I don't see this kind of behavior in my Python apps; so there is likely some difference here. I've got processing that inserts a couple transactional tasks that also insert a lot of additional tasks, but the second level tasks are not inserted transactionaly.
You keep slowly mentioning potentially important details in subsequent posts; it makes it very hard to help you debug. So far we know 1) you are not doing this by directly using the task-queue api (you're using some wrapper/library), 2) you insert tasks in a transaction with no datastore operations, and 3) you transactionaly insert tasks that themselves transactionaly insert tasks. Have you tried to reproduce this problem by directly using the task-queue api? Do the initial tasks modify entities, and if so are they int he same entity group? Are you sure this error is being caused by something with the tasks, and not by fiddling with an array in a loop, or both of the tasks running at the same time working with the same entity group, or some other issue? Robert On Wed, Apr 20, 2011 at 14:48, Matija <[email protected]> wrote: > It is actually ConcurrentModificationException. > If you add tasks to task queue inside transaction and there are one or two > tasks that add also tasks (in same time or in 1 sek window) to task queue > inside transaction, there is very big chance that you will > get ConcurrentModificationException. > > -- > You received this message because you are subscribed to the Google Groups > "Google App Engine" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-appengine?hl=en. > -- You received this message because you are subscribed to the Google Groups "Google App Engine" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-appengine?hl=en.
