6c89577f1d
The following patch renames the rest of q_exc to n_exc which were left when quantum was renamed to neutron. used: find . -name "*.py" -print | xargs sed -i 's/q_exc/n_exc/g' Change-Id: I04041030694b28c400b6c9c552433f2eaa4a37ba Closes-bug: #1287462
109 lines
4.4 KiB
Python
109 lines
4.4 KiB
Python
# vim: tabstop=4 shiftwidth=4 softtabstop=4
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# Copyright 2011 OpenStack Foundation.
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# All Rights Reserved.
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#
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# Licensed under the Apache License, Version 2.0 (the "License"); you may
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# not use this file except in compliance with the License. You may obtain
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# a copy of the License at
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#
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# http://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS, WITHOUT
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# WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the
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# License for the specific language governing permissions and limitations
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# under the License.
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import sqlalchemy
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from sqlalchemy.orm.properties import RelationshipProperty
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from neutron.common import exceptions as n_exc
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from neutron.openstack.common import log as logging
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LOG = logging.getLogger(__name__)
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def paginate_query(query, model, limit, sorts, marker_obj=None):
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"""Returns a query with sorting / pagination criteria added.
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Pagination works by requiring a unique sort key, specified by sorts.
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(If sort keys is not unique, then we risk looping through values.)
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We use the last row in the previous page as the 'marker' for pagination.
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So we must return values that follow the passed marker in the order.
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With a single-valued sort key, this would be easy: sort_key > X.
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With a compound-values sort key, (k1, k2, k3) we must do this to repeat
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the lexicographical ordering:
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(k1 > X1) or (k1 == X1 && k2 > X2) or (k1 == X1 && k2 == X2 && k3 > X3)
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The reason of didn't use OFFSET clause was it don't scale, please refer
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discussion at https://lists.launchpad.net/openstack/msg02547.html
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We also have to cope with different sort directions.
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Typically, the id of the last row is used as the client-facing pagination
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marker, then the actual marker object must be fetched from the db and
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passed in to us as marker.
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:param query: the query object to which we should add paging/sorting
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:param model: the ORM model class
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:param limit: maximum number of items to return
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:param sorts: array of attributes and direction by which results should
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be sorted
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:param marker: the last item of the previous page; we returns the next
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results after this value.
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:rtype: sqlalchemy.orm.query.Query
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:return: The query with sorting/pagination added.
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"""
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if not sorts:
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return query
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# A primary key must be specified in sort keys
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assert not (limit and
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len(set(dict(sorts).keys()) &
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set(model.__table__.primary_key.columns.keys())) == 0)
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# Add sorting
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for sort_key, sort_direction in sorts:
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sort_dir_func = sqlalchemy.asc if sort_direction else sqlalchemy.desc
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try:
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sort_key_attr = getattr(model, sort_key)
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except AttributeError:
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# Extension attribute doesn't support for sorting. Because it
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# existed in attr_info, it will be catched at here
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msg = _("%s is invalid attribute for sort_key") % sort_key
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raise n_exc.BadRequest(resource=model.__tablename__, msg=msg)
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if isinstance(sort_key_attr.property, RelationshipProperty):
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msg = _("The attribute '%(attr)s' is reference to other "
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"resource, can't used by sort "
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"'%(resource)s'") % {'attr': sort_key,
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'resource': model.__tablename__}
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raise n_exc.BadRequest(resource=model.__tablename__, msg=msg)
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query = query.order_by(sort_dir_func(sort_key_attr))
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# Add pagination
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if marker_obj:
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marker_values = [getattr(marker_obj, sort[0]) for sort in sorts]
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# Build up an array of sort criteria as in the docstring
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criteria_list = []
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for i, sort in enumerate(sorts):
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crit_attrs = [(getattr(model, sorts[j][0]) == marker_values[j])
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for j in xrange(i)]
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model_attr = getattr(model, sort[0])
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if sort[1]:
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crit_attrs.append((model_attr > marker_values[i]))
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else:
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crit_attrs.append((model_attr < marker_values[i]))
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criteria = sqlalchemy.sql.and_(*crit_attrs)
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criteria_list.append(criteria)
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f = sqlalchemy.sql.or_(*criteria_list)
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query = query.filter(f)
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if limit:
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query = query.limit(limit)
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return query
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