Anthony has created a working example. That and his analysis is listed
below:
Massimo's solution will work if the query involves only the TaxonomyDetail
table, but your grid involves a join, so it is necessary to include the
tablename (i.e., record.TaxonomyDetail.objectID). But then that will
if you use Massimo's suggested code:
db.TaxonomyDetail.objectID.represent = lambda id, record:A(record.objectID,
_target = _blank, _href = URL('manage_object_super_type',id))
Then in ' manage_object_super_type' do something like this:
def manage_object_super_type(): ##
data =
need just a little more help to get me over the hump
can anyone use the examples of the code and actually make it work?
what happens is that the target record in the target database is never
isolated and displayed, instead all records are displayed.
thanks,
Alex
On Wednesday, July 31, 2013
Should be:
db.TaxonomyDetail.objectID.represent = lambda id, record:A(record.objectID,
_target = _blank, _href = URL('manage_object_super_type',id))
On Tuesday, 30 July 2013 12:25:58 UTC-5, Alex Glaros wrote:
Okay...I have this so far
db.TaxonomyDetail.objectID.represent = lambda
it looks almost right Massimo, but I get this error
'Row' object has no attribute 'objectID'
On Wednesday, July 31, 2013 12:29:48 AM UTC-7, Massimo Di Pierro wrote:
Should be:
db.TaxonomyDetail.objectID.represent = lambda id, record:A(record.
objectID, _target = _blank, _href =
Here is how the current version looks that generates the row has no attrib.
error:
def manage_taxonomy_detail():
db.TaxonomyDetail.objectID.represent = lambda id,
record:A(record.objectID, _target = _blank, _href =
URL('manage_object_super_type',id))
query = ((db.Taxonomy.id ==
Hallo Alex,
On 30 July 2013 19:25, Alex Glaros alexgla...@gmail.com wrote:
Okay...I have this so far
db.TaxonomyDetail.objectID.represent = lambda ObjectSuperType,record
: A(db.TaxonomyDetail.objectID, _target = _blank, _href =
'manage_object_super_type')
But
1. record:
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,
any hints would be appreciated!
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
Here is an example of what I did yesterday. Maybe that will give you a hint:
Field(url, represent = lambda x, record: A('click here for website',
_target = _blank,
_href = x)) ,
In the grid the url column shows a link.
Regards
Johann
--
Because experiencing your loyal
can you please map out a descriptions of the parms and where they come from?
What is the x, record, A?
thanks!
Alex
On Tuesday, July 30, 2013 5:23:50 AM UTC-7, Johann Spies wrote:
Here is an example of what I did yesterday. Maybe that will give you a
hint:
Field(url, represent = lambda
Okay...I have this so far
db.TaxonomyDetail.objectID.represent = lambda ObjectSuperType,record
: A(db.TaxonomyDetail.objectID, _target = _blank, _href =
'manage_object_super_type')
But
1. record: A(db.TaxonomyDetail.objectID displays literally instead of the
field contents
2. Link
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