Uploading files from frontend in Supabase - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select a file from an input element.

Supabase

const file = document.getElementById('fileInput').[1][0];

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

AfilesList

Bfiles

Cselected

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'value' instead of 'files' to get the file list.

Using a non-existent property like 'selected' or 'filesList'.

The files property of an input element contains the list of selected files.

2fill in blank

medium

Complete the code to upload a file to Supabase storage bucket named 'avatars'.

Supabase

const { data, error } = await supabase.storage.from('avatars').upload([1], file);

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

A'user/profile.png'

B'profile.png'

Cfile.name

D'avatars/profile.png'

Attempts:

3 left

💡 Hint

Common Mistakes

Passing the file object as the first argument instead of the path.

Hardcoding the path without using the file name.

The second argument is the file object, and the first is the path where the file will be stored. Using file.name stores the file with its original name.

3fill in blank

hard

Fix the error in the code to handle the upload response correctly.

Supabase

if ([1]) { console.error('Upload failed:', error.message); } else { console.log('Upload succeeded'); }

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

Aerror

B!error

Cdata

D!data

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if (data) instead of if (error).

Using negation incorrectly like !error.

The error object is truthy when the upload fails, so checking if (error) detects failure.

4fill in blank

hard

Fill both blanks to create a public URL for the uploaded file.

Supabase

const { data } = supabase.storage.from('avatars').[1](file.[2]);

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

AgetPublicUrl

Bname

Cpath

DgetUrl

Attempts:

3 left

💡 Hint

Common Mistakes

Using a non-existent method like 'getUrl'.

Using 'path' instead of 'name' for the file property.

The method getPublicUrl generates a public URL, and the file's name is used as the path.

5fill in blank

hard

Fill all three blanks to handle file upload with error checking and URL retrieval.

Supabase

const file = event.target.files[0];
const { data, [1] } = await supabase.storage.from('avatars').upload(file.[2], file);
if ([3]) return console.error('Upload error:', error.message);
const url = supabase.storage.from('avatars').getPublicUrl(file.name).data.publicUrl;

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

Aerror

Bname

Dpath

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'path' instead of 'name' for the file path.

Checking the wrong variable for errors.

Not destructuring the error from the response.

We destructure error from the upload response, use file.name as the path, and check if (error) to handle failures.

Practice

1. What does the supabase.storage.from(bucket).upload(path, file) method do in a frontend app?

easy

A. Downloads a file from the storage bucket

B. Uploads a file to the specified storage bucket at the given path

C. Deletes a file from the storage bucket

D. Lists all files in the storage bucket

Solution

1. Step 1: Understand the method purpose

   The method upload(path, file) is used to send a file from the frontend to the backend storage.

2. Step 2: Identify the action on storage

   Since it uses upload, it means adding or saving a file, not deleting or downloading.

3. Final Answer:

   Uploads a file to the specified storage bucket at the given path -> Option B

4. Quick Check:

   Upload method = Upload file [OK]

Hint: Upload method means sending file to storage [OK]

Common Mistakes:

• Confusing upload with download or delete
• Thinking it lists files instead of uploading
• Ignoring the bucket and path parameters

2. Which of the following is the correct syntax to upload a file named myFile to the bucket avatars at path user1/profile.png using Supabase storage in frontend JavaScript?

easy

A. supabase.storage.from('avatars').upload('user1/profile.png', myFile)

B. supabase.storage.upload('avatars', 'user1/profile.png', myFile)

C. supabase.storage.from('avatars').upload(myFile, 'user1/profile.png')

D. supabase.upload.storage.from('avatars').file('user1/profile.png', myFile)

Solution

1. Step 1: Check method chaining and parameters

   The correct method chain is supabase.storage.from(bucket).upload(path, file).

2. Step 2: Match parameters order

   The first argument to upload is the path string, second is the file object.

3. Final Answer:

   supabase.storage.from('avatars').upload('user1/profile.png', myFile) -> Option A

4. Quick Check:

   Correct method and parameter order = supabase.storage.from('avatars').upload('user1/profile.png', myFile) [OK]

Hint: Remember: from(bucket).upload(path, file) [OK]

Common Mistakes:

• Swapping file and path arguments
• Using upload directly on storage without from()
• Incorrect method chaining order

3. Given this code snippet, what will be logged if the upload succeeds?

const { data, error } = await supabase.storage.from('docs').upload('file.txt', fileInput.files[0]);
if (error) {
  console.log('Upload failed');
} else {
  console.log('Upload succeeded:', data);
}

medium

A. Upload succeeded: { path: 'file.txt', ... }

B. SyntaxError due to await usage

C. Upload succeeded: undefined

D. Upload failed

Solution

1. Step 1: Understand the upload response

   On success, data contains info about the uploaded file, including its path.

2. Step 2: Check error handling

   If error is null, the else block runs and logs success with data.

3. Final Answer:

   Upload succeeded: { path: 'file.txt', ... } -> Option A

4. Quick Check:

   Success logs data object = Upload succeeded: { path: 'file.txt', ... } [OK]

Hint: Success logs data object, error logs failure [OK]

Common Mistakes:

• Assuming data is undefined on success
• Confusing error presence with success
• Thinking await causes syntax error outside async

4. You wrote this code to upload a file but it always logs 'Upload succeeded' even though the file wasn't uploaded. What is the likely mistake?

const { data, error } = supabase.storage.from('images').upload('pic.png', file);
if (error) {
  console.log('Upload failed');
} else {
  console.log('Upload succeeded');
}

medium

A. Wrong bucket name 'images'

B. File variable 'file' is not defined

C. Missing await before supabase.storage.from(...).upload(...)

D. Upload method does not exist on storage

Solution

1. Step 1: Check async usage

   The upload method returns a promise and must be awaited to get data and error.

2. Step 2: Identify missing await effect

   Without await, { data, error } destructures the unresolved Promise, leaving both undefined. Thus if (error) is false, incorrectly logging success.

3. Final Answer:

   Missing await before supabase.storage.from(...).upload(...) -> Option C

4. Quick Check:

   Async calls need await to get results [OK]

Hint: Always await async upload calls [OK]

Common Mistakes:

• Not awaiting the upload promise
• Assuming upload is synchronous
• Ignoring file variable existence

5. You want to upload multiple files selected by the user to the bucket uploads and keep their original filenames. Which approach correctly uploads all files and handles errors properly?

hard

A. Use upload with a fixed filename for all files to overwrite previous uploads

B. Upload all files at once by passing the array to upload method

C. Upload files without awaiting and ignore errors for faster UI

D. Use a loop with await inside to upload each file and check errors after each upload

Solution

1. Step 1: Understand multiple file uploads

   Supabase storage upload handles one file at a time, so loop over files.

2. Step 2: Handle async and errors properly

   Use await inside the loop to wait for each upload and check for errors to inform the user.

3. Final Answer:

   Use a loop with await inside to upload each file and check errors after each upload -> Option D

4. Quick Check:

   Upload files one by one with await and error check [OK]

Hint: Upload files one by one with await and check errors [OK]

Common Mistakes:

• Passing array directly to upload method
• Not awaiting uploads causing race conditions
• Overwriting files by using same filename