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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