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[IMP] improve runbot scheduler algorithm

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Now it is very simple: it always builds sticky branch if any, otherwise
last pending job.
This commit is contained in:
Gery Debongnie 2014-07-04 14:47:58 +02:00
parent 969f28f126
commit 4e1b44e3aa
1 changed files with 10 additions and 24 deletions
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34
runbot/runbot.py
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@ -280,35 +280,21 @@ class runbot_repo(osv.osv):
# launch new tests
testing = bo.search_count(cr, uid, dom + [('state', '=', 'testing')])
pending = bo.search_count(cr, uid, dom + [('state', '=', 'pending')])
while testing < repo.testing:
# select sticky build if any
while testing < repo.testing and pending > 0:
# pending_ids = bo.search(cr, uid, dom + [('state', '=', 'pending')])
# find sticky pending build if any, otherwise, last pending (by id, not by sequence) will do the job
pending_ids = bo.search(cr, uid, dom + [('state', '=', 'pending'), ('branch_id.sticky', '=', True)], limit=1)
if not pending_ids:
pending_ids = bo.search(cr, uid, dom + [('state', '=', 'pending')], order="id desc")
pending = bo.browse(cr, uid, pending_ids[0])
pending.schedule()
# select the next build to process
pending_ids = bo.search(cr, uid, dom + [('state', '=', 'pending')])
if pending_ids:
pending = bo.browse(cr, uid, pending_ids[0])
else:
break
# gather information about currently running builds
running_ids = bo.search(cr, uid, dom + [('state', '=', 'running')])
running_len = len(running_ids)
running_max = 0
if running_ids:
running_max = bo.browse(cr, uid, running_ids[0]).sequence
# determine if pending one should be launched
if running_len < repo.running or pending.sequence >= running_max:
pending.schedule()
else:
break
# compute the number of testing job again
# compute the number of testing and pending jobs again
testing = bo.search_count(cr, uid, dom + [('state', '=', 'testing')])
pending = bo.search_count(cr, uid, dom + [('state', '=', 'pending')])
# terminate and reap doomed build
build_ids = bo.search(cr, uid, dom + [('state', '=', 'running')])