Powershell parenthesis in pattern

Asked Jul 14 '17 at 18:46

Active Jul 14 '17 at 19:06

Viewed 1,205 times

I need to search for a pattern in a file and change the value of the version. Here's the pattern :

[file: FileVersion("1.2.3.$VAR$")]

In my Powershell script, I have a variable with the Pattern inside :

$FileVersionPattern   = '[file: FileVersion("[0-9]*.[0-9]*.[0-9]*.$VAR$")]'

(Get-Content "$filePath")    -replace($FileVersionPattern, ('[file: FileVersion("' + $Major + '.' + $Minor + '.' + $Release + '.$VAR$")]'))   | Set-Content  "$filePath"

So the return should be ($Major = 5 $Minor = 5 $Release = 5) :

[file: FileVersion("5.5.5.$VAR$")]

It seems that the $VAR$ and the parenthesis is causing my script to fail.

edited Jul 14 '17 at 19:06

Soviut

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asked Jul 14 '17 at 18:46

Emile

What are you trying to do with `'[file: FileVersion("[0-9]*.[0-9]*.[0-9]*.$VAR$")]'` with `$VAR`? Is that a variable you're using in your pattern? If so using single quotes is going to make that a literal string. – Jason Snell Jul 14 '17 at 18:49
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Note that to match a `[`, you need to escape it, `\[`. – Wiktor Stribiżew Jul 14 '17 at 18:50
The $VAR$ is not useful for the Powershell script. I'll try with the \ – Emile Jul 14 '17 at 18:52
Same with a dot, to match a literal `.`, you need to escape it. And you need to escape `(` and `)` and `$`. The pattern should be `'\[file: FileVersion$"[0-9]*\.[0-9]*\.[0-9]*\.\$VAR\$"$]'` – Wiktor Stribiżew Jul 14 '17 at 18:58
`\[file:\ FileVersion$"[0-9]+\.[0-9]+\.[0-9]+\.\$VAR\$"$]` – Mathias R. Jessen Jul 14 '17 at 19:03
Spaces do not have to be escaped, but I guess `[regex]::Escape` is necessary here. Anyway, it is a dupe by all means. – Wiktor Stribiżew Jul 14 '17 at 19:04

Powershell parenthesis in pattern

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