Show if =visible when
No dia 04/06/2013 21:51, "Niphlod" <[email protected]> escreveu:
> it works, but a few hiccups with the examples posted....
>
> def index1():
> """ shows bb only if aa is checked """
> db.define_table('thing', Field('aa','boolean'),Field('bb'))
> db.thing.bb.show_if = db.thing.aa==True
> form = SQLFORM(db.thing)
> return locals()
>
> def index2():
> """ shows bb only when aa is not set to "x" """
> db.define_table('thing', Field('aa'),Field('bb'))
>
> db.thing.aa.requires=IS_IN_SET(('x','y','z'))
> db.thing.bb.show_if =
> db.thing.aa!='x'
> form = SQLFORM(db.thing)
> return locals()
>
> def index3():
> """ shows bb only when one types "x" or "y" in aa"""
> db.define_table('thing', Field('aa'),Field('bb'))
> db.thing.bb.show_if = db.thing.aa.belongs(('x','y'))
> form = SQLFORM(db.thing)
> return locals()
>
> def index4():
> """ shows bb only when one types "abcd" in aa"""
> db.define_table('thing', Field('aa'),Field('bb'))
> db.thing.bb.show_if = db.thing.aa.contains('abcd')
> form = SQLFORM(db.thing)
> return locals()
>
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>
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