I've added another represent HumanLanguage "name" to represent
HumanLanguage "ID" in the Word table, and it works correctly. Same table,
same everything, so I will see if there's any differences in what I'm doing.
Alex
On Friday, February 22, 2013 10:08:04 AM UTC-8, Alex Glaros wrote:
>
> 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 (result of a query) 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> 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> 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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