There is a grandparent (HumanLanguage), parent (DictionaryType), child
(Word), relationship. If I go straight from the grandparent to the child,
"represent" works.
(First I click on HumanLangages, then I click on Word)
If I go from the parent to the child, I get this message: <type
'exceptions.TypeError'> <lambda>() takes exactly 2 arguments (1 given)
(First I click on HumanLangages, then I click on DictionaryType, then, I
click on Word)
If I reduce the number of parms to db.Word.dictionaryTypeID.represent =
lambda id: db.DictionaryType(id).dictionaryName, then it works if I go from
the parent to the child, but if I go from the grandparent to the child, I
get <type 'exceptions.TypeError'> <lambda>() takes exactly 1 argument (2
given).
If a join was allowed, it would solve the problem but I think they are not
supported in smartgrid.
Any ideas would be much appreciated,
Thanks,
Alex
On Friday, February 22, 2013 5:24:48 AM UTC-8, Jim S wrote:
>
> Depends on whether or not you want that behavior globally or not.
> Remember, your model gets executed on every request. So, if you want that
> behavior globally I'd put it there. If you just need it for one instance,
> put it in the controller.
>
> -Jim
>
>
> On Fri, Feb 22, 2013 at 12:20 AM, Alex Glaros <alexg...@gmail.com<javascript:>
> > wrote:
>
>> It worked Jim
>>
>> thanks so much for going through the process of writing and testing the
>> code.
>>
>> I'm new to web2py, is there any preference as to where that statement
>> goes: controller or model?
>>
>> much appreciated,
>>
>> Alex
>>
>>
>> On Thu, Feb 21, 2013 at 8:31 PM, Jim S <j...@qlf.com <javascript:>>wrote:
>>
>>> 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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