django concat string with comma in template

Question

I am newbie in django. How can I concat string in a for loop in django template

{% for lead in project.leaders %}
    {% if forloop.counter == 1 %}
        {% lead_member = lead.0 %}
     {% else %}
        {% lead_member = ','.lead.0 %}
     {% endif %}
{{ lead_member }}
{% endfor %}

Finally my lead_member should be test1,test2,test3....

what is happening now (my current code)

 {% for lead in project.leaders %}
    {{ lead.0}}
 {% endfor %}

and the output is test1test2test3.... but i want to make same as test1,test2,test3....

In what language would this make sense? You can't assign in Django templates, but even if you could, repeatedly assigning to the same variable wouldn't create a list. And you don't have any commas anywhere, which was supposed to be the whole point of the question. — Daniel Roseman, Sep 17 '15 at 19:13
check now my question revised. this is demo code. I want to achieve that what mentioned in last line — Rejoanul Alam, Sep 17 '15 at 19:15
Why do you need to concat at all? the template is just rendering a html page so the output will be the same — Sayse, Sep 17 '15 at 19:19
@Sayse your solution will concat extra `comma` if there are only one data. Somewhat confused and couldn't understand what is unclear in my question?? I just want to concat a comma after each data — Rejoanul Alam, Sep 17 '15 at 19:27

Brandon Taylor · Accepted Answer · 2015-09-18T14:23:35.800

8

Try this. it works

{% for lead in project.leaders %}
    {{ lead.0 }}{% if not forloop.last %}, {% endif %}
{% endfor %}

There's no need to assign anything, nor do you need that type of complexity by using assignment tags. To keep your templating stupid-simple, you could always do this in your view, or even at the model level:

# don't step on the `join` built-in
from django.template.defaultfilters import join as join_filter

class Project(models.Model):

    @property
    def leaders(self):
        return join_filter(self.objects.values_list('some_field', flat=True), ', ')

Then all you have to do in the template is:

{{ project.leaders }}

edited Sep 18 '15 at 14:23

answered Sep 17 '15 at 19:28

Brandon Taylor

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Thanks. Tried in `views` not worked. However first solution worked. – Rejoanul Alam Sep 17 '15 at 19:42
I have made a complex data structure (dictionary) using `dict(zip(k,a))`. I can show you if you want – Rejoanul Alam Sep 17 '15 at 19:44
Sure, send me link to a Gist or something – Brandon Taylor Sep 17 '15 at 19:45
https://gist.github.com/rejoan/d35e11adc0886ac3ae51 I need to comma separated `username` for `lm` as well as `nm`. No need to hard work for this as problem is already solved – Rejoanul Alam Sep 17 '15 at 19:48
Dear Lord. I have no idea what's going on with that Gist :) – Brandon Taylor Sep 17 '15 at 19:49
Yeah, what I was saying was that the Gist is very complicated for doing something as simple as concatenating values. – Brandon Taylor Sep 17 '15 at 19:57

score 0 · Answer 2 · edited May 23 '17 at 11:44

It's hard to understand your question, but I hope i did it. There is a number of related questions such as String-concatination, How to concatenate in django

It's possible to create first string, concatinate it with comma and new string for every iteration. You are also able to make smth like ','.join(list_of_strings) on your server side before rendering. You can also join your list in templating by {{ list|join:", " }}.

django concat string with comma in template

2 Answers2