use the links argument
links = [
...........
dict(header='additional buttons', body=lambda row: BUTTON(.....) if
row.needs_button == 0 else ''
)
]
On Thursday, June 11, 2015 at 12:45:29 AM UTC+2, LoveWeb2py wrote:
>
> I should have been more specific. I meant I want to create a SQLFORM.grid
> button and only have it show if a row.id value is present
>
> On Wednesday, June 10, 2015 at 5:42:33 PM UTC-4, LoveWeb2py wrote:
>>
>> I'd like to create a SQLFORM button, but only show the button if a value
>> is present in the cell. Is this possible? I'm thinking of creating a
>> function that will check to see if each record has a value present in the
>> field
>>
>> for example:
>>
>> define_table('my_sqldata',
>> Field('name',
>> Field('address',
>> migrate=False)
>>
>> if my_sqldata.address != '' then show icon in SQLFORM else: show nothing
>>
>> Is this possible?
>>
>
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