Saving uploaded files in Flask - Mini Project: Build & Apply

Saving uploaded files

📖 Scenario: You are building a simple web app where users can upload a file. You want to save the uploaded file on the server so you can use it later.

🎯 Goal: Create a Flask app that accepts a file upload from a user and saves the file to a folder on the server.

📋 What You'll Learn

Create a Flask app with a route to upload files

Set a folder path to save uploaded files

Use Flask's request object to get the uploaded file

Save the uploaded file to the specified folder

💡 Why This Matters

🌍 Real World

Uploading and saving files is common in web apps for user profile pictures, documents, or media content.

💼 Career

Understanding file uploads is essential for backend web developers working with Flask or similar frameworks.

1

Set up Flask app and upload folder

Import Flask and request from flask. Create a Flask app called app. Set a variable UPLOAD_FOLDER to the string 'uploads/'.

Flask

from flask import Flask, request

# Your code here

Hint

Use Flask(__name__) to create the app. Set UPLOAD_FOLDER as a string with the folder name.

2

Create upload route and HTML form

Add a route /upload to the app that accepts GET and POST methods. Inside the route function upload_file, if the request method is GET, return a simple HTML form string with a file input named file and a submit button.

Flask

from flask import Flask, request

app = Flask(__name__)
UPLOAD_FOLDER = 'uploads/'

@app.route('/upload', methods=['GET', 'POST'])
def upload_file():
    # Your code here

Hint

Use @app.route with methods=['GET', 'POST']. Return a form with enctype='multipart/form-data' and a file input named file.

3

Get uploaded file from request

Inside the upload_file function, after the GET block, add code for POST method. Get the uploaded file from request.files using the key 'file' and store it in a variable called uploaded_file.

Flask

from flask import Flask, request

app = Flask(__name__)
UPLOAD_FOLDER = 'uploads/'

@app.route('/upload', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'GET':
        return '''
            
                
                
            
        '''
    uploaded_file = request.files['file']  # Your code here

Hint

Use request.files['file'] to get the uploaded file object.

4

Save the uploaded file

Still inside the upload_file function, after getting uploaded_file, save it to the folder UPLOAD_FOLDER using the save() method. Use uploaded_file.filename to get the original filename. Return a string 'File saved successfully' after saving.

Flask

from flask import Flask, request

app = Flask(__name__)
UPLOAD_FOLDER = 'uploads/'

@app.route('/upload', methods=['GET', 'POST'])
def upload_file():
    if request.method == 'GET':
        return '''
            
                
                
            
        '''
    uploaded_file = request.files['file']
    # Save the file here
    # Your code here

Hint

Use uploaded_file.save() with the path combining UPLOAD_FOLDER and uploaded_file.filename. Return a success message string.

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