Saving uploaded files in Flask - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ component_behavior

intermediate

What happens when this Flask code saves an uploaded file?

Consider this Flask route that handles a file upload and saves it:

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

@app.route('/upload', methods=['POST'])
def upload():
    file = request.files['file']
    file.save('/tmp/' + file.filename)
    return 'File saved'

What will happen 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 = request.files['file']
    file.save('/tmp/' + file.filename)
    return 'File saved'

A The file will be saved in the /tmp directory with the name example.txt

B The file will be saved in the current directory with the name example.txt

C The code will raise a KeyError because 'file' is not in request.files

D The file will be saved but with a random generated filename

💡 Hint

Look at the file.save line and the path used.

The file.save method saves the uploaded file to the path given. Here, it saves to /tmp/ plus the original filename, so the file will be saved as /tmp/example.txt.

📝 Syntax

intermediate

Which option correctly saves an uploaded file in Flask?

Which of the following Flask code snippets correctly saves an uploaded file to the 'uploads' folder?

A

file = request.files['file']
file.save('uploads/' + file.filename)

B

file = request.files.get('file')
file.save('uploads')

C

file = request.files['file']
file.save('uploads')

D

file = request.files['file']
file.save('uploads/' + file)

💡 Hint

Remember that file.save() needs a full path including filename.

Option A correctly saves the file by combining the folder path and the filename. Options B and C miss the filename, causing errors. Option A tries to concatenate a file object, which is invalid.

🔧 Debug

advanced

Why does this Flask file upload code raise an error?

Given this Flask route:

@app.route('/upload', methods=['POST'])
def upload():
    file = request.files['file']
    file.save('uploads/' + file.filename)
    return 'Saved'

When a user uploads a file, the server raises a FileNotFoundError. Why?

Flask

@app.route('/upload', methods=['POST'])
def upload():
    file = request.files['file']
    file.save('uploads/' + file.filename)
    return 'Saved'

A The file object does not have a filename attribute

B The 'uploads' directory does not exist on the server

C The request method is not POST

D The file is too large to save

💡 Hint

Check if the folder path exists before saving files.

Flask's file.save() does not create directories. If the 'uploads' folder is missing, saving the file raises FileNotFoundError.

❓ state_output

advanced

What is the value of file.filename after upload?

In Flask, after a user uploads a file named photo.png, what is the value of file.filename in this code?

file = request.files['file']

Flask

file = request.files['file']

A An empty string ''

B 'file'

C The full file path on the user's computer

D 'photo.png'

💡 Hint

Think about what file.filename represents in Flask uploads.

The file.filename attribute contains the original name of the uploaded file, here 'photo.png'. It does not include the full path from the user's computer.

🧠 Conceptual

expert

Which option best prevents filename conflicts when saving uploaded files in Flask?

When saving uploaded files in Flask, how can you avoid overwriting existing files with the same name?

A Always save files with their original filename in the same folder

B Overwrite existing files with the new upload to keep latest version

C Generate a unique filename using a UUID and save the file with it

D Save files only in memory without writing to disk

💡 Hint

Think about how to keep filenames unique to avoid conflicts.

Using a UUID or timestamp to create a unique filename prevents overwriting existing files. Saving with original names risks conflicts. Overwriting loses data. Saving only in memory does not persist files.

Practice

1. In Flask, which object do you use to access files uploaded by a user in a form?

easy

A. request.args

B. request.files

C. request.data

D. request.form

Solution

1. Step 1: Understand Flask request objects

   Flask provides different objects to access data sent by the client. request.files holds uploaded files.

2. Step 2: Identify correct object for files

   request.form is for form fields, request.data is raw data, and request.args is for URL query parameters.

