File upload processing in Flask - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What is the output when uploading a file with this Flask route?

Consider this Flask route that handles file uploads. What will the response be if a user uploads a file named example.txt?

Flask

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

@app.route('/upload', methods=['POST'])
def upload_file():
    file = request.files.get('file')
    if file:
        filename = file.filename
        return f'File {filename} uploaded successfully'
    return 'No file uploaded'

A File None uploaded successfully

B No file uploaded

C Internal Server Error

D File example.txt uploaded successfully

💡 Hint

Check how the file is accessed from the request and what attribute is used for the filename.

The request.files.get('file') retrieves the uploaded file object. If it exists, file.filename returns the original filename. So the response includes the uploaded filename.

📝 Syntax

intermediate

Which option correctly saves an uploaded file in Flask?

Given a Flask route receiving a file upload, which code snippet correctly saves the file to the uploads/ folder?

Flask

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

@app.route('/upload', methods=['POST'])
def upload_file():
    file = request.files.get('file')
    if file:
        # Save file here
        pass
    return 'No file uploaded'

A save(file, 'uploads/' + file.filename)

B file.save('uploads/' + file.filename)

C file.write('uploads/' + file.filename)

D file.store('uploads/' + file.filename)

💡 Hint

Look for the method provided by Flask's FileStorage object to save files.

The save() method is the correct way to save an uploaded file to disk in Flask. Other methods like write or store do not exist on the file object.

🔧 Debug

advanced

Why does this Flask file upload code raise a KeyError?

Examine this Flask route snippet. Why does it raise a KeyError when no file is uploaded?

Flask

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

@app.route('/upload', methods=['POST'])
def upload_file():
    file = request.files['file']
    if file:
        return f'Uploaded {file.filename}'
    return 'No file uploaded'

A Because <code>request.files['file']</code> raises KeyError if 'file' key is missing

B Because <code>file.filename</code> is None

C Because the route does not allow POST method

D Because <code>request.files</code> is always empty

💡 Hint

Consider how dictionary access works in Python when a key is missing.

Accessing request.files['file'] directly raises a KeyError if the key 'file' is not present. Using request.files.get('file') avoids this error by returning None instead.

❓ state_output

advanced

What is the value of filename after this Flask upload code runs?

Given this Flask route, what will be the value of filename if the user uploads a file named photo.png?

Flask

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

@app.route('/upload', methods=['POST'])
def upload_file():
    file = request.files.get('file')
    filename = file.filename if file else 'No file'
    return filename

A photo.png

B No file

C None

D file

💡 Hint

Check the conditional expression assigning filename.

If a file is uploaded, file is not None, so filename is set to file.filename, which is the uploaded file's name.

🧠 Conceptual

expert

Which Flask configuration is essential to securely handle file uploads?

When building a Flask app that accepts file uploads, which configuration helps prevent security risks by limiting the size of uploaded files?

A app.config['SECRET_KEY'] = 'mysecret'

B app.config['UPLOAD_FOLDER'] = '/tmp/uploads'

C app.config['MAX_CONTENT_LENGTH'] = 16 * 1024 * 1024

D app.config['DEBUG'] = True

💡 Hint

Think about how to limit the size of incoming data to avoid large file uploads.

The MAX_CONTENT_LENGTH setting limits the maximum size of incoming request data, preventing users from uploading files larger than the specified size. Other options configure upload folder, debugging, or security keys but do not limit upload size.

Practice

1. What does request.files in Flask represent when handling file uploads?

easy

A. A dictionary-like object containing uploaded files

B. A list of all form fields submitted

C. A string with the file path on the server

D. A function to save files automatically

Solution

1. Step 1: Understand request.files role

   It holds files sent by the client in a dictionary-like object where keys are input names.

2. Step 2: Differentiate from other request parts

   Unlike form fields in request.form, request.files specifically contains uploaded files.

