thanks!
it works!
2013/12/28 Massimo Di Pierro <massimo.dipie...@gmail.com>
> try replace
>
> Field.Virtual('anexos', lambda row: "<a href=''>Teste_%s</a>" % row.parte.
> id)
>
> with
>
> Field.Virtual('anexos', lambda row: XML("<a href=''>Teste_%s</a>" % row.
> parte.id))
> or
> Field.Virtual('anexos', lambda row: A('Teste_%s'" % row.parte.id
> ,_href=''))
>
>
> but you may also want to consider using this instead:
>
> db.parte.id.label = "anexos"
> db.parte.id.represent = lambda id,row: lambda row: A('Teste_%s'" % id
> ,_href=''))
>
> On Friday, 27 December 2013 12:23:08 UTC-6, Diego Tostes wrote:
>>
>> Hi,
>>
>> I am trying to create a virtualfield as a link like the this code:
>>
>> http://pastebin.com/c2wZqECH
>>
>> But when i create a SQL.grid this virtualfield is displayed as a string.
>>
>> Is it possible to create a link ?
>>
>> Rgds,
>>
>> Diego
>>
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