How to return value from async method in python?

Question

I have the following flask app:

async def run():
   conn = await asyncpg.connect(db_url)
   values = await conn.fetch('''SELECT ... FROM ... WHERE ...;''')
   await conn.close()

@app.route('/')
def test():
    loop = asyncio.get_event_loop()
    res = loop.run_until_complete(run())
    return json.dumps([dict(r) for r in res]) 

if __name__ == '__main__':
    app.run()

When I run this code I got TypeError: 'NoneType' object is not iterable. How to return my values converted to JSON?

in this code, is it okay that test() function is not async? — Wenuka
– Wenuka, Commented Feb 1, 2024 at 8:16

Pit · Accepted Answer · 2017-03-16 11:35:38Z

30

You need to return your values in your run function for them to be available in test:

async def run():
   conn = await asyncpg.connect(db_url)
   values = await conn.fetch('''SELECT ... FROM ... WHERE ...;''')
   await conn.close()
   return values

answered Mar 16, 2017 at 11:35

Pit

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3 Comments

user6611771 Over a year ago

Thanks, is it necessary to return value after await conn.close() or I can return values and then await conn.close() ?

Pit Over a year ago

return exits the function, so the await would not be called. Maybe try yield values before calling await conn.close().

Pit Over a year ago

On second thought: what you probably want to do is loop.create_task(conn.close()) and still use return values. This will schedule the closing of the connection without waiting for it in your function.

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