Ticket #3163: meta-option-create_db_table.diff

django/core/management.py

@@ -474 +474 @@
                 print "Processing %s.%s model" % (app_name, model._meta.object_name)
             if model._meta.db_table in table_list:
                 continue
+            if model._meta.create_db_table==False:
+                continue
             sql, references = _get_sql_model_create(model, seen_models)
             seen_models.add(model)
             created_models.add(model)

django/db/models/options.py

@@ -13 +13 @@
 
 DEFAULT_NAMES = ('verbose_name', 'db_table', 'ordering',
                  'unique_together', 'permissions', 'get_latest_by',
-                 'order_with_respect_to', 'app_label')
+                 'order_with_respect_to', 'app_label', 'create_db_table')
 
 class Options(object):
     def __init__(self, meta):
@@ -33 +33 @@
         self.has_auto_field = False
         self.one_to_one_field = None
         self.parents = []
+        self.create_db_table = True
 
     def contribute_to_class(self, cls, name):
         cls._meta = self

docs/model-api.txt

@@ -928 +928 @@
 Here's a list of all possible ``Meta`` options. No options are required. Adding
 ``class Meta`` to a model is completely optional.
 
+``create_db_table``
+-------------------
+
+If you don't want this model to be created as a table in your database (when 
+calling ``manage.py syncdb``), set this to ``False``.
+
+    create_db_table = False
+
+If this isn't given, Django will assume ``True`` and ``manage.py syncdb`` will
+create this model as a table in your database.
+
+This is useful, when you are using DB views which you create via
+raw SQL but still want to be able to use the view as a model from within your 
+application.
+
 ``db_table``
 ------------