create a new variable for each loop in a foreach-loop

Question

How can I put $org into an array together with $count?

Like this example array:

$myArray = @{
  1="SampleOrg";
  2="AnotherSampleOrg"
}

Another example:

$myArray = @{
  $count="$org";
  $count="$org"
}

Example foreach:

$count=0;get-organization | foreach {$count++; $org = $_.Name.ToString();write-host $count  -nonewline;write-host " $org"}
$answer = read-host "Select 1-$count"

The above will display:

1 SampleOrg
2 AnotherSampleOrg

Select 1-2:

What I would like to do afterwards is to put the array to use in a switch.

Example:

switch ($answer)
   {
     1 {$org=myArray[1]} #<-- or whatever that corresponds to "SampleOrg"
     2 {$org=myArray[2]} #<-- or whatever that corresponds to "AnotherSampleOrg"
   }

I'm not sure if I understand you correctly, but IMO you just need to add a $myArray.Add($count, $org) to your foreach-loop. EDIT: And you have to initialize your array somewhere before the loop: $myArray = @{} — Ocaso Protal
– Ocaso Protal, Commented Aug 8, 2011 at 12:49
Your solution was working superb! $myArray = @{};$count=0;get-organization | foreach {$count++; $org = $_.Name.ToString();write-host $count -nonewline;write-host " $org";$myArray.Add($count, $org)} — Niklas J. MacDowall
– Niklas J. MacDowall, Commented Aug 8, 2011 at 13:32

Ocaso Protal · Accepted Answer · 2011-08-08 14:36:37Z

4

You have to initialize your hashtable somewhere before the loop:

$myArray = @{}

and add a

$myArray.Add($count, $org)

to your foreach-loop.

EDIT: For the discussion about hastable/array see the whole thread ;) I just kept the name of the variable from the original posting

edited Aug 8, 2011 at 14:36

answered Aug 8, 2011 at 13:34

Ocaso Protal

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4 Comments

Niklas J. MacDowall Over a year ago

End-result:

$myArray = @{};$count=0;get-organization | foreach {$count++; $org = $_.Name.ToString();write-host $count  -nonewline;write-host " $org";$myArray.Add($count, $org)}

Scott Weinstein Over a year ago

$myArray = @{} is not an array

Niklas J. MacDowall Over a year ago

Hashtables is a type of array. (Associative arrays)

JasonMArcher Over a year ago

Hashtables are a type of collection, but not a type of Array.

Scott Weinstein · Accepted Answer · 2011-08-08 17:19:03Z

4

Looks like you're confusing arrays and Hashtables. Arrays are ordered, and indexed by an numeric value. Hashtables are associative, and indexed by any value that has equality defined.

This is array syntax

$arr = @(1,2,3)

and this is Hashtable syntax

$ht = @{red=1;blue=2;}

For your question, the following will work

$orgs = @(get-organization | % { $_.Name })

this will create a 0 based array, mapping int -> OrgName, so

$orgs[$answer]

will get the correct name. Or if you're using 1 based indexing

$orgs[$answer-1]

Note, I removed the switch, as there's no reason for it.

edited Aug 8, 2011 at 17:19

answered Aug 8, 2011 at 12:55

Scott Weinstein

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5 Comments

JNK Over a year ago

won't that create an array with a nested hash table instead of the opposite?

Scott Weinstein Over a year ago

@JNK - yes, exactly. As the index seems to be an integer, an array is the most apprpriate data structure

JNK Over a year ago

gotcha. The $answer index threw me off - $answer is an int not a key! It looks like he wants to use a 1-based index too, so he should look out for the 0.

Niklas J. MacDowall Over a year ago

In the posted solution there is no prefix number, i.e 1 SampleOrg Also, when I get the value of $orgs[1] I get Name and Value and not a single variable.

Niklas J. MacDowall Over a year ago

If the $answer=1 which corresponds to AnotherSampleOrg Then in the switch case: 1 {$org=orgs[$answer]} this will set $orgs to: Name Value ---- ----- SampleOrg AnotherSampleOrg But I want to be set to only AnotherSampleOrg.

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