3. Final Answer:

   A dictionary-like object containing uploaded files -> Option A

4. Quick Check:

   Files = dictionary-like object [OK]

Hint: Remember: files come in a dictionary, not a list or string [OK]

Common Mistakes:

• Confusing request.files with request.form
• Expecting a list instead of a dictionary-like object
• Thinking it automatically saves files

2. Which of the following is the correct way to check if a file was uploaded in Flask?

easy

A. if request.files.filename == ''

B. if 'file' in request.files and request.files['file'].filename != ''

C. if request.form['file'] != None

D. if request.files['file'] is None

Solution

1. Step 1: Check file presence in request.files

   Use 'file' in request.files to confirm the file input exists.

2. Step 2: Verify filename is not empty

   Check request.files['file'].filename != '' to ensure a file was selected.

3. Final Answer:

   if 'file' in request.files and request.files['file'].filename != '' -> Option B

4. Quick Check:

   Check key and filename not empty [OK]

Hint: Check key exists and filename is not empty string [OK]

Common Mistakes:

• Checking request.form instead of request.files
• Using is None instead of checking filename
• Comparing filename directly to None

3. Given this Flask code snippet:

file = request.files['photo']
filename = secure_filename(file.filename)
file.save(f"uploads/{filename}")
return f"Saved {filename}"

What will be the output if a user uploads a file named my cat.png?

medium

A. Saved mycat.png

B. Saved my cat.png

C. Saved my_cat.png

D. Error due to spaces in filename

Solution

1. Step 1: Understand secure_filename effect

   This function replaces spaces with underscores and removes unsafe characters.

2. Step 2: Apply to 'my cat.png'

   The filename becomes 'my_cat.png' after sanitization.

3. Final Answer:

   Saved my_cat.png -> Option C

4. Quick Check:

   Spaces replaced by underscores in filename [OK]

Hint: Remember secure_filename replaces spaces with underscores [OK]

Common Mistakes:

• Assuming spaces remain in filename
• Expecting filename unchanged
• Thinking spaces cause save error

4. What is wrong with this Flask file upload code?

file = request.files['file']
file.save(file.filename)
return 'File saved!'

medium

A. It does not check if a file was uploaded

B. It saves file to a wrong directory

C. It should use request.form instead of request.files

D. It uses an unsafe filename directly

Solution

1. Step 1: Identify filename safety issue

   Using file.filename directly can allow unsafe paths or characters.

2. Step 2: Recommend using secure_filename

   To avoid security risks, filenames must be sanitized before saving.

3. Final Answer:

   It uses an unsafe filename directly -> Option D

4. Quick Check:

   Always sanitize filenames before saving [OK]

Hint: Always use secure_filename before saving files [OK]

Common Mistakes:

• Not checking if file exists before saving
• Saving files without sanitizing filename
• Confusing request.form with request.files

5. You want to allow users to upload only PNG images and save them safely in Flask. Which code snippet correctly implements this?

hard

A. file = request.files['image']
if file and file.filename.endswith('.png'):
filename = secure_filename(file.filename)
file.save(f"uploads/{filename}")
return 'PNG saved'
else:
return 'Invalid file'

B. file = request.files['image']
if file.filename:
file.save(file.filename)
return 'File saved'
else:
return 'No file'

C. file = request.files.get('image')
if file and file.filename:
filename = secure_filename(file.filename)
file.save(f"uploads/{filename}")
return 'File saved'
else:
return 'No file uploaded'

D. file = request.files['image']
if file and file.filename.endswith('.png'):
file.save(file.filename)
return 'PNG saved'
else:
return 'Invalid file'

Solution

1. Step 1: Check file presence and extension

   file = request.files['image'] if file and file.filename.endswith('.png'): filename = secure_filename(file.filename) file.save(f"uploads/{filename}") return 'PNG saved' else: return 'Invalid file' checks if file exists and filename ends with '.png' to allow only PNG files.

2. Step 2: Use secure_filename before saving

   It sanitizes the filename to prevent security issues before saving.

3. Step 3: Return appropriate messages

   Returns confirmation if saved, else an invalid file message.

4. Final Answer:

   Option A correctly checks extension and uses secure_filename -> Option A

5. Quick Check:

   Check extension + secure filename + save [OK]

Hint: Check extension and sanitize filename before saving [OK]

Common Mistakes:

• Not sanitizing filename before saving
• Not checking file extension properly
• Saving files without checking if uploaded