问题
I'm developing an application with Google App Engine and stumbled across the following scenario, which can perhaps be described as "MVP-lite".
When modeling many-to-many relationships, the standard property to use is the ListProperty. Most likely, your list is comprised of the foreign keys of another model.
However, in most practical applications, you'll usually want at least one more detail when you get a list of keys - the object's name - so you can construct a nice hyperlink to that object. This requires looping through your list of keys and grabbing each object to use its "name" property.
Is this the best approach? Because "reads are cheap", is it okay to get each object even if I'm only using one property for now? Or should I use a special property like tipfy's JsonProperty to save a (key, name) "tuple" to avoid the extra gets?
回答1:
Though datastore reads are comparatively cheaper datastore writes, they can still add significant time to request handler. Including the object's names as well as their foreign keys sounds like a good use of denormalization (e.g., use two list properties to simulate a tuple - one contains the foreign keys and the other contains the corresponding name).
If you decide against this denormalization, then I suggest you batch fetch the entities which the foreign keys refer to (rather than getting them one by one) so that you can at least minimize the number of round trips you make to the datastore.
回答2:
When modeling one-to-many (or in some cases, many-to-many) relationships, the standard property to use is the ListProperty.
No, when modeling one-to-many relationships, the standard property to use is a ReferenceProperty, on the 'many' side. Then, you can use a query to retrieve all matching entities.
Returning to your original question: If you need more data, denormalize. Store a list of titles alongside the list of keys.
来源：https://stackoverflow.com/questions/4246074/use-a-listproperty-or-custom-tuple-property-in-app-engine