Have you tried adding a format to your tables and defining your tables
using objects instead of 'reference...'?
I'm no web2py expert by any means, but I had a similar situation and when I
changed the reference for my foreign id to db.other_table instead of
'reference other_table,' the format defined for that table was used instead
of the id.
Something like this:
db.define_table('HumanLanguage',
Field('languageName','string'),
Field('forumLocations','string'),
Field('comments','string'),
format = '%(languageName)s'), # define a format
auth.signature)
db.HumanLanguage.languageName.requires = IS_NOT_EMPTY()
##
## Define DictionaryReferenceModel and WordReferenceModel above here
## DictionaryReferenceModel, WordReferenceModel should have a format
##
## 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',db.HumanLanguage ), # changed to object
Field('DictionaryReferenceModelID', db.DictionaryReferenceModel), #
changed to object
format = '%(dictionaryName)s', # define format
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'))
db.define_table('Word',
Field('wordName','string'),
Field ('definition','string'),
Field('languageID', db.HumanLanguage ), # changed to object
Field('dictionaryTypeID',db.DictionaryType), # changed to object
Field('wordReferenceModelID',db.WordReferenceModel), # changed to object
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')))
On Friday, February 22, 2013 2:48:19 PM UTC-6, Alex Glaros wrote:
>
> thanks for offering to take a look.
>
> Screen shots here:
> https://docs.google.com/document/d/1e517YRN8FK08vqsVWmp1UuZ0LLevCNlkrhWyCkBQ7uQ/edit?usp=sharing
>
> Alex
>
> On Friday, February 22, 2013 12:16:39 PM UTC-8, Jim S wrote:
>>
>> Sorry, I'm kinda lost in what you're trying to do. Can you show it in
>> code or screen shots?
>>
>> -Jim
>>
>>
>> On Fri, Feb 22, 2013 at 12:08 PM, Alex Glaros <alexg...@gmail.com> 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 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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