Django objects.all() method return my list in string format (with sqlite)

Question

Sorry if my question sounds stupid as i am new to Django.

I have a view in Django like this :

def myview(request):

list = Ref.objects.all()

return render_to_response('annonces.html', {'list': list} ,RequestContext(request))

Ref is a small sqlite database where i store things scraped from the web.

In my django template, i want to loop on a field of my list object (hurl) which contains a list of links (images) in order to show it on my web page. I've tried this :

   {% for value in list %}
       {% for link in value.hurl %}
           <li><img src="{{ link }}"></li>
       {% endfor %}
   {% endfor %}

Unfortunately, it doesn't work. The object list convert everything to string so the loop in my template create a line for each character in the field hurl.

Would you know how i can get my object to render hurl as a list instead of a string ?

Here is an example of value.hurl :

u"['http://3.visuels.poliris.com/thumbnails/3/3/6/5/3365ddc7-3f27.jpg', 'http://1.visuels.poliris.com/thumbnails/1/2/b/6/12b6c286-372f.jpg', 'http://b.visuels.poliris.com/thumbnails/b/4/f/3/b4f32a1c-3159.jpg']"

Here is the model. Not sure if URLField is correct for hurl.

class Ref(models.Model):

ref = models.IntegerField(max_length=10, primary_key='True')
title = models.CharField(max_length=500)
hurl = models.URLField()
url = models.URLField()
prix = models.CharField(max_length=500)

Thanks Gilles

That's a string, not a list. So when you iterate over it, you get the what you are seeing now. — Burhan Khalid
– Burhan Khalid, Commented Sep 11, 2013 at 21:21
yes that's why i'm asking how i can get a list and not a string. — Gilles
– Gilles, Commented Sep 11, 2013 at 21:28

Thomas Orozco · Accepted Answer · 2013-09-11 21:23:22Z

1

You should change your object model and store the multiple values for hurl as another object.

That object should have a ForeignKey to your Ref object:

class Ref(models.Model):
    # snip

class Link(models.Model):
    url = models.CharField(...)
    ref = models.ForeignKey("Ref")

Then, in your template, do:

{% for value in list %}
    {% for link in value.link_set %}
        <li><img src="{{ link }}"></li>
    {% endfor %}
{% endfor %}

For better performance, in your view, do:

list = Ref.objects.prefetch_related("link_set").all()

answered Sep 11, 2013 at 21:23

Thomas Orozco

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2 Comments

Gilles Over a year ago

Ok Thank you, i'll try and let you know! I will amend my scrappy code and export images in the Link table. It looks like i will have one line per image in the link table then ?

Gilles Over a year ago

It works ! I've slightly changed your list though. I have : {% for link in value.pic_set.all %} <li><img src="{{ link.pic }}"></li>. Otherwise, the template fails. Thank you very much for your help

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