That works perfectly! Thank you very much
On Monday, 19 November 2018 14:38:55 UTC, Anthony wrote:
>
> With joins, you have to specify the ID field explicitly:
>
> SQLFORM.grid(..., field_id=db.applics1.id)
>
> Anthony
>
> On Monday, November 19, 2018 at 4:48:14 AM UTC-5, Francis Windram wrote:
>>
>> Hi All,
>>
>> I'm hitting an issue with the selectable field in SQLFORM.grid.
>>
>> I have a controller including a grid generated from a join on three
>> tables as follows:
>>
>> applics_fields = ['first_name', 'last_name', 'email', 'rank', 'nerc_elig'
>> , 'cv', 'coverletter', 'refletter1',
>> 'refletter2']
>> proposals_fields = ['project_title', 'case_partner']
>> pi_fields = ['first_name', 'last_name']
>> fields_to_select = [db.applics1[field] for field in applics_fields] +
>> \
>> [db.proposals[field] for field in proposals_fields
>> ] + \
>> [db.PIs[field] for field in pi_fields]
>>
>> grid = SQLFORM.grid((db.applics1.project == db.proposals.id) & (db.
>> proposals.PI_name == db.PIs.id),
>> csv=False,
>> deletable=False,
>> create=False,
>> details=False,
>> editable=False,
>> fields=fields_to_select,
>> headers={'PIs.last_name': 'PI Last Name',
>> 'PIs.first_name': 'PI First Name'},
>> selectable=[('Download Public PDFs', lambda ids:
>> zip_pdfs(ids, track=1))],
>> paginate=False)
>>
>> This generates the correct grid in the view, however when I select the
>> records I want to pass to zip_pdfs(), the ids passed to the function are
>> those of the PI table, not of the applics1 table.
>>
>> Is there a way to specify from which table it pulls the ids to be
>> selected and passed to zip_pdfs()?
>>
>> Thanks,
>> Francis
>>
>
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