[Grok-dev] Re: Recipe - How to create a vocabulary that allows the display of non-ASCII characters

[Philipp von Weitershausen]

Sebastian Ware wrote:
> I just want to pitch my idea of what a recipe might look like. In this 
> case showing how to create a vocabulary that displays unicode titles.
> 
> Mvh Sebastian
> 
> ** Recipe - How to create a vocabulary that allows the display of 
> non-ASCII characters **
> Keywords: Recipe, Vocabulary, Unicode, Global Utility
> 
> How to create a vocabulary that displays non-ASCII titles. Useful for 
> use with zope.schema.Choice fields. The title of each term that is to be 
> displayed can be Unicode strings.
> 
> [code:vocabulary.py]
> from zope import schema
> from zope.schema.vocabulary import SimpleVocabulary
> from hurry import query
> 
> class VocabularySource(object):
>     
>     def __call__(self, context):
>         # Get a list of objects (I am using hurry.query to search a 
> catalog)
>         theQuery = query.Eq(('workflow_catalog', 'workflow_state'), 
> interfaces.PUBLISHED)
>         result = query.query.Query().searchResults(theQuery)
>         # For each object, add it as a list of terms 
> (zope.schema.vocabulary.SimpleTerm)
>         # creating each term with SimpleVocabulary.createTerm(value, 
> token, title). This
>         # allows the title to be unicode, the token has to be ASCII (and 
> can be the
>         # same as value).
>         theTerms = []
>         for item in result:
>             theTerms.append(SimpleVocabulary.createTerm(item.__name__,
>                                                         item.iso_code,
>                                                         item.title))
>         return  SimpleVocabulary(theTerms)
> 
> # Register the vocabulary as a global utility.
> grok.global_utility(VocabularySource,
>                     provides=schema.interfaces.IVocabularyFactory,
>                     name=u'Published Objects')
> [/code]

This is weird. You're registering this as a utility providing 
IVocabularyFactory, but the object doesn't actually implement it. This 
can actually be implemented much more easily:

   class MyVocabularyFactory(grok.GlobalUtility):
       grok.implements(IVocabularyFactory)
       grok.name(u'Published Objects')

       def __call__(self, context):
           ...

No further registration necessary.


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