Hi thanks for the reply.
What I was trying to do in the original code was return the data to a div
not to a specified page, so what you are saying is correct.
The part about the table, is that it seems to only iterate over the first
record, maybe the code is different for tables, because it has to iterate
for each row and data.
Not sure how to solve at the moment, I'm new to web2py, I'm use to MVC in
Php with codeigniter, so its a bit off a learning curve, especially with
jquery and ajax.
With codeigniter I could send back, Jquery back through with html, which
was then embedded into the page and would just work.
So on to the next part of learning python and web2py.
Thanks for your reply.
On Sunday, 22 March 2015 20:45:18 UTC, Anthony wrote:
>
> Just to clarify the original problem, when a controller action returns a
> dictionary (as in your original code), web2py assumes it should then look
> for an execute a view. If a dedicated view does not exist for the action,
> it will default to generic.html (though this is disabled by default, except
> for local requests in the welcome app). So, you should either create a view
> that generates the HTML you want based on what is returned in the
> dictionary, or you should not return a dictionary.
>
> Instead of a dictionary, you can directly return HTML as a string (which
> is what your new code does) by applying the .xml() method to the UL helper.
> However, you don't actually need to apply the .xml() method, because if you
> return an HTML helper object, web2py will automatically serialize it to a
> string for you.
>
> As for returning an HTML table rather than a UL, there's no reason you
> couldn't do that instead. Just create a TABLE object and return that.
>
> Anthony
>
> On Saturday, March 21, 2015 at 8:17:06 PM UTC-4, Garry Smith wrote:
>>
>> Hi with your help I have manage to get the links working using the
>> following code:-
>>
>> def artist():
>>
>> rows = db().select(db.media.artist,db.media.id, groupby =
>> db.media.artist)
>>
>> links = map(lambda row:A(row.artist,
>> callback=URL('album',vars=dict(id=row.id)), target='album'), rows)
>>
>> test = UL([LI(link) for link in links]).xml()
>>
>> return test
>>
>> This now returns the links in a list.
>> Would off like it in a table, but it only returns first record, not sure
>> how I could itterate to return the table rows.
>>
>> Thanks again
>>
>> On Saturday, 21 March 2015 18:19:02 UTC, Garry Smith wrote:
>>>
>>> Hi
>>> I'm trying to return the some links to a div. I have made links with
>>> some information from the database and using the A htlm helper.
>>> The problem I am having is getting the data back correctly, I am getting
>>> the links back correctly, but I am being sent a full web page with the
>>> links, which is then put in the div.
>>>
>>> *VIEW*
>>> this is link which is sent to the artist page, which gets all the
>>> distinct artist.
>>>
>>> {{=A('click' , callback=URL('artist'), target="artist")}}
>>>
>>> *CONTROLLER*
>>>
>>> This is the controller that makes the links and returns to the div as
>>> specified in above link.
>>>
>>> def artist():
>>> rows = db().select(db.media.artist,db.media.id, groupby =
>>> db.media.artist)
>>>
>>> links = []
>>>
>>> for row in rows:
>>> a = str(row.id)
>>> b = 'album?id='
>>> c = b+a
>>> l = row.artist
>>> f = A(l , callback=URL(c), target="album")
>>> links.append(f)
>>>
>>> return dict(links=links)
>>>
>>> I have tried diffrent ways to solve this, but to no avail, all I want is
>>> to return the links to div.
>>>
>>> What i am trying to do is rewrite a media player, which I wrote in php a
>>> few years ago, but I am having few issues writing a returning links on the
>>> sly.
>>>
>>> Thanks in advance for any help.
>>>
>>>
>>>
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
---
You received this message because you are subscribed to the Google Groups
"web2py-users" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
For more options, visit https://groups.google.com/d/optout.
