Merge branch 'fix-multiple-ref-prefix' into 'master'

Extract the longest-matching ref from a commit path when multiple matches occur

### What does this MR do?

This MR extracts the longest-matching ref from a commit path. In cases when there are multiple refs that prefix the path, the ref name is ambiguous. Using the heuristic that the longest-matching ref seems like a sensible default.

### Why was this MR needed?

Suppose there is a branch named `release/app` and a tag named `release/app/v1.0.0`. Suppose `README.md` exists in the root directory.

Let's suppose the path passed in is `release/app/v1.0.0/README.md`. There are two possible ways to interpret the ref and path:

1. ref = `release/app`, path = `v1.0.0/README.md`
2. ref = `release/app/v1.0.0`, path = `README.md`

The crux of the issue is that there is ambiguity which one is correct; both could be real possibilities. In the current implementation of `extract_ref`, GitLab gets confused and tries neither: it uses ref = `release` and path = `app/v1.0.0/README.md`. Since the file does not exist, it returns 404.

### What are the relevant issue numbers?

Closes #1839

See merge request !859

1 files changed, 6 insertions, 2 deletions

diff --git a/lib/extracts_path.rb b/lib/extracts_path.rb
index 6e4ed01e079..3f420553d42 100644
--- a/lib/extracts_path.rb
+++ b/lib/extracts_path.rb
@@ -55,12 +55,16 @@ module ExtractsPath
       valid_refs = @project.repository.ref_names
       valid_refs.select! { |v| id.start_with?("#{v}/") }
 
-      if valid_refs.length != 1
+      if valid_refs.length == 0
         # No exact ref match, so just try our best
         pair = id.match(/([^\/]+)(.*)/).captures
       else
+        # There is a distinct possibility that multiple refs prefix the ID.
+        # Use the longest match to maximize the chance that we have the
+        # right ref.
+        best_match = valid_refs.max_by(&:length)
         # Partition the string into the ref and the path, ignoring the empty first value
-        pair = id.partition(valid_refs.first)[1..-1]
+        pair = id.partition(best_match)[1..-1]
       end
     end