I understand what you're saying, but that's all I want to do. I don't
want to remove it from the datastore, just from the list, so I can
perform some temporary operations on somelist -anotherObject

This makes sense to me now...time to revise a bunch of code. Thanks
for the help.

On Oct 20, 3:36 pm, djidjadji <[EMAIL PROTECTED]> wrote:
> If you want to remove one item from the fetched list then
> somelist = theObject.subCategories.fetch(1000)
> somelist.remove(anotherObject)
> will not work because the 'anotherObject' is another python
> representation of the object in the datastore, fetch creates NEW
> instances for the objects it retrieves. Thus the remove will not work.
> somelist = [theObject.subCategories]
> will result in a list which has as its first element a Query object.
> somelist = theObject.subCategories.fetch(1000)
> already returns a list with object instance.
>
> 2008/10/20 Alex Vartan <[EMAIL PROTECTED]>:
>
>
>
> > get will only return a single object as far as I know. I want to fetch
> > all of them. So must I use
> > somelist = theObject.subCategories.fetch()
> > or will somelist = [theObject.subCategories] work as well, explicitly
> > assigning somelist as a list.
>
> > On Oct 20, 9:33 am, djidjadji <[EMAIL PROTECTED]> wrote:
> >> This should work, use get() from the Query object
>
> >> somelist = theObject.subCategories.get()
> >> somelist.remove(anotherObject)
>
> >> 2008/10/20 Alex Vartan <[EMAIL PROTECTED]>:
>
> >> > Hi,
>
> >> > Simple question, but want to make sure I get this right. The docs say:
>
> >> > "collection_name is the name of the property to give to the referenced
> >> > model class whose value is a Query for all entities that reference the
> >> > entity. If no collection_name is set, then modelname_set (with the
> >> > name of the model in lowercase letters and "_set" added) is used."
>
> >> > I've got code that reads like this in one of my classes:
>
> >> >        parentCategory =
> >> > db.SelfReferenceProperty(collection_name='subCategories')
>
> >> > Then I need to make a copy of the resulting list, but remove an item
> >> > for temporary processing:
>
> >> > somelist = theObject.subCategories
> >> > somelist.remove(anotherObject)
>
> >> > WIll this work, without having to call a fetch? Since the docs make it
> >> > seem like "theObject.subCategories" is actually an object of type
> >> > Query.
> >> > Thanks,
> >> > Alex
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