Write-Host cannot parse variable's property

Question

It's quite weird that Write-Host doesn't seem to parse variable's property.

PS C:\Users\x> $v=@{item1="kkk";item2="aaa"}
PS C:\Users\x> $v.Keys|%{$v.Item($_)}
kkk
aaa
PS C:\Users\x> Write-Host "$v.Count elements"
System.Collections.Hashtable.Count elements
PS C:\Users\x> $v.Count
2
PS C:\Users\x> $v
Name                           Value
----                           -----
item1                          kkk
item2                          aaa

You could see, $v is a hashtable and

$v.Count

prints 2. But why does

Write-Host "$v.Count"

print out System.Collections.Hashtable.Count? This is not what I expected.

I don't have a solid answer for why it does it, but it's a behavior that I've noticed a lot - you can include a simple variable in a string, but not members with dot notation. If you want to have that work, you have to put it in a single variable first ($count = $v.count), use string formatting (Write-Host ("{0} elements" -f $v.Count)) or provide it as separate elements (Write-Host $v.Count "elements") — squid808
– squid808, Commented May 18, 2015 at 2:30

user4003407 · Accepted Answer · 2015-05-18 02:38:38Z

7

Text representation of $v is System.Collections.Hashtable ([string]$v or "$v"). And "$v.Count elements" means {text representation of $v}.Count elements not {text representation of $v.Count} elements, so for me it is expected that you get System.Collections.Hashtable.Count elements as result.

Write-Host is not responsible for expanding double quoted strings. It done by PowerShell before it invoke Write-Host.

"$v.Count" 
prints 2

For me it prints System.Collections.Hashtable.Count. 2 printed by $v.Count not by "$v.Count".

You should use "$($v.Count) elements" to get expected result.

answered May 18, 2015 at 2:38

user4003407

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1 Comment

Immanuel Kant Over a year ago

Thanks for your explanation! This extra $ solves my problem(interpretation)!

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