It worked!
thanks Mark and everyone that looked at this.
Alex Glaros
On Monday, February 25, 2013 1:30:21 PM UTC-8, Mark wrote:
>
> The grandparent, parent, child relationships maybe too complex for
> smartgrid, or maybe a bug in smartgrid? Try to use the IF statement to
> differentiate the relationships:
>
> if 'DictionaryType.languageID' in request.args:
> db.Word.dictionaryTypeID.represent = lambda id:
> db.DictionaryType(id).dictionaryName
> else:
> db.Word.dictionaryTypeID.represent = lambda s,r: s.dictionaryName
>
>
> On Friday, February 22, 2013 1:08:04 PM UTC-5, 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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>>>>>
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