable from an existing array. and if it existis, wanted to updated the $array. really stuck and need some help to give me some example that I can learn upon.

populating my hash from csv.
$hash = @{ }
import-csv data.csv | foreach {hash.add($_tSamaccountname,gSamaccountname)
this imports a name mike and key michael

then i have an $array that contains mike,mark.
how do I loop thru this array looking if the value is in my $hash, then update the array from the key.

So my $array should now be michael,mark.

Windows Server PowerShell
Windows Server PowerShell
Windows Server: A family of Microsoft server operating systems that support enterprise-level management, data storage, applications, and communications.PowerShell: A family of Microsoft task automation and configuration management frameworks consisting of a command-line shell and associated scripting language.
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Accepted answer
  1. Rich Matheisen 45,906 Reputation points

    Replace line #1 in the code below to load the hash with your CSV.

    $x= [PSCustomObject]@{tSamaccountname='mike';gSamaccountname='michael'}, [PSCustomObject]@{tSamaccountname='mark';gSamaccountname='marcus'}
    [System.Collections.ArrayList]$a = @('mike','mark')
    $h = @{}
    $x |
            $h.($_.tSamaccountname) = $_.gSamaccountname
    for ($i =0; $i -lt $a.count; $i++){
            if ($h.ContainsKey($a[$i])){
                $a.item($i) = $h.($a[$i])

1 additional answer

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  1. Ian Xue (Shanghai Wicresoft Co., Ltd.) 34,271 Reputation points Microsoft Vendor


    If the tSamaccountname is mike and the gSamaccountname is michael, you can update the array like this

    $hash = @{ }  
    import-csv d:\temp\data.csv | foreach {  
    $array = 'mike','mark'  
    for($i=0; $i -lt $array.count; $i++){  
            $array[$i] = $hash.($array[$i])  

    Best Regards,
    Ian Xue


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