Wouldn't a LOAD/Ajax combo work better in this case and then just include
it in your page?
On Monday, March 23, 2015 at 8:41:18 AM UTC-4, Garry Smith wrote:
>
> 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.
>>>>
>>>>
>>>>
--
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