Etract filename in a variable

Will allow file testing.

boto-updater.py

@@ -2,7 +2,7 @@ import boto3
 import os.path
 
 # Arguments
-folder_to_sync = './test/data/'
+folder_to_sync = './test/data'
 bucket_target = 'my-bucket'
 
 # Main
@@ -10,15 +10,22 @@ s3 = boto3.client('s3', use_ssl=False, endpoint_url="http://172.17.0.2:9000", aw
 
 # Get file list
 # Source : https://stackoverflow.com/questions/3207219/how-do-i-list-all-files-of-a-directory
-listOfFiles = [f for f in os.listdir(folder_to_sync) if os.path.isfile(folder_to_sync + f)]
+file_list = os.listdir(folder_to_sync)
 print("Local : ", folder_to_sync)
-print(listOfFiles)
+print(file_list)
 
 # Upload
 print("Uploading ...")
-for f in listOfFiles:
-    print(f)
-    res = s3.upload_file(folder_to_sync + f, bucket_target, f)
+for file_name in file_list:
+    print(file_name, end=' ')
+
+    file_path = folder_to_sync + '/' + file_name
+
+    if os.path.isfile(file_path):
+        res = s3.upload_file(file_path, bucket_target, file_name)
+        print('ok')
+    else:
+        print('(Skipped)')
 
 print("Done")