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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