I'm having a single redis server serve multiple clients. Sometimes server needs to run a lua script that takes around a minute to complete. However during that time other clients get response error:
redis.exceptions.ResponseError: BUSY Redis is busy running a script. You can only call SCRIPT KILL or SHUTDOWN NOSAVE
Is there in built way to tell client to just keep waiting or retrying this response?
Redis(socket_timeout=9999) doesn't seem to have an effect on this.
I couldn't find a proper way to do so I went with a bit of a hack:
from redis import Redis
class MyRedis(Redis):
lua_retry_time = 120
# override execute to retry busy errors
def execute_command(self, *args, **options):
wait_time = 0
if not self.lua_retry_time:
return super().execute_command(*args, **options)
while wait_time < self.lua_retry_time:
try:
return super().execute_command(*args, **options)
except ResponseError as e:
if 'busy redis is busy' not in ''.join(e.args).lower():
raise e
if wait_time == 0: # only print once
print('Redis is busy, waiting up to 120 seconds...')
time.sleep(2)
wait_time += 2
return super().execute_command(*args, **options)
Every command in redis-py package goes through execute_command method this this should cover everything.
This will keep retrying for 120 seconds every 2 seconds.