Compare similar values from hashtable with loop in Powershell

Question

I have 2 hash tables :

[hashtable]$Localisation = @{
"Macdo" = "OU=France,OU=Paris";
"BurgerKing" = "OU=USA,OU=LA";
"Quick" = "OU=Japan,OU=Tokyo";
}

[hashtable]$Profil = @{
"Big Mac" = "Macdo";
"Whooper" = "BurgerKing";
"Burger" = "Quick, BurgerKing, Macdo";
"Fries" = "BurgerKing, Macdo";
"Coke" = "Quick, Macdo";
"HappyMeal" = "Macdo";
}

I need to get this kind of result:

"Big Mac" = "OU=France,OU=Paris"
"Whooper" = "OU=USA,OU=LA";
"Burger" = "OU=Japan,OU=Tokyo, OU=USA,OU=LA, OU=France,OU=Paris"
"Fries" = "OU=USA,OU=LA, OU=France,OU=Paris";
"Coke" = "OU=Japan,OU=Tokyo, OU=France,OU=Paris";
"HappyMeal" = "OU=France,OU=Paris";

or

Big Mac      =  OU=France,OU=Paris
Whooper      =  OU=USA,OU=LA
Burger       =  OU=Japan,OU=Tokyo, 
                OU=USA,OU=LA, 
                OU=France,OU=Paris
Fries        =  OU=USA,OU=LA,
                OU=France,OU=Paris
Coke         =  OU=Japan,OU=Tokyo, 
                OU=France,OU=Paris
HappyMeal    =  OU=France,OU=Paris

I tried :

$tempLoca = @()

foreach ($value in $Profil.values) {
    if($Localisation.Contains($value)) {
        $tempLoca = $Localisation.Contains($value),$Profil.key
    }
}

But I get :

$tempLoca 
OU=France,OU=Paris

With my code I have only the last value. I don't know if I need to put my values in array or in hashtable (because they are multiple similar values).

Do you have an idea? Thanks

Check out this answer. It's similar case so you should be able to adapt the solution to your needs. — Robert Dyjas
– Robert Dyjas, Commented Aug 13, 2018 at 16:30
Actually it might be the same case as it's also your question... — Robert Dyjas
– Robert Dyjas, Commented Aug 13, 2018 at 16:45

Theo · Accepted Answer · 2018-08-13 15:37:40Z

3

Try this:

[hashtable]$Localisation = @{
"Macdo" = "OU=France,OU=Paris";
"BurgerKing" = "OU=USA,OU=LA";
"Quick" = "OU=Japan,OU=Tokyo";
}

[hashtable]$Profil = @{
"Big Mac" = "Macdo";
"Whooper" = "BurgerKing";
"Burger" = "Quick, BurgerKing, Macdo";
"Fries" = "BurgerKing, Macdo";
"Coke" = "Quick, Macdo";
"HappyMeal" = "Macdo";
}


$tempLoca = @()
foreach ($key in $Profil.Keys) {
    $locals = ($Profil.$key -split ',') | ForEach-Object { $_.Trim() }
    $result = @()
    foreach ($item in $locals) {
        if($Localisation.ContainsKey($item)) {
            $result += $Localisation.$item
        }
    } 
    $tempLoca += '"{0}" = "{1}"' -f $key, ($result -join '; ')
}

$temploca

It will output

"Big Mac" = "OU=France,OU=Paris"
"HappyMeal" = "OU=France,OU=Paris"
"Burger" = "OU=Japan,OU=Tokyo; OU=USA,OU=LA; OU=France,OU=Paris"
"Whooper" = "OU=USA,OU=LA"
"Fries" = "OU=USA,OU=LA; OU=France,OU=Paris"
"Coke" = "OU=Japan,OU=Tokyo; OU=France,OU=Paris"

Note that I combine the OU values from the $Localisation hash with a semicolon ; to distinct them from the values themselves. If that is not what you want, just replace ($result -join '; ') with ($result -join ', ')

answered Aug 13, 2018 at 15:37

Theo

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1 Comment

P0werSh3ell Over a year ago

Thanks for the explication ! It's great !

user6811411 · Accepted Answer · 2018-08-13 17:57:57Z

3

An IMO more PowerShell like way, building a PSCustomObject and grouping it:

$ProfileLocalisation = ForEach ($key in $Profil.Keys) {
    ForEach ($local in ($Profil.$key -split ',').Trim() ) {
        [PSCustomObject]@{
            Profile = $key
            Localisation = $Localisation.$local
        }
    }
}
$ProfileLocalisation

Sample output:

Profile   Localisation
-------   ------------
Big Mac   OU=France,OU=Paris
HappyMeal OU=France,OU=Paris
Burger    OU=Japan,OU=Tokyo
Burger    OU=USA,OU=LA
Burger    OU=France,OU=Paris
Whooper   OU=USA,OU=LA
Fries     OU=USA,OU=LA
Fries     OU=France,OU=Paris
Coke      OU=Japan,OU=Tokyo
Coke      OU=France,OU=Paris

And grouped:

$ProfileLocalisation | Group-Object Profile | ForEach-Object {
    [PSCustomObject]@{
        Profile = $_.Name
        Localisations = ($_.Group.Localisation -join ';')
    }
}

Profile   Localisations
-------   -------------
Big Mac   OU=France,OU=Paris
HappyMeal OU=France,OU=Paris
Burger    OU=Japan,OU=Tokyo;OU=USA,OU=LA;OU=France,OU=Paris
Whooper   OU=USA,OU=LA
Fries     OU=USA,OU=LA;OU=France,OU=Paris
Coke      OU=Japan,OU=Tokyo;OU=France,OU=Paris

answered Aug 13, 2018 at 17:57

user6811411

1 Comment

P0werSh3ell Over a year ago

I can't use the first solution : [System.String[] doesn't contain method "Trim". I use Powershell 2.0 And for the second (Grouped). Nothing happen

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