FastAPI - File Handling
You want to save multiple uploaded files to disk in FastAPI. Which code snippet correctly saves all files to a folder named 'uploads'?
You want to save multiple uploaded files to disk in FastAPI. Which code snippet correctly saves all files to a folder named 'uploads'?
hard🚀 Application Q8 of 15
Step-by-Step Solution
Solution:
Step 1: Read file contents asynchronously
Use await file.read() to read file bytes asynchronously from UploadFile.Step 2: Write bytes to disk in binary mode
Open the target file in 'wb' mode and write the bytes to save the file correctly.Final Answer:
for file in files: contents = await file.read() with open(f'uploads/{file.filename}', 'wb') as f: f.write(contents) -> Option DQuick Check:
Async read + binary write saves files correctly [OK]
Quick Trick: Use await file.read() and write in 'wb' mode [OK]
Common Mistakes:
MISTAKES- Opening file in 'r' mode for writing
- Calling file.read() without await
- Assuming UploadFile has save() method
Master "File Handling" in FastAPI
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