don't use URL at all, just do:
('TaC', False, A('TaC',
_href='https://static.domain.com/files/Terms_and_Conditions.pdf',
_target="_blank", vars=dict(attachment=True)))
If it gives an error call the .xml() directly on A:
('TaC', False, A('TaC',
_href='https://static.domain.com/files/Terms_and_Conditions.pdf',
_target="_blank", vars=dict(attachment=True)).xml())
2018-02-24 12:18 GMT-05:00 'Annet' via web2py-users <[email protected]>:
> I have the following link in a view:
>
> {{=A('TaC',
> _href='https://static.domain.com/files/Terms_and_Conditions.pdf',
> _target="_blank", vars=dict(attachment=True))}}
>
> I want to convert it to a menu item
>
> ('TaC', False, URL(???))
>
> How do I convert the _href to a URL()
>
>
> Kind regards
>
> Annet
>
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