Check stack is a path

stack_orchestrator/constants.py

@@ -0,0 +1,16 @@
+# Copyright © 2023 Vulcanize
+
+# This program is free software: you can redistribute it and/or modify
+# it under the terms of the GNU Affero General Public License as published by
+# the Free Software Foundation, either version 3 of the License, or
+# (at your option) any later version.
+
+# This program is distributed in the hope that it will be useful,
+# but WITHOUT ANY WARRANTY; without even the implied warranty of
+# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+# GNU Affero General Public License for more details.
+
+# You should have received a copy of the GNU Affero General Public License
+# along with this program.  If not, see <http:#www.gnu.org/licenses/>.
+
+stack_file_name = "stack.yml"

stack_orchestrator/repos/setup_repositories.py

@@ -25,6 +25,7 @@ import click
 import importlib.resources
 from pathlib import Path
 import yaml
+from stack_orchestrator.constants import stack_file_name
 from stack_orchestrator.util import include_exclude_check
 
 
@@ -238,9 +239,12 @@ def command(ctx, include, exclude, git_ssh, check_only, pull, branches, branches
 
     repos_in_scope = []
     if stack:
+        # Check if the stack argument is a path or a plain string.
+        stack_file_path = Path(stack).joinpath(stack_file_name)
+        if not stack_file_path.exists():
             # In order to be compatible with Python 3.8 we need to use this hack to get the path:
             # See: https://stackoverflow.com/questions/25389095/python-get-path-of-root-project-structure
-        stack_file_path = Path(__file__).absolute().parent.parent.joinpath("data", "stacks", stack, "stack.yml")
+            stack_file_path = Path(__file__).absolute().parent.parent.joinpath("data", "stacks", stack, stack_file_name)
         with stack_file_path:
             stack_config = yaml.safe_load(open(stack_file_path, "r"))
             # TODO: syntax check the input here