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Project update not working with 1.49 anymore

MarianMarian
in Bug Reports
Hi, we are using a script that auto uploads new projects to our 4 servers. Today I updated one of them to 1.49 for testing and the upload does not work anymore. It is done via POST call, uploading a zip.
Being logged in with a browser works just fine, but not with the script anymore while the upload on the other 3 servers work like a charm (1.46 works).

I compared both post calls with firefox dev tools and could not find any difference, I am a bit out of ideas.

What could the issue be?


This is the python code we are using.

 def uploadToServer(self):
       
        print("{}: Uploading project to server...".format(self.server_name))
       
        upload_url = 'https://{}:{}/project/{}/import/1'.format(server, self.server_port, self.server_uuid)
        upload_post_data = {'a': 'uploadZIP', 'name': os.path.basename(self.zip_file)}

        with open(self.zip_file, 'rb') as upload_file: # open file in read binary mode
            post_result = requests.post(upload_url, data = upload_post_data, files={'': upload_file}, headers= {'x-api-key': self.server_api_key})
            post_result = json.loads(post_result.text)
           
            if "ok" in post_result and post_result['ok'] == True:
                print("{}: Upload successful.".format(self.server_name))

            else:
                print("{}: Upload failed: {}".format(self.server_name, post_result['error'] if "error" in post_result else post_result))
                print(post_result)
                return False

        return True

Comments

  • RepetierRepetier
    There had been a change in upload handling. The file being uploaded should always have parameter "filename" the one containing the file, but in past only one file per upload was supported, so first was always taken. Now some functions need multiple uploads so name matters.

    Can't say how you tell python to assign the file that parameter name, but since I don't see you using it anywere I guess that is why it fails.
  • MarianMarian
    Thanks, that was a quick fix. Weird thing is that I tried that before and it didn't work. But I may have addes another error then. For anyone interested:

        def uploadToServer(self):
           
            print("{}: Uploading project to server...".format(self.server_name))
           
            upload_url = 'https://{}:{}/project/{}/import/1'.format(server, self.server_port, self.server_uuid)
            upload_post_data = {'a': 'uploadZIP', 'name': os.path.basename(self.zip_file)}

            with open(self.zip_file, 'rb') as upload_file: # open file in read binary mode
                post_result = requests.post(upload_url, data = upload_post_data, files={'filename': upload_file}, headers= {'x-api-key': self.server_api_key})
                post_result = json.loads(post_result.text)
               
                if "ok" in post_result and post_result['ok'] == True:
                    print("{}: Upload successful.".format(self.server_name))

                else:
                    print("{}: Upload failed: {}".format(self.server_name, post_result['error'] if "error" in post_result else post_result))
                    return False

            return True


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