Escape string in PowerShell Regex to a regular string - regex

Question

I have a string that contains some regex in, $(=?, Now this string is a password that I need to pass for some application that I'm building.
The code that I'm trying to use is:
$x = 'GIWs#K?hks2v&HKXb$S9=HK*AZN=i!(S?7'
[Regex]::Escape($x)
I've already tried the method with [Regex]::Escape() and it doesn't meet my requirements because I'm trying to insert the string as a password and it replacing the Regex with \.
Perhaps after I'm doing the [Regex]::Escape() should I try to delete the \ that I'm getting from the result of the command?
After running the [Regex]::Escape() this is the result I'm getting when printing the output:
GIWs#K\?hks2v&HKXb=HK\*AZN=i!\(S\?7
I'm trying to achieve the string without the ' \ ' characters but with the Escape function:
GIWs#K?hks2v&HKXb=HK*AZN=i!(S?7

Answer 1

This is not an answer because I don't know what the problem actually is. However, there are some inherent problems with your current attempt to handle the password string. If you use double quotes ("") around a string, PowerShell will interpolate the string inside the quotes. So any alphanumeric characters following an unescaped $, will be considered a variable name during interpolation. If that variable has no value, $variable will be replaced with a null value. You can see this behavior below:
"rt4837s$GT=\"
rt4837s=\
You should use single quotes ('') when quoting string literals (characters that will be left as is). PowerShell will not attempt interpolation when unescaped single quote pairs are encountered unless there is quote nesting. See below:
'rt4837s$GT=\'
rt4837s$GT=\
If you need a regex escaped string, the same rules apply from above and you should use single quotes.
[regex]::escape('dfaseryh$S9=r??*')
dfaseryh\$S9=r\?\?\*
If for any reason, you need to access that string later without the escape characters, then you can use the regex method Unescape().
[regex]::unescape('dfaseryh\$S9=r\?\?\*')
dfaseryh$S9=r??*
Practical Example of Using Regex Replace:
$OriginalString = 'Username = Anonymous; Password = <password>'
$regexReplace = [regex]::Escape('<password>')
$Password = 'GIWs#K?hks2v&HKXb$S9=HK*AZN=i!(S?7'
$OriginalString -replace $regexReplace,($Password -replace '\$','$$$$')
# Output
Username = Anonymous; Password = GIWs#K?hks2v&HKXb$S9=HK*AZN=i!(S?7
In the code above, $OriginalString is just an ordinary string that can be retrieved from any command or set by a coder. It contains a string <password> that we want to replace with a complex password string GIWs#K?hks2v&HKXb$S9=HK*AZN=i!(S?7.
$Password contains the complex password. Since we only care about replacing <password> and are choosing to use regex replace operator -replace, we need a valid regex expression for matching <password>. There is a caveat here though. When using -replace, the $ in the replacement string is used to prefix capture group names. So there can be cases where the literal string has an unintentional replacement. Capture group 0 is always there if there is a match. So $0 will always cause issues without proper escaping. It is probably best to just escape $ regardless.
For the regex match, we use [regex]::Escape('<password>') since we are unsure if <> are special in regex. If there are no special characters, then the string within the regex expression will not be modified. If it does contain special characters, they will be escaped with \.
As a result, <password> is replaced with GIWs#K?hks2v&HKXb$S9=HK*AZN=i!(S?7.
A recap of the syntax is as follows:
'String With Something You Want to Replace' -replace 'Regex Expression to Match String You Want to Replace','Replacement That Is a Literal String With Escaped $'