I'm looking over a block of code I've used (sourced from another question) and I haven't been able to figure out what the .
in .{process
represents in this snippet (comments removed):
Get-ItemProperty $path |
.{process{ if ($_.DisplayName -and $_.UninstallString) { $_ } }} |
Select-Object DisplayName, Publisher, InstallDate, DisplayVersion, HelpLink, UninstallString |
Sort-Object DisplayName
I know that %
is For-EachObject
and ?
is shorthand for Where
or Where-Object
, but the question remains:
What is .
shorthand for?
The pipeline character in Windows PowerShell is the vertical bar (also called the pipe: | ). On most U.S. keyboards, it is found on the key with the backslash.
$? Contains the execution status of the last command. It contains True if the last command succeeded and False if it failed. For cmdlets and advanced functions that are run at multiple stages in a pipeline, for example in both process and end blocks, calling this.
The Splatting Operator To create an array, we create a variable and assign the array. Arrays are noted by the "@" symbol.
The assignment by addition operator += either increments the value of a variable or appends the specified value to the existing value. The action depends on whether the variable has a numeric or string type and whether the variable contains a single value (a scalar) or multiple values (a collection).
.
is the dot sourcing operator, which runs a script in the current scope rather than a new scope like call operator (i.e. &
).
That second segment invokes a script block and in that script block defines an advanced function. The advanced function iterates each item in the pipeline and selectively passes it along.
This is not really an idiomatic use. What this script is trying to achieve could be done in a simpler, more readable way by using Where-Object
(often shortened to where
or ?
):
Get-ItemProperty $path | where { $_.DisplayName -and $_.UninstallString }
.
is the dot source operator. I've never seen it used quite this way, but it's identical to using &
(the call operator) in this context.
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