3. Final Answer:

   request.files -> Option B

4. Quick Check:

   Uploaded files = request.files [OK]

Hint: Uploaded files are always in request.files [OK]

Common Mistakes:

• Using request.form to get files
• Trying to access files from request.args
• Confusing request.data with files

2. Which of the following is the correct way to save an uploaded file named file to a folder called uploads in Flask?

easy

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

B. file.save(file.filename)

C. save(file, 'uploads/')

D. request.save(file, 'uploads/')

Solution

1. Step 1: Use the file object's save method

   The uploaded file object has a save() method that takes the full path to save the file.

2. Step 2: Provide correct path including folder and filename

   Concatenate the folder path and the file's filename to save it in the right place.

3. Final Answer:

   file.save('uploads/' + file.filename) -> Option A

4. Quick Check:

   Use file.save(path) with full path [OK]

Hint: Use file.save('folder/' + file.filename) to save files [OK]

Common Mistakes:

• Calling save() without a path
• Using save() as a standalone function
• Not including folder path in save()

3. What will happen when this Flask code runs after a file upload?

file = request.files['photo']
file.save('/tmp/' + file.filename)
return 'Saved!'

medium

A. Nothing is saved because request.files['photo'] is empty by default

B. The file is saved but the filename is lost, causing an error

C. The code raises an error because save() needs a full absolute path

D. The uploaded file is saved in /tmp/ with its original filename and 'Saved!' is returned

Solution

1. Step 1: Access the uploaded file from request.files

   The code gets the file named 'photo' from the uploaded files correctly.

2. Step 2: Save the file to /tmp/ folder with original filename

   Calling save() with '/tmp/' + filename saves the file in that folder.

3. Final Answer:

   The uploaded file is saved in /tmp/ with its original filename and 'Saved!' is returned -> Option D

4. Quick Check:

   save() stores file at given path, returns confirmation [OK]

Hint: save() stores file at given path, returns confirmation [OK]

Common Mistakes:

• Assuming save() needs absolute path always
• Thinking request.files['photo'] is empty without upload
• Confusing save() behavior with form data saving

4. Identify the error in this Flask code snippet for saving an uploaded file:

file = request.files['file']
file.save(file.filename)

medium

A. Missing folder path in save() can cause files to save in wrong location

B. Using request.files instead of request.form causes error

C. file.save() requires a file object, not a filename string

D. file.filename is not a valid attribute of the file object

Solution

1. Step 1: Check the save() method argument

   The code saves the file using only the filename, without specifying a folder path.

2. Step 2: Understand file saving location

   Without a folder path, the file saves in the current working directory, which may be unintended or insecure.

3. Final Answer:

   Missing folder path in save() can cause files to save in wrong location -> Option A

4. Quick Check:

   Always include folder path in save() [OK]

Hint: Always include folder path in save() to control file location [OK]

Common Mistakes:

• Not specifying folder path in save()
• Confusing request.files with request.form
• Assuming file.filename is invalid

5. You want to securely save an uploaded file in Flask to the uploads folder, avoiding unsafe filenames. Which code snippet correctly achieves this?

from werkzeug.utils import secure_filename
file = request.files['file']
filename = secure_filename(file.filename)
file.save(f'uploads/{filename}')

hard

A. The code will fail because secure_filename() returns a file object, not a string

B. secure_filename() is unnecessary; file.save() cleans filenames automatically

C. This code safely saves the file with a cleaned filename in uploads folder

D. You must manually create the uploads folder before saving, or it errors

Solution

1. Step 1: Use secure_filename to clean the filename

   The function secure_filename() removes unsafe characters to prevent security issues.

2. Step 2: Save the file with the cleaned filename in the uploads folder

   Using file.save() with the cleaned filename inside the uploads folder saves the file securely.

3. Final Answer:

   This code safely saves the file with a cleaned filename in uploads folder -> Option C

4. Quick Check:

   Use secure_filename() before saving uploaded files [OK]

Hint: Use secure_filename() to avoid unsafe file names [OK]

Common Mistakes:

• Skipping secure_filename and risking unsafe filenames
• Assuming save() cleans filenames automatically
• Not creating uploads folder before saving