Add this line before creating your smartgrid:
db.Word.dictionaryTypeID.represent = lambda s,r: s.dictionaryName
Does that help? It worked for the trimmed down model I made...
-Jim
On Thursday, February 21, 2013 9:16:39 PM UTC-6, Alex Glaros wrote:
>
> db.define_table('HumanLanguage',Field('languageName','string'),Field('forumLocations','string'),Field('comments','string'),
>
> auth.signature)
> db.HumanLanguage.languageName.requires = IS_NOT_EMPTY()
>
> db.define_table('Word',Field('wordName','string'), Field ('definition',
> 'string'), Field('languageID','reference
> HumanLanguage'),Field('dictionaryTypeID','reference
> DictionaryType'),Field('wordReferenceModelID','reference
> WordReferenceModel'), Field('comments','string'), auth.signature)
> db.Word.languageID.requires = IS_IN_DB(db, 'HumanLanguage.id',
> '%(languageName)s',zero=T('choose one'))
> db.Word.dictionaryTypeID.requires = IS_IN_DB(db, 'DictionaryType.id',
> '%(dictionaryName)s',zero=T('choose one'))
> db.Word.wordName.requires = IS_NOT_EMPTY()
> db.Word.wordReferenceModelID.requires = IS_NULL_OR(IS_IN_DB(db,
> 'WordReferenceModel.id', '%(wordID)s',zero=T('choose one')))
>
> ## Uses English language as standard connector between all languages.
> WordID below points to the English version of the word that is the
> standard. The reason "Is_Null" clause is there is because the first time
> the word is encountered, it won't be in the English dictionary
> db.define_table('WordReferenceModel',Field('wordID','reference
> Word'),Field('dictionaryTypeID','reference DictionaryType'),
> Field('picture', 'upload', default=''),Field('comments','string'),
> auth.signature)
> db.WordReferenceModel.wordID.requires = IS_NOT_EMPTY()
> db.WordReferenceModel.wordID.requires = IS_IN_DB(db, 'Word.id',
> '%(wordName)s',zero=T('choose one'))
> db.WordReferenceModel.dictionaryTypeID.requires = IS_IN_DB(db,
> 'DictionaryType.id', '%(dictionaryName)s',zero=T('choose one'))
> ## dictionary_type_query = (db.DictionaryType.dictionaryName=='English')
> ## /* need this too for wordReferenceModel */
>
> ## Dictionary type means what category of dictionary is it? Medical,
> computer,etc. There is one for each language.
> db.define_table('DictionaryType',Field('dictionaryName','string'),Field('comments','string'),
>
> Field('languageID','reference HumanLanguage'),
> Field('DictionaryReferenceModelID', 'reference
> DictionaryReferenceModel'), auth.signature)
> db.DictionaryType.dictionaryName.requires = IS_NOT_EMPTY()
> db.DictionaryType.languageID.requires = IS_IN_DB(db, 'HumanLanguage.id',
> '%(languageName)s',zero=T('choose one'))
> db.DictionaryType.DictionaryReferenceModelID.requires = IS_IN_DB(db,
> 'DictionaryReferenceModel.id', '%(DictionaryTypeID)s',zero=T('choose one'))
>
> ## Uses English dictionary type as standard connector between all
> dictionary types. DictionaryType.id points to the English DictionaryType.id
> db.define_table('DictionaryReferenceModel',
> Field('DictionaryTypeID','reference
> DictionaryType'),Field('comments','string'),
> auth.signature)
> db.DictionaryReferenceModel.DictionaryTypeID.requires = IS_NOT_EMPTY()
>
> db.define_table('Synonyms',Field('synonymName','string'),Field('wordID','reference
>
> Word'),Field('comments','string'),
> auth.signature)
> db.Synonyms.synonymName.requires = IS_NOT_EMPTY()
> db.Synonyms.wordID.requires = IS_NOT_EMPTY()
> db.Synonyms.wordID.requires = IS_IN_DB(db, 'Word.id',
> '%(Word)s',zero=T('choose one'))
>
> db.define_table('PublicComments',Field('wordID','reference
> Word'),Field('comments','string'),
> auth.signature)
> db.PublicComments.comments.requires = IS_NOT_EMPTY()
> db.PublicComments.wordID.requires = IS_NOT_EMPTY()
> db.PublicComments.wordID.requires = IS_IN_DB(db, 'Word.id',
> '%(wordName)s',zero=T('choose one'))
>
>
> On Thursday, February 21, 2013 6:10:56 PM UTC-8, Jim S wrote:
>>
>> Can you show the model code?
>>
>> -Jim
>>
>> On Thursday, February 21, 2013 7:35:52 PM UTC-6, Alex Glaros wrote:
>>>
>>> Instead of *db.Word.dictionaryTypeID* displaying (which is a foreign
>>> key in db.Word), I’d like a value from the foreign table to appear, i.e.,
>>> DictionaryType. dictionaryName.
>>>
>>>
>>> In the example below, when user cascades down to the Word table, it only
>>> shows db.Word.dictionaryTypeID but I’d like to add the corresponding
>>> DictionaryType. dictionaryName value to it.
>>>
>>>
>>> def search_lang():
>>>
>>> grid = SQLFORM.smartgrid(db.HumanLanguage,
>>> linked_tables=['Word','DictionaryType'], fields = [db.HumanLanguage.id,
>>> db.HumanLanguage.languageName, db.Word.id, db.Word.wordName,
>>> db.Word.definition, db.DictionaryType.dictionaryName,
>>>
>>> *db.Word.dictionaryTypeID*],
>>>
>>> user_signature=False)
>>>
>>> return dict(grid=grid)
>>>
>>>
>>> id<http://127.0.0.1:8000/tech_dictionary/default/search_lang/HumanLanguage/Word.languageID/1?keywords=&order=Word.id>
>>>
>>>
>>> Wordname<http://127.0.0.1:8000/tech_dictionary/default/search_lang/HumanLanguage/Word.languageID/1?keywords=&order=Word.wordName>
>>>
>>>
>>> Definition<http://127.0.0.1:8000/tech_dictionary/default/search_lang/HumanLanguage/Word.languageID/1?keywords=&order=Word.definition>
>>>
>>>
>>> Dictionarytypeid<http://127.0.0.1:8000/tech_dictionary/default/search_lang/HumanLanguage/Word.languageID/1?keywords=&order=Word.dictionaryTypeID>
>>> 1 beaker glass jar 1
>>>
>>>
>>> Want to replace the "1" under Dictionarytypeid with value from foreign
>>> table.
>>>
>>> Thanks,
>>>
>>> Alex Glaros
>>>
>>
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