Gracefully print out debug messages even outside app context

The flask.current_app object depends on the application context
existing. In the case that no such context existing, then we still log
out information, gracefully with this commit.

2 files changed, 12 insertions, 4 deletions

diff --git a/gn3/commands.py b/gn3/commands.py
index 7a9ba67..d5215f8 100644
--- a/gn3/commands.py
+++ b/gn3/commands.py
@@ -3,6 +3,7 @@ commands"""
 import sys
 import json
 import pickle
+import logging
 import tempfile
 import subprocess
 
@@ -161,7 +162,8 @@ def run_cmd(cmd: str, success_codes: Tuple = (0,), env: Optional[str] = None) ->
     out = str(results.stdout, 'utf-8')
     if results.returncode not in success_codes:  # Error!
         out = str(results.stderr, 'utf-8')
-        current_app.logger.debug(out)
+        (# We do not always run this within an app context
+            current_app.logger.debug if current_app else logging.debug)(out)
     return {"code": results.returncode, "output": out}
 
 def run_async_cmd(
diff --git a/gn3/debug.py b/gn3/debug.py
index acc2402..3c88bd3 100644
--- a/gn3/debug.py
+++ b/gn3/debug.py
@@ -2,11 +2,17 @@
 import logging
 from flask import current_app
 
-logger = logging.getLogger(__name__)
+__this_module_name__ == __name__
+
+def getLogger():
+    return (
+        logging.getLogger(__name__)
+        if not bool(current_app)
+        else current_app.logger)
 
 def __pk__(*args):
     value = args[-1]
     title_vals = " => ".join(args[0:-1])
-    current_app.logger.setLevel(logging.DEBUG) # Force debug level since we assume we are using it!
-    current_app.logger.debug("%s: %s", title_vals, value)
+    logger = getLogger()
+    logger.debug("%s: %s", title_vals, value)
     return value