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
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
try like this:
$replacedString = $origString -replace "(IF /I `"%)(.+)(:.+%`"==`")(.+`")(.+)","if ( $`$`$2 -match `"^`$4 ) {`$5 }"
来源:https://stackoverflow.com/questions/9401713/powershell-replace-regex-and-dollar-sign-woes