5d37139374
BUG= Review URL: https://webrtc-codereview.appspot.com/1166004 git-svn-id: http://webrtc.googlecode.com/svn/trunk@3627 4adac7df-926f-26a2-2b94-8c16560cd09d
134 lines
4.7 KiB
Python
134 lines
4.7 KiB
Python
#!/usr/bin/env python
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#-*- coding: utf-8 -*-
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# Copyright (c) 2012 The WebRTC project authors. All Rights Reserved.
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#
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# Use of this source code is governed by a BSD-style license
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# that can be found in the LICENSE file in the root of the source
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# tree. An additional intellectual property rights grant can be found
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# in the file PATENTS. All contributing project authors may
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# be found in the AUTHORS file in the root of the source tree.
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"""Loads build status data for the dashboard."""
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from google.appengine.ext import db
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def _status_not_ok(status):
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return status not in ('OK', 'warnings')
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def _all_ok(statuses):
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return filter(_status_not_ok, statuses) == []
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def _get_first_entry(iterable):
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if not iterable:
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return None
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for item in iterable:
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return item
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class BuildStatusLoader:
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""" Loads various build status data from the database."""
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def __init__(self):
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pass
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@staticmethod
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def load_build_status_data():
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"""Returns the latest conclusive build status for each bot.
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The statuses OK, failed and warnings are considered to be conclusive.
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The algorithm looks at the 100 most recent status entries, which should
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give data on roughly the last five revisions if the number of bots stay
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around 20 (The number 100 should be increased if the number of bots
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increases significantly). This should give us enough data to get a
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conclusive build status for all active bots.
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With this limit, the algorithm will adapt automatically if a bot is
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decommissioned - it will eventually disappear. The limit should not be
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too high either since we will perhaps remember offline bots too long,
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which could be confusing. The algorithm also adapts automatically to new
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bots - these show up immediately if they get a build status for a recent
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revision.
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Returns:
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A list of BuildStatusData entities with one entity per bot.
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"""
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build_status_entries = db.GqlQuery('SELECT * '
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'FROM BuildStatusData '
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'ORDER BY revision DESC '
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'LIMIT 100')
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bots_to_latest_conclusive_entry = dict()
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for entry in build_status_entries:
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if entry.status == 'building':
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# The 'building' status it not conclusive, so discard this entry and
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# pick up the entry for this bot on the next revision instead. That
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# entry is guaranteed to have a status != 'building' since a bot cannot
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# be building two revisions simultaneously.
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continue
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if bots_to_latest_conclusive_entry.has_key(entry.bot_name):
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# We've already determined this bot's status.
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continue
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bots_to_latest_conclusive_entry[entry.bot_name] = entry
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return bots_to_latest_conclusive_entry.values()
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@staticmethod
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def load_last_modified_at():
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build_status_root = db.GqlQuery('SELECT * '
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'FROM BuildStatusRoot').get()
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if not build_status_root:
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# Operating on completely empty database
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return None
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return build_status_root.last_updated_at
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@staticmethod
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def compute_lkgr():
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""" Finds the most recent revision for which all bots are green.
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Returns:
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The last known good revision (as an integer) or None if there
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is no green revision in the database.
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Implementation note: The data store fetches stuff as we go, so we won't
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read in the whole status table unless the LKGR is right at the end or
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we don't have a LKGR. Bots that are offline do not affect the LKGR
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computation (e.g. they are not considered to be failed).
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"""
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build_status_entries = db.GqlQuery('SELECT * '
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'FROM BuildStatusData '
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'ORDER BY revision DESC ')
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first_entry = _get_first_entry(build_status_entries)
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if first_entry is None:
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# No entries => no LKGR
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return None
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current_lkgr = first_entry.revision
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statuses_for_current_lkgr = [first_entry.status]
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for entry in build_status_entries:
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if current_lkgr == entry.revision:
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statuses_for_current_lkgr.append(entry.status)
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else:
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# Starting on new revision, check previous revision.
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if _all_ok(statuses_for_current_lkgr):
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# All bots are green; LKGR found.
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return current_lkgr
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else:
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# Not all bots are green, so start over on the next revision.
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current_lkgr = entry.revision
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statuses_for_current_lkgr = [entry.status]
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if _all_ok(statuses_for_current_lkgr):
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# There was only one revision and it was OK.
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return current_lkgr
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# There are no all-green revision in the database.
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return None
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