not sure what the function syntax should be. I want to select and display
one record.
Here is the code so far, which invokes a compile error "invalid syntax"
db.TaxonomyDetail.objectID.represent = lambda id,
record:A(record.TaxonomyDetail.objectID, _target = "_blank", _href =URL(r =
request, C='default/manage_object_super_type', f=select, vars = dict(id =
str(ObjectSuperType.objectID)))
On Sunday, July 28, 2013 4:55:53 AM UTC-7, Alex Glaros wrote:
>
> How do I create a hyperlink back to parent record ObjectSuperType.id from
> displayed field db.TaxonomyDetail.objected below?
>
> The link would take user to the controller that displays parent table
> ObjectSuperType. How do I pass the parms to the other controller?
>
> Here is the controller for the detail record. Look for field
> db.TaxonomyDetail.objectID. I’d like to list its value, and also link from
> it to the parent ObjectSuperType.id.
>
> def manage_taxonomy_detail():
> query = ((db.Taxonomy.id == db.TaxonomyData.taxonomyID) &
> (db.TaxonomyDetail.taxonomyDataID==db.TaxonomyData.id))
> taxonomyList =
> SQLFORM.grid(query,create=True,editable=True,deletable=True,
> details=True,links_in_grid=True,
> paginate=10,
> fields=[db.Taxonomy.taxonomyLongName,
> db.TaxonomyData.taxonomyDataName, db.TaxonomyDetail.objectID,
> db.TaxonomyDetail.partyID])
> return dict(taxonomyList = taxonomyList)
>
> Here is the db.TaxonomyDetail table:
>
> db.define_table('TaxonomyDetail', ## Allows user to assign a taxonomy
> field to any object
> Field('objectID', 'reference ObjectSuperType'),
> Field('taxonomyID','reference Taxonomy'),
> Field('taxonomyDataID','reference TaxonomyData'),
> Field('partyID','reference Party', label='Data Owner'),
> Field('publicAccessLevel','reference PublicAccessLevelLookupTable'),
> Field('taxonomyDetailComments', 'string')) ## by person entering the data
> ## ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> db. TaxonomyDetail.objectID.requires = IS_IN_DB(db, 'ObjectSuperType.id',
> '%(objectDisplayName)s',zero=T('choose one'))
> db. TaxonomyDetail.objectID.represent = lambda id,row:
> db.ObjectSuperType(id).objectDisplayName
> db. TaxonomyDetail.taxonomyID.requires = IS_IN_DB(db, 'Taxonomy.id',
> '%(taxonomyLongName)s',zero=T('choose one'))
> db. TaxonomyDetail.taxonomyDataID.requires = IS_IN_DB(db,
> db.TaxonomyData.id, '%(taxonomyDataName)s',zero=T('choose one'))
> db. TaxonomyDetail.publicAccessLevel.requires = IS_IN_DB(db,
> db.PublicAccessLevelLookupTable.publicAccessCode,
> '%(publicAccessDescription)s',zero=T('choose one'), orderby=
> db.PublicAccessLevelLookupTable.id)
> db. TaxonomyDetail.partyID.requires = IS_IN_DB(db, db.Party.id,
> '%(displayName)s',zero=T('choose one'))
> db. TaxonomyDetail.partyID.represent = lambda id,row:
> db.Party(id).displayName
>
> Here is the controller for the parent ObjectSuperType. How do I pass the
> parms to it?
>
> def manage_object_super_type(): ##
> grid = SQLFORM.smartgrid(db.ObjectSuperType),
> return dict(grid=grid)
>
> thanks,
>
> Alex Glaros
>
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