# Target URL url = "http://example.com/upload"

# Check if the file was uploaded successfully if response.status_code == 200: print("File uploaded successfully") else: print("Upload failed") The root cause of this vulnerability lies in the FileUpload class, specifically in the save() method. The method does not perform adequate validation on the uploaded file, allowing an attacker to bypass security checks. Code Review A code review of the FileUpload class reveals the following:

import os from werkzeug.utils import secure_filename