fix(uploadservice): preserve path for S3 upload with subdirs

`path.basename(filename)` strips as well the relative path from the filename, which breaks the ability to upload to S3 buckets with subdirectories included via slashes in the filename input.

This commit takes the logic from the base upload_file method to preserve the correct relative path if subdirectories shall be included, stripping away possible leading slashes, which was another issue that could appear.

Additionally, the leading slashes are now `lstrip`ped from the path, instead of using a while loop, simplifying code, and the debug message now shows the path the file is uploaded to.

Signed-off-by: MichaIng <micha@dietpi.com>

Author: MichaIng

motioneye/uploadservices.py

@@ -61,10 +61,7 @@ def upload_file(self, target_dir, filename, camera_name):
         if target_dir:
             target_dir = os.path.realpath(target_dir)
             rel_filename = os.path.realpath(filename)
-            rel_filename = rel_filename[len(target_dir) :]
-
-            while rel_filename.startswith('/'):
-                rel_filename = rel_filename[1:]
+            rel_filename = rel_filename[len(target_dir) :].lstrip('/')
 
             self.debug(f'uploading file "{target_dir}/{rel_filename}" to {self}')
 
@@ -1221,10 +1218,20 @@ def upload_file(self, target_dir, filename, camera_name):
             aws_secret_access_key=self._secret_key,
         )
 
+        if target_dir:
+            target_dir = os.path.realpath(target_dir)
+            rel_filename = os.path.realpath(filename)
+            rel_filename = rel_filename[len(target_dir) :].lstrip('/')
+
+        else:
+            rel_filename = os.path.basename(filename)
+
         # Uploads the given file using a managed uploader, which will split up
         # large files automatically and upload parts in parallel.
-        self.debug(f'uploading file "{filename}" to S3 bucket "{self._bucket}"')
-        s3.upload_file(filename, self._bucket, os.path.basename(filename))
+        self.debug(
+            f'uploading file "{filename}" to S3 bucket "{self._bucket}" path "{rel_filename}"'
+        )
+        s3.upload_file(filename, self._bucket, rel_filename)
 
     def test_access(self):
         try: