Fixed #28863 -- Fixed filter on annotation that contains Q.

django/db/models/sql/where.py

@@ -26,6 +26,7 @@ class WhereNode(tree.Node):
     contains_aggregate attribute.
     """
     default = AND
+    resolved = False
 
     def split_having(self, negated=False):
         """
@@ -108,7 +109,7 @@ class WhereNode(tree.Node):
                 # around the inner SQL in the negated case, even if the
                 # inner SQL contains just a single expression.
                 sql_string = 'NOT (%s)' % sql_string
-            elif len(result) > 1:
+            elif len(result) > 1 or self.resolved:
                 sql_string = '(%s)' % sql_string
         return sql_string, result_params
 
@@ -181,6 +182,11 @@ class WhereNode(tree.Node):
     def is_summary(self):
         return any(child.is_summary for child in self.children)
 
+    def resolve_expression(self, *args, **kwargs):
+        clone = self.clone()
+        clone.resolved = True
+        return clone
+
 
 class NothingNode:
     """A node that matches nothing."""

tests/expressions/tests.py

@@ -73,6 +73,15 @@ class BasicExpressionsTests(TestCase):
             ],
         )
 
+    @unittest.skipIf(connection.vendor == 'oracle', "Oracle doesn't support using boolean type in SELECT")
+    def test_filtering_on_annotate_that_uses_q(self):
+        self.assertEqual(
+            Company.objects.annotate(
+                num_employees_check=ExpressionWrapper(Q(num_employees__gt=3), output_field=models.BooleanField())
+            ).filter(num_employees_check=True).count(),
+            2,
+        )
+
     def test_filter_inter_attribute(self):
         # We can filter on attribute relationships on same model obj, e.g.
         # find companies where the number of employees is greater