Re: How to pass a string to a custom filter?

[Mirto Silvio Busico]

Thanks; but I was not clear.
I'll try to explain: the code is not intended to really create the 
breadcrumbs. The code tries to identify the problem.

The problem is that the statement:
    parti=aurl.split('/')
doesn't generate a list

Moreover the statement:
    aurl.count('/') (with the supposed url that is 
"http://localhost/zsite/ancore/";) gives the value 0 (I expected 4)

Worst, I used your suggestion and now the bc assignement line is
    bc= u' - %s - len=%d char= %s %s' % 
(parti[0],len(parti),aurl.count('/'),(isinstance(aurl, basestring)))
and isinstance(aurl, basestring) id evaluated as True (so aurl is a 
string; but split('/') doesn't work)

I don't know what to look for: I receive a string, the string contains 
'/', the string.split('/') doesn't generate the expected 5 elements.
Obvoiusly, the same code pasted in the shell (python manage.py shell) 
works as expected.

Any Hint?

Thanks
    Mirto

P.S. BTW the urls are different from development framework and production:

    * runserver: http://localhost:8000/ancore/
    * apache:     http://localhost/zsite/ancore/


bruno desthuilliers ha scritto:
> On 14 oct, 16:34, Mirto Silvio Busico <[EMAIL PROTECTED]> wrote:
>   
>> Hi all,
>> I'm trying to pass a string to a custom tag to create custom breadcrumbs.
>>
>> I'm using as a base the snippet athttp://www.djangosnippets.org/snippets/656/
>> and using the documentation 
>> fromhttp://docs.djangoproject.com/en/dev/howto/custom-template-tags/
>>
>> But it seems that I'm not able to pass the URL as a string.
>>
>> In ... zsite/ancore/templatetags/mytags.py (zsite is the django project
>> directory) I have:
>> ==============================================================
>> #!/usr/bin/python
>> # encoding: utf-8
>> from django import template
>> from django.utils.safestring import mark_safe
>> from django.utils.translation import ugettext
>> from django.template.defaultfilters import stringfilter
>> from types import *
>>
>> register = template.Library()
>>
>> @register.filter(name='dovesono')
>> @stringfilter
>> def dovesono(aurl):
>>
>>    bc = u''
>>    parti=aurl.split('/')
>>
>>    bc= u' - %s - len=%d char= %s %s' %
>> (parti[0],len(parti),aurl.count('/'),(type(aurl) == StringTypes))
>>     
>
> StringTypes is a tuple (<type 'str'>, <type 'unicode'>). So you want :
>   type(aurl) in StringTypes
>
> which is better written:
>   isinstance(aurl, basestring)
>
> but anyway, given your example use case, this test is more than
> probably more than useless...
>
>   
>>    for i in range(len(parti)):
>>       bc=bc+' &raquo '+parti[i]
>>     
>
> The canonical way to iterate over a sequence is:
>
>      for elt in sequence:
>         do_whatever_with(elt)
>
> but you don't need such a low-level construct here - just use
> str.join(seq):
>
>      bc = u' &raquo '.join(parti)
>
> Note that since request.path is an absolute path (iow with leading
> slash), and may have a trailing one too, you may want to get rid of
> both before splitting the request.path, ie:
>
>     parti = aurl.strip('/').split('/')
>
>     # code here
>
>   
>>    return mark_safe(bc)
>> ==============================================================
>>
>> In the calling template I have:
>> ==============================================================
>> {% block breadcrumbs %}
>>     {% load mytags %}
>>     <div class="breadcrumbs">
>>     {{ request.get_full_path|dovesono }}
>>     
>
> request.get_full_path will also gives you the querystring part if
> there's one. I think you want request.path instead.
>
>   
>>     </div>
>> {% endblock %}
>> ==============================================================
>>
>> and the generated html is:
>> ==============================================================
>>
>>         <div class="breadcrumbs">
>>      -  - len=1 char= 0 False &raquo
>>     </div>
>>
>> ==============================================================
>>
>> So (type(aurl) == StringTypes) is evaluated False
>>
>> What I'm doing wrong?
>> What is the method to pass the actual url to a filter as a string?
>>     
>
> It is already one. The bug is elsewhere.
>
> HTH
>
>
> >
>
>   


-- 

_________________________________________________________________________
Busico Mirto Silvio
Consulente ICT
cell. 333 4562651
email [EMAIL PROTECTED]


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