You want to return a list of users from MongoDB in FastAPI, but each user's _id must be a string, not ObjectId. Which code snippet correctly does this?
hard🚀 Application Q15 of 15
FastAPI - Database Integration
You want to return a list of users from MongoDB in FastAPI, but each user's _id must be a string, not ObjectId. Which code snippet correctly does this?
Ausers = await db.users.find().to_list(100)
return [{**u, '_id': str(u['_id'])} for u in users]
Busers = await db.users.find().to_list(100)
return [u for u in users if str(u['_id'])]
Cusers = await db.users.find().to_list(100)
return [str(u) for u in users]
Dusers = await db.users.find().to_list(100)
return [u['_id'] for u in users]
Step-by-Step Solution
Solution:
Step 1: Fetch users as list of dicts
await db.users.find().to_list(100) returns list of user dicts with ObjectId _id fields.
Step 2: Convert each user's _id to string
Use list comprehension to copy each dict and replace _id with str(u['_id']).
Final Answer:
users = await db.users.find().to_list(100)
return [{**u, '_id': str(u['_id'])} for u in users] -> Option A
Quick Check:
Convert ObjectId to string per user [OK]
Quick Trick:Use dict unpacking and str() on _id in list comprehension [OK]
Common Mistakes:
MISTAKES
Returning ObjectId directly without conversion
Trying to convert whole user dict to string
Filtering users instead of converting _id
Master "Database Integration" in FastAPI
9 interactive learning modes - each teaches the same concept differently
