Serving uploaded files in Flask - Cheat Sheet & Quick Revision

Recall & Review

beginner

What Flask function is commonly used to serve uploaded files to users?

The send_from_directory() function is used to safely serve uploaded files from a specific folder to users.

intermediate

Why should you avoid serving uploaded files directly from the upload folder without precautions?

Serving files directly can expose sensitive files or allow execution of harmful files. Using send_from_directory() helps control access and prevents security risks.

beginner

How do you specify the folder from which Flask serves uploaded files?

You pass the folder path as the first argument to send_from_directory(), for example: send_from_directory('uploads', filename).

beginner

What HTTP method is typically used when a user requests to download or view an uploaded file?

The GET method is used because the user is requesting to retrieve and view or download the file.

beginner

What is a simple Flask route example to serve an uploaded file named filename from the uploads folder?

<pre>from flask import send_from_directory

@app.route('/uploads/&lt;filename&gt;')
def uploaded_file(filename):
    return send_from_directory('uploads', filename)</pre>

Which Flask function should you use to serve a file from a specific folder?

A redirect()

B render_template()

C send_from_directory()

D url_for()

send_from_directory() is designed to serve files from a folder safely.

What is the main security reason to use send_from_directory() instead of serving files directly?

A It compresses files automatically

B It prevents unauthorized file access

C It changes file names

D It encrypts files

send_from_directory() helps prevent unauthorized access by controlling which files are served.

In Flask, which HTTP method is used to request an uploaded file for download or viewing?

A GET

B DELETE

C PUT

D POST

GET is used to retrieve resources like files.

What argument do you pass first to send_from_directory()?

A The filename

B The file size

C The file extension

D The folder path

The first argument is the folder path where the file is located.

Which of these is a correct Flask route to serve a file named filename from the uploads folder?

A @app.route('/uploads/<filename>') def file(filename): return send_from_directory('uploads', filename)

B @app.route('/uploads') def file(): return send_from_directory('uploads')

C @app.route('/uploads/<filename>') def file(filename): return render_template(filename)

D @app.route('/uploads/<filename>') def file(filename): return redirect('uploads/' + filename)

Option A correctly uses send_from_directory() with the folder and filename.

Explain how to serve an uploaded file safely in Flask and why it is important.

Think about how Flask controls file access and protects the server.

Describe the steps to create a Flask route that allows users to download files they uploaded.

Focus on the route setup and the function used to send files.

Practice

1. What is the main Flask function used to serve uploaded files from a specific folder?

easy

A. send_from_directory

B. render_template

C. redirect

D. url_for

Solution

1. Step 1: Understand serving files in Flask

   Flask provides a function specifically to serve files from a folder safely, which is send_from_directory.

2. Step 2: Identify the correct function

   render_template is for HTML templates, redirect sends users to a different URL, and url_for builds URLs. Only send_from_directory serves files.

3. Final Answer:

   send_from_directory -> Option A

4. Quick Check:

   Serving files = send_from_directory [OK]

Hint: Serving files? Use send_from_directory [OK]

Common Mistakes:

• Confusing send_from_directory with render_template
• Trying to serve files with redirect
• Using url_for to serve files directly

2. Which of the following is the correct syntax to serve a file named example.txt from the folder uploads in Flask?

easy

A. redirect('uploads/example.txt')

B. send_file('uploads/example.txt')

C. send_from_directory('uploads', 'example.txt')

D. url_for('uploads/example.txt')

Solution

1. Step 1: Recall the function signature

   send_from_directory takes two arguments: the folder name and the filename.

2. Step 2: Match the syntax

   send_from_directory('uploads', 'example.txt') correctly calls send_from_directory('uploads', 'example.txt'). Options A, B, and D use incorrect functions or formats.

