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Fixed #23605 -- Fixed nested subquery regression

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Added relabeled_clone() method to sql.Query to fix the problem. It
manifested itself in rare cases where at least double nested subquery's
filter condition might target non-existing alias.

Thanks to Trac alias ris for reporting the problem.
This commit is contained in:
Anssi Kääriäinen 2014-10-07 16:07:46 +03:00 • committed by Tim Graham
parent 21e21c7bc2
commit 5c481db295
5 changed files with 77 additions and 3 deletions
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21
django/db/models/sql/query.py
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@ -106,6 +106,10 @@ class Query(object):
# type they are. The key is the alias of the joined table (possibly
# the table name) and the value is JoinInfo from constants.py.
self.alias_map = {}
# Sometimes the query contains references to aliases in outer queries (as
# a result of split_exclude). Correct alias quoting needs to know these
# aliases too.
self.external_aliases = set()
self.table_map = {} # Maps table names to list of aliases.
self.join_map = {}
self.default_cols = True
@ -240,6 +244,7 @@ class Query(object):
obj.model = self.model
obj.alias_refcount = self.alias_refcount.copy()
obj.alias_map = self.alias_map.copy()
obj.external_aliases = self.external_aliases.copy()
obj.table_map = self.table_map.copy()
obj.join_map = self.join_map.copy()
obj.default_cols = self.default_cols
@ -303,6 +308,11 @@ class Query(object):
obj._setup_query()
return obj
def relabeled_clone(self, change_map):
clone = self.clone()
clone.change_aliases(change_map)
return clone
def get_aggregation(self, using, force_subq=False):
"""
Returns the dictionary with the values of the existing aggregations.
@ -774,7 +784,9 @@ class Query(object):
ident = (change_map.get(ident[0], ident[0]),) + ident[1:]
self.join_map[ident] = aliases
for old_alias, new_alias in six.iteritems(change_map):
alias_data = self.alias_map[old_alias]
alias_data = self.alias_map.get(old_alias)
if alias_data is None:
continue
alias_data = alias_data._replace(rhs_alias=new_alias)
self.alias_refcount[new_alias] = self.alias_refcount[old_alias]
del self.alias_refcount[old_alias]
@ -801,6 +813,9 @@ class Query(object):
data = data._replace(lhs_alias=change_map[lhs])
self.alias_map[alias] = data
self.external_aliases = {change_map.get(alias, alias)
for alias in self.external_aliases}
def bump_prefix(self, outer_query):
"""
Changes the alias prefix to the next letter in the alphabet in a way
@ -999,6 +1014,9 @@ class Query(object):
value = value()
elif hasattr(value, 'resolve_expression'):
value = value.resolve_expression(self, reuse=can_reuse)
# Subqueries need to use a different set of aliases than the
# outer query. Call bump_prefix to change aliases of the inner
# query (the value).
if hasattr(value, 'query') and hasattr(value.query, 'bump_prefix'):
value = value._clone()
value.query.bump_prefix(self)
@ -1562,6 +1580,7 @@ class Query(object):
lookup = lookup_class(Col(query.select[0].col[0], pk, pk),
Col(alias, pk, pk))
query.where.add(lookup, AND)
query.external_aliases.add(alias)
condition, needed_inner = self.build_filter(
('%s__in' % trimmed_prefix, query),