正则表达式 - 替换在Powershell中

Question

I am trying to read a .sln file and extract the strings that contain the path to the .csproj within my solution.

The lines that contain the information that I am looking for look like this:

Project("{FAE04EC0-301F-11D3-BF4B-00C04F79EFBC}") = "Project", "Project\Project.csproj", "{0DB516E6-4358-499D-BFBF-408F50A44E14}"

So, this is what I am trying:

$projectsInFile = Select-String "$slnFile" -pattern '^Project'
$csprojectsNames = $projectsInFile -replace ".+= `"(\S*) `""

Now, $csprojectsName contain the information that I am looking for, but also the rest of the string.

Just like this:

Project\Project.csproj", "{0DB516E6-4358-499D-BFBF-408F50A44E14}"

What is the best way to retrieve the name of the .csproj file without needing to manually cut the rest of the string?

Thank you

Answer 1

What you can do is capture the entire string and use a capture group in your replacement string thereby dropping the unneeded parts.

$csprojectsNames = $projectsInFile -replace '.+= "(\S*) "(.*?)",.*"','$2'

The second capture group is the data inbetween the quotes that follow = "Project", "....." . Since it is the second capture group we replace the entire with that group '$2' . Using single quotes ensure that PowerShell does not try to expand a variable.

Better approach

You might just be able to use [^"]*?\\.csproj in select-string directly without having to do a secondary parse. That will match everything before .csproj that is not a quote so it wont gooble up too much.

Answer 2

You can use a group to capture the file path and then use the value of the group in as the replacement value. For instance:

$csprojectsNames = $projectsInFile -replace 'Project\(.*?\) = "Project", "(.*?)"', '$1'