On 1 December 2017 at 14:32, hw <hw at gc-24.de> wrote: > Gordon Messmer wrote: >> >> On 12/01/2017 08:49 AM, hw wrote: >>> >>> # time foo >>> real 43m39.841s >>> user 15m31.109s >>> sys 0m44.136s >>> >>> >>> Almost 30 minutes have disappeared, but it actually took about that long, >>> so what happened? >> >> >> >> I may misunderstand your question, but >> >> "time" is provided by the bash shell. It may be provided by a command if >> you are using a different shell. When the command following the "time" >> keyword completes, bash will print the amount of elapsed time (the amount of >> time that passed between the command's start and its exit), the amount of >> time the command was using the CPU and not in a sleep state, and the amount >> of time the kernel was using the CPU to service requests from the command. >> >> So your "foo" application was in a sleep state for around 30 minutes of >> the 44 minutes that passed between when you started it and when it finished. > > > Hm. Foo is a program that imports data into a database from two CVS files, > using a connection for each file and forking to import both files at once. > > So this would mean that the database (running on a different server) takes > almost two times as much as foo --- which I would consider kinda > excruciatingly > long because it´s merely inserting rows into two different tables after they > were > prepared by foo and then processes some queries to convert the data. > > The queries after importing may take like 3 or 5 minutes. About 4.5 million > rows > are being imported. > > Would you consider about 20 minutes for importing as long? That depends on a lot of things.. from drive speed to drive layout to database to network congestion to... without that information the question is not answerable. > > _______________________________________________ > CentOS mailing list > CentOS at centos.org > https://lists.centos.org/mailman/listinfo/centos -- Stephen J Smoogen.