3. Final Answer:

   send_from_directory('uploads', 'example.txt') -> Option C

4. Quick Check:

   Correct syntax = send_from_directory(folder, filename) [OK]

Hint: Use send_from_directory(folder, filename) syntax [OK]

Common Mistakes:

• Using send_file with folder and filename as one string
• Using redirect to serve files
• Using url_for incorrectly for file serving

3. Given this Flask route to serve uploaded files:

from flask import Flask, send_from_directory
app = Flask(__name__)

@app.route('/uploads/<filename>')
def uploaded_file(filename):
    return send_from_directory('uploads', filename)

What happens when a user visits /uploads/photo.png?

medium

A. The server returns a 404 error because the route is incorrect.

B. The server returns the filename as plain text.

C. The server redirects the user to the home page.

D. The file 'photo.png' is sent from the 'uploads' folder to the user.

Solution

1. Step 1: Analyze the route and function

   The route captures the filename from the URL and calls send_from_directory with 'uploads' folder and the filename.

2. Step 2: Understand the behavior on URL visit

   Visiting /uploads/photo.png triggers the function to send the file 'photo.png' from the 'uploads' folder to the user.

3. Final Answer:

   The file 'photo.png' is sent from the 'uploads' folder to the user. -> Option D

4. Quick Check:

   Route sends file from folder = The file 'photo.png' is sent from the 'uploads' folder to the user. [OK]

Hint: Route with filename uses send_from_directory to serve file [OK]

Common Mistakes:

• Assuming route returns text instead of file
• Thinking route redirects instead of serving file
• Believing route is invalid and causes 404

4. Identify the error in this Flask code snippet meant to serve uploaded files:

from flask import Flask, send_from_directory
app = Flask(__name__)

@app.route('/files/<filename>')
def files(filename):
    return send_from_directory(filename, 'uploads')

medium

A. The folder and filename arguments are reversed in send_from_directory.

B. The route decorator syntax is incorrect.

C. The function name 'files' is a reserved keyword.

D. send_from_directory cannot be used inside a route.

Solution

1. Step 1: Check send_from_directory arguments

   The function call uses send_from_directory(filename, 'uploads'), but the first argument should be the folder, second the filename.

2. Step 2: Confirm correct argument order

   The correct call is send_from_directory('uploads', filename). The route and function name are valid, and send_from_directory is allowed in routes.

3. Final Answer:

   The folder and filename arguments are reversed in send_from_directory. -> Option A

4. Quick Check:

   send_from_directory(folder, filename) order matters [OK]

Hint: Folder first, filename second in send_from_directory [OK]

Common Mistakes:

• Swapping folder and filename arguments
• Thinking function names like 'files' are reserved
• Believing send_from_directory can't be used in routes

5. You want to serve uploaded files securely from the folder user_uploads. Which Flask route setup correctly serves files while preventing directory traversal attacks?

hard

A. @app.route('/files/<path:filename>')
def serve_file(filename):
return send_from_directory('user_uploads', filename)

B. @app.route('/files/<filename>')
def serve_file(filename):
return send_from_directory('user_uploads', filename)

C. @app.route('/files/<filename>')
def serve_file(filename):
return send_file(f'user_uploads/{filename}')

D. @app.route('/files/<filename>')
def serve_file(filename):
return open(f'user_uploads/{filename}').read()

Solution

1. Step 1: Understand directory traversal risks

   Allowing path:filename lets users include slashes, risking access outside the folder.

2. Step 2: Choose safe route and function

   @app.route('/files/<filename>')
   def serve_file(filename):
       return send_from_directory('user_uploads', filename)

   uses <filename> (no slashes) and send_from_directory, which safely serves files and prevents traversal.

3. Step 3: Evaluate other options

   @app.route('/files/<path:filename>')
   def serve_file(filename):
       return send_from_directory('user_uploads', filename)

   allows paths, risking traversal.

   @app.route('/files/<filename>')
   def serve_file(filename):
       return send_file(f'user_uploads/{filename}')

   uses send_file with string path, less safe.

   @app.route('/files/<filename>')
   def serve_file(filename):
       return open(f'user_uploads/{filename}').read()

   reads file content directly, unsafe and breaks response.

4. Final Answer:

   Route with <filename> and send_from_directory for safe serving -> Option B

5. Quick Check:

   Use send_from_directory with filename only to prevent traversal [OK]

Hint: Use <filename> (no path) with send_from_directory for safety [OK]

Common Mistakes:

• Using allowing directory traversal
• Using send_file with raw paths risking security
• Reading file content directly instead of sending file