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

[Sebastian Ware]

Works like a charm :)

Mvh Sebastian

24 sep 2007 kl. 18.23 skrev 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.
>
>
> -- 
> http://worldcookery.com -- Professional Zope documentation and  
> training