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.