Ok figured out I needed to change that to "as_trees". However the database
suggestion you gave is causing a
"IntegrityError: FOREIGN KEY constraint failed"
error immediately upon trying to put any data into any field. For example
if I put Animals in the title and leave the other field "parent_id" blank
it gives me the error. If I put a number in parent_id with the title an
error, and if I just put a number in the parent_id field also an error.
Any ideas?
Thanks
On Tuesday, January 21, 2014 9:45:27 AM UTC-6, Massimo Di Pierro wrote:
>
> First of all notice that if you use web2py2.8.2 you can do this:
>
> db.define_table('store_catalog',
> Field('title'),
> Field('slug',requires=IS_SLUG(),compute=lambda row:
> IS_SLUG.urlify(row.title)),
> Field('parent_id','reference store_catalog'))
>
> def menu_rec(items): return
> [(x.title,None,URL('action',args=x.slug),menu_rec(x.children)) for x in
> items or []]
> response.menu = menu_rec(db(db.store_catalog).select().as_tree())
>
> And you get a recursive menu with arbitrary depth and a web interface to
> edit it with slugs.
>
> On Tuesday, 21 January 2014 00:11:59 UTC-6, James Burke wrote:
>>
>> Hi,
>>
>> I managed to get one working a little while ago doing this. This has been
>> adapted for your situation, may require some tweaking:
>>
>> _tables.py - model
>> db.define_table('store_catelog')
>>
>>
>> db.define_table('store_catelog',
>> Field('title', 'string'),
>> Field('description', 'text'),
>> Field('parent_catelog', 'reference store_catelog', default=None),
>> format='%(title)s',
>> redefine=True)
>>
>> menu.py - model
>> response.menu = []
>>
>> categories = db(db.store_catelog.parent == None).select(orderby=db.
>> store_catelog.id)
>>
>> for category in categories:
>> response.menu+=[ (T(category.title), True if active_page == str(
>> category.title.lower()) else False, URL('default','page/%s' % (category.
>> title.lower())), [
>> (T(i.title), True if active_page == str(i.id) else False, URL(
>> 'default','page/%s' % (i.id)), []) for i in db((db.stored_category.parent
>> == category.id)).select(db.store_category.ALL) if i.parent == category.id
>> ]) ]
>>
>> The query returns any categories without a parent and adds them to the
>> menu. While doing so looks for any subcategories using the category as a
>> parent.
>>
>> Hope that helps you out.
>>
>> -James
>>
>> On Tuesday, January 21, 2014 9:33:50 AM UTC+13, americ...@gmail.comwrote:
>>>
>>>
>>> Hi all,
>>>
>>> I have been trying for a little over 2 weeks to figure this out...
>>>
>>> I'm trying to generate a menu that drops down, I have been able to
>>> statically create it by overwriting sample app that has the same drop down
>>> menu like the web2py.com site.
>>>
>>> I have this in my db.py:
>>> db.define_table('store_catalog',
>>> Field('maincategory', 'string'),
>>> Field('subcategory', 'string'),
>>> Field('description', 'text'))
>>>
>>>
>>> in my menu.py I have gotten this so far:
>>> response.menu=[]
>>> response.menu.append([T('Catalog'), False, '',
>>> [(T('%s' % menucatchoice['maincategory']), False, 'link',
>>> [(T('%s' % menucatchoice['subcategory']), False,
>>> 'link'),]) for menucatchoice in rows ] ])
>>>
>>> It gets me a drop down menu except that for each subcategory it repeats
>>> adding it to the main category. Let's say there is only 1 main category and
>>> 2 subs that go after that
>>> Catalog (this just shows with the caret next to it as intended)
>>> MainCategory
>>> Sub
>>> Sub
>>> What I have does put the "Catalog" for the first but, what I get is:
>>> MainCategory
>>> (blank)
>>> MainCategory(don't want this it's being repeated)
>>> Subcategory
>>> MainCategory(and this one is also a repeat)
>>> Subcategory
>>>
>>> I have tried to break out the response.menu with so many different
>>> .appends it's not funny. I have also tried adding the "[-1]". This was
>>> the closest I have gotten to what I want it to look like. I'm at an
>>> absolute loss on this, any and all help would be greatly appreciated.
>>>
>>
--
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