Reputation: 416
I have written a simple scraping program using asyncio. Here are my code snippets:
loop = asyncio.get_event_loop()
task = loop.create_task(conSpi.parse(arguments.url))
value = loop.run_until_complete(asyncio.wait([task]))
loop.close()
I want to print the result being returned in value
. Rather than printing the variable's value, it prints something like this:
{<Task finished coro=<ConcurrentSpider.parse() done,
defined at /home/afraz/PycharmProjects/the-lab/concurrentspider.py:28> result=3>}
How can I get the result only and not get rest printed?
Upvotes: 19
Views: 26726
Reputation: 57445
In Python versions >= 3.7, I'd suggest asyncio.run
:
import asyncio
async def do_task(duration):
await asyncio.sleep(duration)
return duration
if __name__ == "__main__":
result = asyncio.run(do_task(3))
print(result) # => 3
If you have multiple tasks to run in parallel:
import asyncio
async def do_task(duration):
await asyncio.sleep(duration)
return duration
async def do_multiple_tasks():
return await asyncio.gather(do_task(3), do_task(1), do_task(2))
if __name__ == "__main__":
result = asyncio.run(do_multiple_tasks())
print(result) # => [3, 1, 2]
Upvotes: 3
Reputation: 6529
The simplest approach is to write
value = loop.run_until_complete(task)
That only works if you want to wait on one task. If you need more than one task, you'll need to use asyncio.wait correctly. It returns a tuple containing completed and pending futures. By default though, the pending futures will be empty because it waits for all futures to complete.
So something like
done, pending = loop.run_until_complete(asyncio.wait( tasks))
for future in done:
value = future.result() #may raise an exception if coroutine failed
# do something with value
Upvotes: 27