Refine how we detect namespace packages

Previously we used a hand crafted approach to detect namespace packages, however we should rely on ``importlib`` to detect them for us.

Fix #12112

changelog/12112.bugfix.rst

@@ -0,0 +1 @@
+Improve namespace packages detection when :confval:`consider_namespace_packages` is enabled, covering more situations (like editable installs).

doc/en/reference/reference.rst

@@ -1279,8 +1279,7 @@ passed multiple times. The expected format is ``name=value``. For example::
    Controls if pytest should attempt to identify `namespace packages <https://packaging.python.org/en/latest/guides/packaging-namespace-packages>`__
    when collecting Python modules. Default is ``False``.
 
-   Set to ``True`` if you are testing namespace packages installed into a virtual environment and it is important for
+   Set to ``True`` if the package you are testing is part of a namespace package.
-   your packages to be imported using their full namespace package name.
 
    Only `native namespace packages <https://packaging.python.org/en/latest/guides/packaging-namespace-packages/#native-namespace-packages>`__
    are supported, with no plans to support `legacy namespace packages <https://packaging.python.org/en/latest/guides/packaging-namespace-packages/#legacy-namespace-packages>`__.

src/_pytest/pathlib.py

@@ -771,19 +771,11 @@ def resolve_pkg_root_and_module_name(
     pkg_path = resolve_package_path(path)
     if pkg_path is not None:
         pkg_root = pkg_path.parent
-        # https://packaging.python.org/en/latest/guides/packaging-namespace-packages/
         if consider_namespace_packages:
-            # Go upwards in the hierarchy, if we find a parent path included
+            for candidate in (pkg_root, *pkg_root.parents):
-            # in sys.path, it means the package found by resolve_package_path()
+                if _is_namespace_package(candidate):
-            # actually belongs to a namespace package.
-            for parent in pkg_root.parents:
-                # If any of the parent paths has a __init__.py, it means it is not
-                # a namespace package (see the docs linked above).
-                if (parent / "__init__.py").is_file():
-                    break
-                if str(parent) in sys.path:
                     # Point the pkg_root to the root of the namespace package.
-                    pkg_root = parent
+                    pkg_root = candidate.parent
                     break
 
         names = list(path.with_suffix("").relative_to(pkg_root).parts)
@@ -795,6 +787,35 @@ def resolve_pkg_root_and_module_name(
     raise CouldNotResolvePathError(f"Could not resolve for {path}")
 
 
+def _is_namespace_package(module_path: Path) -> bool:
+    # If the path has na __init__.py file, it means it is not
+    # a namespace package:.
+    # https://packaging.python.org/en/latest/guides/packaging-namespace-packages.
+    if (module_path / "__init__.py").is_file():
+        return False
+
+    module_name = module_path.name
+
+    # Empty module names break find_spec.
+    if not module_name:
+        return False
+
+    # Modules starting with "." indicate relative imports and break find_spec, and we are only attempting
+    # to find top-level namespace packages anyway.
+    if module_name.startswith("."):
+        return False
+
+    spec = importlib.util.find_spec(module_name)
+    if spec is not None and spec.submodule_search_locations:
+        # Found a spec, however make sure the module_path is in one of the search locations --
+        # this ensures common module name like "src" (which might be in sys.path under different locations)
+        # is only considered for the module_path we intend to.
+        # Make sure to compare Path(s) instead of strings, this normalizes them on Windows.
+        if module_path in [Path(x) for x in spec.submodule_search_locations]:
+            return True
+    return False
+
+
 class CouldNotResolvePathError(Exception):
     """Custom exception raised by resolve_pkg_root_and_module_name."""