File upload forms in Flask - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the Flask class.

Flask

from flask import [1]

Drag options to blanks, or click blank then click option'

Arender_template

BRequest

CFlask

Dredirect

Attempts:

3 left

💡 Hint

Common Mistakes

Importing Request instead of Flask

Using lowercase flask instead of Flask

The Flask class is imported from the flask module to create the app.

2fill in blank

medium

Complete the code to get the uploaded file from the form.

Flask

file = request.files.get('[1]')

Drag options to blanks, or click blank then click option'

Afile

Bfilename

Cupload

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'filename' which is a property of the file object, not the form field name

Using 'upload' or 'data' which are not the form field names

The key used to get the uploaded file from the form is the name attribute of the file input, commonly 'file'.

3fill in blank

hard

Fix the error in saving the uploaded file securely.

Flask

filename = secure_filename(file.[1])
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))

Drag options to blanks, or click blank then click option'

Afilename

Bname

Cfile

Dpath

Attempts:

3 left

💡 Hint

Common Mistakes

Using file.name which does not exist

Using file.file which is incorrect

Using file.path which is not an attribute

The uploaded file object has a 'filename' attribute that holds the original file name.

4fill in blank

hard

Fill both blanks to check if the request method is POST and a file is present.

Flask

if request.method == '[1]' and '[2]' in request.files:

Drag options to blanks, or click blank then click option'

APOST

Bfile

CGET

Dupload

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for 'GET' method instead of 'POST'

Checking for wrong file field name in request.files

To handle file uploads, the request method must be 'POST' and the file field name (usually 'file') must be in request.files.

5fill in blank

hard

Fill all three blanks to create a Flask route that accepts file uploads and saves the file.

Flask

@app.route('/upload', methods=['[1]'])
def upload_file():
    if request.method == '[2]' and '[3]' in request.files:
        file = request.files['file']
        filename = secure_filename(file.filename)
        file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
        return 'File uploaded successfully'
    return 'Upload a file'

Drag options to blanks, or click blank then click option'

APOST

BGET

Cfile

DPUT

Attempts:

3 left

💡 Hint

Common Mistakes

Using GET or PUT instead of POST in route methods

Checking for wrong request method

Using wrong file field name

The route must allow POST method to upload files. The request method is checked for 'POST' and the file field name is 'file'.

Practice

1. What must you set in the HTML form tag to allow file uploads in a Flask app?

easy

A. method="get"

B. enctype="multipart/form-data"

C. action="upload"

D. type="file"

Solution

1. Step 1: Understand form encoding for file uploads

   File uploads require the form to send data in a special format that supports files, which is "multipart/form-data".

2. Step 2: Identify the correct form attribute

   The form tag must have the attribute enctype="multipart/form-data" to handle file uploads correctly.

3. Final Answer:

   enctype="multipart/form-data" -> Option B

4. Quick Check:

   Form enctype = multipart/form-data [OK]

Hint: File uploads always need enctype multipart/form-data [OK]

Common Mistakes:

• Using method="get" instead of post
• Forgetting enctype attribute
• Setting type="file" on form tag instead of input
• Confusing action URL with encoding

2. Which Flask code correctly accesses an uploaded file named 'photo' from a form?

easy

A. file = request.form['photo']

B. file = request.args['photo']

C. file = request.data['photo']

D. file = request.files['photo']

Solution

1. Step 1: Identify where uploaded files are stored in Flask

   Uploaded files are stored in request.files, not in request.form or request.args.

2. Step 2: Access the file by its form field name

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

3. Final Answer:

   file = request.files['photo'] -> Option D

4. Quick Check:

   Uploaded files = request.files [OK]

Hint: Use request.files to get uploaded files, not request.form [OK]

Common Mistakes:

• Using request.form to get files
• Using request.args which is for URL params
• Trying to access request.data as a dict
• Misspelling 'files' or field name

3. What will the following Flask code print if no file is uploaded?

file = request.files.get('document')
if file and file.filename != '':
    print('File received')
else:
    print('No file uploaded')

medium

A. File received

B. KeyError exception

C. No file uploaded

D. Empty string printed

Solution

1. Step 1: Understand request.files.get behavior when no file sent

   request.files.get('document') returns None if no file was uploaded with that name.

2. Step 2: Evaluate the if condition

   Since file is None, the condition 'file and file.filename != ""' is False, so the else block runs.

3. Final Answer:

   No file uploaded -> Option C

4. Quick Check:

   No file means else prints [OK]

Hint: Check file is not None and filename not empty to confirm upload [OK]

Common Mistakes:

• Assuming request.files['document'] raises error if missing
• Not checking filename for empty string
• Expecting 'File received' even if no file sent
• Confusing request.files.get with request.files[]

4. Identify the error in this Flask file upload snippet:

file = request.files['file']
filename = secure_filename(file)
save_path = os.path.join('uploads', filename)
file.save(save_path)

medium

A. secure_filename should be called on file.filename, not file

B. file.save() requires a file object, not a path

C. os.path.join cannot join strings

D. request.files['file'] returns a string, not a file object

Solution

1. Step 1: Check usage of secure_filename

   secure_filename expects a string filename, but file is a file object, so file.filename should be passed.

2. Step 2: Confirm correct file saving steps

   After getting a safe filename, join path and call file.save(path) correctly.

3. Final Answer:

   secure_filename should be called on file.filename, not file -> Option A

4. Quick Check:

   Pass filename string to secure_filename [OK]

Hint: Use secure_filename(file.filename), not file itself [OK]

Common Mistakes:

• Passing file object to secure_filename
• Confusing file.save argument type
• Misusing os.path.join with non-string
• Assuming request.files returns string

5. You want to allow users to upload multiple images at once and save only PNG files securely. What issue does this Flask code snippet have?

files = request.files.getlist('images')
for file in files:
    if file and file.filename.endswith('.png'):
        filename = secure_filename(file.filename)
        file.save(os.path.join('uploads', filename))

hard

A. file.filename.endswith('.png') is case-sensitive and may miss '.PNG'

B. getlist() returns a single file, so loop will fail

C. secure_filename should be called after saving the file

D. This code correctly filters PNG files and saves them securely

Solution

1. Step 1: Understand getlist() usage

   request.files.getlist('images') correctly returns a list of uploaded files for the field 'images'.

2. Step 2: Check file extension filtering

   Using file.filename.endswith('.png') is case-sensitive and will miss files with uppercase extensions like '.PNG'.

3. Step 3: Confirm secure_filename usage

   secure_filename is correctly called before saving the file.

4. Final Answer:

   file.filename.endswith('.png') is case-sensitive and may miss '.PNG' -> Option A

5. Quick Check:

   Check file extension case-insensitively [OK]

Hint: Check file extensions case-insensitively for uploads [OK]

Common Mistakes:

• Assuming getlist returns one file
• Calling secure_filename after saving
• Ignoring uppercase file extensions
• Not checking if file is None before saving