Windows – Remove bracket characters in filenames using Powershell

Unless you use -LiteralPath, square brackets cause real problems with character escaping. The problem comes from PowerShell de-escaping the strings multiple times internally and using special characters for pattern matching. Getting PowerShell to correctly recognize a literal square bracket in a path string is complex.

If you're using single-quote strings, a bracket needs two backticks to escape it. If you're using double-quote strings, a bracket needs four backticks to escape it.

If you're looking for a file named MyFile[1].txt, you need to use either:

'MyFile``[1``].txt'

or:

"MyFile````[1````].txt"

Yes, it's a pain. To see why it does this, you have to know what's going on. It's easy to do that by working backwards.

Let's say you want to get a file literally named [ab].txt.

The wildcard pattern matching that Get-ChildItem does means that if it gets [ab].txt as the path, then it will look for files named a.txt and b.txt. So, if we want to match a literal [ab].txt, we have to escape our brackets with the escape character: the backtick. That gives us this as the actual string of characters we want Get-ChildItem to use for the filespec:

`[ab`].txt

However, we have to pass this filespec as a string. That means Get-ChildItem will escape those backticks, but that's not what we want! We want literal backticks. So, we escape the backticks with backticks in our string to make sure Get-ChildItem uses the right filespec:

'``[ab``].txt'

If we want to use double-quoted strings, then we have to escape each backtick again, as the double-quoted string will de-escape the string. And that's how you end up with this:

"````[ab````].txt"

This is why so many PowerShell functions that take filespecs have the -LiteralPath option.

A lot of it will depend on how consistent your data will be. There will be a multitude of methods. You could do it with a split:

$test = '24.;128.32.;128,6;0.128.68.;128,1.;0.76.;;;;;0,1.;0.4.;0,2.;0.5.;0,4.;;;0.76.;128,;;;;;;;;172.30.1.1,172.30.1.228,'
$partIWant = $test.Split(';')[$($test.Split(';').Count-1)]
$partIWant

You could use a regex as well:

$partIWant = $test -replace '.+(\d+.\d+\.\d+\.\d+\.\d+,\d+\.\d+\.\d+\.\d+,)', '$1'
$partIWant

You will need to pick the method that you want to use and craft it to the specific data that you will be working with.