Python脚本自动上传反馈至Web服务：无需手动转字典的实现问询

Python Script to Auto-Upload Feedback Files to Django REST API #

First, make sure you have the requests library installed—if not, run this command in your terminal:

pip install requests

Here's the complete script that follows your requirements:

import os
import requests

def upload_feedback():
    # Define paths and API endpoint
    feedback_directory = '/data/feedback'
    api_endpoint = 'http://35.193.233.100/feedback'
    
    # Loop through all items in the feedback directory
    for file_name in os.listdir(feedback_directory):
        full_file_path = os.path.join(feedback_directory, file_name)
        
        # Skip directories, only process text files
        if not os.path.isfile(full_file_path):
            continue
            
        try:
            # Read and parse the content of each feedback file
            with open(full_file_path, 'r') as file:
                # Clean up lines: remove whitespace and empty lines
                cleaned_lines = [line.strip() for line in file if line.strip()]
                
                # Validate we have all required fields
                if len(cleaned_lines) < 4:
                    print(f"Skipping {file_name}: Not enough content (needs title, name, date, feedback)")
                    continue
                    
                # Auto-generate the feedback dictionary
                feedback_dict = {
                    'title': cleaned_lines[0],
                    'name': cleaned_lines[1],
                    'date': cleaned_lines[2],
                    'feedback': '\n'.join(cleaned_lines[3:])  # Handles multi-line feedback
                }
            
            # Send the POST request to the Django API
            response = requests.post(api_endpoint, json=feedback_dict)
            
            # Print results for debugging
            if response.status_code == 201:
                print(f"✅ Successfully uploaded {file_name}! Status: {response.status_code}")
            else:
                print(f"❌ Failed to upload {file_name}. Status: {response.status_code} | Response: {response.text}")
                
        except Exception as error:
            print(f"⚠️ Error processing {file_name}: {str(error)}")

if __name__ == "__main__":
    upload_feedback()

Breakdown of How It Works: #

• File Listing: Uses os.listdir() to get all items in the target directory, then os.path.isfile() to ignore any subdirectories.
• Auto-Dictionary Creation: Reads each file line by line, maps the first three lines to title, name, and date, and joins any remaining lines into the feedback field (supports multi-line feedback text).
• API Upload: Uses requests.post() with the json parameter—this automatically converts the dictionary to JSON, so no manual conversion is needed.
• Error Handling: Includes checks for incomplete files and general exceptions, plus prints status codes and server responses to help troubleshoot issues.

How to Run: #

1. Save the script as feedback_uploader.py
2. Execute it with:

python feedback_uploader.py

You’ll get clear output for each file showing success or failure. A status code of 201 means the feedback was successfully created on the server.

内容的提问来源于stack exchange，提问作者karim reda