PowerShell: -replace, regex and ($) dollar sign woes

Vippy picture Vippy · Feb 22, 2012 · Viewed 8.4k times · Source

I am in the process of converting thousands of lines of batch code into PowerShell. I'm using regex to help with this process. The problem is part of the code is:

$`$2

When replaced it just shows $2 and doesn't expand out the variable. I've also used single quotes for the second portion of replace instead of escaping the variables, same result.

$origString = @'
IF /I "%OPERATINGSYSTEM:~0,6%"=="WIN864" SET CACHE_OS=WIN864
...many more lines of batch code
'@

$replacedString = $origString -replace "(IF /I `"%)(.+)(:.+%`"==`")(.+`")(.+)","if ( $`$2 -match `"^`$4 ) {`$5 }"

$replacedString

Answer

stej picture stej · Feb 22, 2012

You could try something like this:

$origString -replace "(IF /I `"%)(.+)(:.+%`"==`")(.+`")(.+)",'if ($$$2 -match "^$4" ) {$5 }'

Note the $$$2. This evaluates to $ and content of $2.


Some code to show you the differences. Try it yourself:

'abc' -replace 'a(\w)', '$1'
'abc' -replace 'a(\w)', "$1"  # "$1" is expanded before replace to ''
'abc' -replace 'a(\w)', '$$$1'
'abc' -replace 'a(\w)', "$$$1" #variable $$ and $1 is expanded before regex replace
                               #$$ and $1 don't exist, so they are expanded to ''

$$ = 'xyz'
$1 = '123'
'abc' -replace 'a(\w)', "$$$1`$1" #"$$$1" is expanded to 'xyz123', but `$1 is used in regex