[Bug] "cwd" in docfx.json (in metadata src element) is broken in 2.71.1

[jskeet]

Describe the bug

This is a regression in 2.71.1. When a source is specified in the docfx.json metadata using "cwd", the specified file is not found.

To Reproduce
Steps to reproduce the behavior (see the zip file as an alternative to doing all of this by hand):

CwdIssue.zip

• Create a "Src" directory with a C# project (in its own directory) under it
• Create a docfx.json file which specifies the C# project relative to Src, using a cwd of "Src"
• Run "dotnet docfx metadata" and see the warnings below, with no output

Using .NET Core SDK 6.0.415
No files are found with glob pattern SampleLibrary/SampleLibrary.csproj, excluding <none>, under directory "C:\Users\jonskeet\Issues\docfx\CwdIssue"
warning: No .NET API project detected.
Creating output...
warning: No .NET API detected for .


Build succeeded with warning.

    2 warning(s)
    0 error(s)

Expected behavior

Appropriate metadata generated

Context (please complete the following information):

• OS: Windows
• Docfx version: 2.71.1

Additional context

The exact same example works fine in 2.71.0.

Looking through the file history, it looks like this commit was to blame. Changing

The commit was labelled as a refactor, but deleted a property - that's not refactoring. Additionally, this change was not mentioned in the release notes (and IMO should never have happened in a patch release even if it had been

As a side note: docfx is the dependency I dread upgrading more than any other due to things like this. I really appreciate the tool - but I'm frustrated by the amount of time I've spent over the years changing my build process to deal with breaking changes :(

[yufeih]

Apologies for the regression in 2.71.1. As a temporary solution, would it be ok to change cwd to src in your docfx.json? Thanks for your input and indeed we need a better change management.

[jskeet]

I'll need to find out all the places we generate docfx.json (it's a complex project) - but I'll probably try to do that.