Python file upload with urllib3: Making POST Requests simplified

Python urllib3 : Exercise-17 with Solution

Write a Python program that makes a POST request to upload a file to a server using urllib3.

Sample Solution:

Python Code :

import urllib3
def upload_file(url, file_path):
        # Create a PoolManager for managing the connection pool
        http = urllib3.PoolManager()

        # Open the file in binary mode for reading
        with open(file_path, 'rb') as file:
            # Read the content of the file
            file_content = file.read()

            # Make a POST request with the file content
            response = http.request(
                headers={'Content-Type': 'application/octet-stream'}

            # Check if the request was successful (status code 200)
            if response.status == 200:
                print("File upload successful:")
                print(f"Error: Unable to upload file. Status Code: {response.status}")

    except Exception as e:
        print(f"Error: {e}")

# Define the URL for file upload
upload_url = 'https://www.example.com/upload'  # Replace with the actual upload URL

# Define the local file path to be uploaded
file_to_upload = 'test.txt'  # Replace with the actual file path

# Make the POST request to upload the file
upload_file(upload_url, file_to_upload)

Sample Output:

Error: Unable to upload file. Status Code: 404


Here's a brief explanation of the above Python urllib3 library code:

  • Importing the urllib3 Module:
    • Import the "urllib3" module, which is a powerful HTTP client for Python.
  • Define the upload_file Function:
    • Define a function named "upload_file()" that takes two parameters - 'url' (the URL for file upload) and 'file_path' (the local file path to be uploaded).
  • Create a PoolManager:
    • Creates a PoolManager object (http) to manage the connection pool for making HTTP requests.
  • Open File in Binary Mode:
    • Opens the file specified by 'file_path' in binary mode for reading ('rb').
  • Read File Content:
    • Reads the content of the file and stores it in the 'file_content' variable.
  • Make a POST Request:
    • Makes a POST request to the specified "url" with the file content as the request body.
    • Sets the "Content-Type" header to indicate an octet-stream (binary) file.
  • Check Response Status:
    • Check if the response status code is 200.
    • If successful, print a success message along with the response data.
    • If unsuccessful, print an error message with the status code.
  • Exception Handling:
    • Handles any exceptions that might occur during the execution of the code and prints an error message.
  • Defines Upload URL and Local File Path:
    • Defines the URL for file upload ('upload_url') and the local file path to be uploaded ('file_to_upload').
  • Makes the POST Request:
    • Calls the "upload_file()" function with the specified upload URL and file path to initiate the file upload process.


Flowchart: Python file upload with urllib3: Making POST Requests simplified.

Python Code Editor :

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous: Implementing Retry Mechanism in Python urllib3 for Resilient HTTP Requests.
Next: Python urllib3: Demonstrate Cookie persistence for seamless Web requests.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.

Follow us on Facebook and Twitter for latest update.