I\'ve seen a number of example scripts online that use this. Most recently, I saw it in a script on automating TFS:
[string] $fields = \"Title=$($taskTitle);
The syntax helps with evaluating the expression inside it.
$arr = @(1,2,3)
$msg1 = "$arr.length"
echo $msg1 # prints 1 2 3.length - .length was treated as part of the string
$msg2 = "$($arr.length)"
echo $msg2 # prints 3
You can read more at http://ss64.com/ps/syntax-operators.html
To complement Amith George's helpful answer with more background information:
From what I can tell,
$($taskTitle)seems to be equivalent to$taskTitle.
Indeed, in the context of "...", an expandable string (interpolating string):
You do NOT need $(...) with a simple variable reference such as $taskTitle or $env:HOME
${taskTitle} or ${env:HOME} - i.e., {...} around the identifier - so as to disambiguate the variable name from subsequent characters in the string.You DO need $(...) for anything else:
"count is: $($var.Count)" "path prefix: $($var + '/')""file names: $(Get-ChildItem *.txt | Select-Object -ExpandProperty Name)"In short:
$(...) inside "..." is needed for anything other than simple variable references and allows you to embed entire statements inside "..."; as usual, when the string is evaluated, the $(...) part is replaced with the (stringified) output from the embedded statement(s).
If you don't want to think about when $(...) is and isn't needed, you can choose to always use it (e.g., $($taskTitle)), but note that it's cumbersome to type and visually "noisy".
$($var) is not the same as that of $var / ${var}, namely if $var is a collection (implementing [System.Collections.IEnumerable]) that happens to contain only a single item - see PetSerAl's comments below.Unless the referenced variable's / embedded statement's value already is a string, it is stringified using the .NET .ToString() method, with the notable twist that types that support culture-sensitive stringification are stringified with the invariant culture, which, loosely speaking, is like US-English format; e.g., "$(1.2)" always yields 1.2, even in cultures where , is the decimal mark; see this answer of mine for more.
Documentation:
The official name for $(...) is the subexpression operator, as (tersely) documented in Get-Help about_Operators, though the explanation there doesn't discuss the operator's specific use in the context of expandable strings.
Conversely, Get-Help about_Quoting_Rules, which discusses string literals including expandable strings, shows examples of $(...) use only in the context of expandable strings.