Powershell variable expansion when calling other programs

Question

I have a small problem trying to unzip a file using the 7za command-line utility in Powershell.

I set the $zip_source variable to the zip file's path and the $unzip_destination to the desired output folder.

However the command-line usage of 7za needs arguments specified like this:

7za x -y <zip_file> -o<output_directory>

So my current call looks like this:

& '7za' x -y "$zip_source" -o$unzip_destination

Due to the fact that there can be no space between -o and the destination it seems that PowerShell will not expand the $unzip_destination variable, whereas $zip_source is expanded.

Currently the program simply extracts all the files into the root of C:\ in a folder named $unzip_destination. Setting different types of quotes around the variable won't work neither:

-o"$unzip_destination" : still extracts to C:\$unzip_destination
-o'$unzip_destination' : still extracts to C:\$unzip_destination
-o $unzip_destination  : Error: Incorrect command line

Is there any way to force an expansion before running the command?

score 6 · Answer 1 · answered May 07 '12 at 14:49

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Try this:

& '7za' x -y "$zip_source" "-o$unzip_destination"

answered May 07 '12 at 14:49

jon Z

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Works as well - I guess I should have kept on experimenting with quotes. – BergmannF May 07 '12 at 14:52

score 3 · Accepted Answer · edited Nov 02 '18 at 13:40

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try like this:

-o $($unzip_destination)

^{Editor's note: This solution only works with a space after -o (in which case just -o $unzip_destination would do) - if you remove it, the command doesn't work as intended.

This approach is therefore not suitable for appending a variable value directly to an option name, as required by the OP.}

edited Nov 02 '18 at 13:40

mklement0

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answered May 07 '12 at 14:45

CB.

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There can be no space between -o and the unzip_destination – jon Z May 07 '12 at 14:51
When I leave out the space after the -o that works - thanks. Does `$()` start some kind of subshell in PowerShell (sorry I am new to Windows scripting from a Unix-background). – BergmannF May 07 '12 at 14:51
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$() is called a subexpression, and it causes everything inside the parenths to be returned as one value: $("Hello, " + "world!") returns literally: Hello, world! These are useful when wanting the output of some expression to be translated to a string in the middle of another string. – SpellingD May 07 '12 at 21:01
Removing the space does _not_ work - unless you enclose the whole argument in `"..."`, but then you can just use jonZ's simpler answer. Good hint, @SpellingD; worth adding that, in the interest of robustness, `$(...)` as part of a string should always be used with explicit double-quoting, because with an unquoted token it doesn't always work - as is the case here, because the token starts with `-`; see https://github.com/PowerShell/PowerShell/issues/6467 – mklement0 Nov 02 '18 at 13:52

score 0 · Answer 3 · edited Oct 10 '12 at 20:02

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This should work:

& '7za' x -y $zip_source -o${unzip_destination}

edited Oct 10 '12 at 20:02

Alfabravo

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answered Jul 30 '12 at 14:20

jjg.denis.robert

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Perhaps surprisingly, that would only work if the token didn't start with `-`. – mklement0 Nov 02 '18 at 13:30

Powershell variable expansion when calling other programs

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