Question

How do I display all the elements in a list with a line break in a Django template?
For example, with a list
ls = ['bin', 'boot', 'data', 'data2'] I would like to display the elements in the format

bin
boot
data
data2

I have tried:

'\n'.join(ls)

This returns each word with spaces:

b i n b o o t d a t a d a t a 2

The relevant code from my views.py:

   result1 = subprocess.Popen(['ls', '/home/zurelsoft/R'], stdout=subprocess.PIPE).communicate()[0]
        result = ''.join(result1)
        return render_to_response('thanks.html', {'res':result, 'res1':command}, context_instance=RequestContext(request))

The corresponding section of the template:

<td>{{ res }} </td>

Answer 1

Modify your view to use the splitlines() function to split the output into a list of lines:

def myview(request):
    result = subprocess.Popen(['ls', '/home/zurelsoft/R'], stdout=subprocess.PIPE).communicate()[0].splitlines()
    return render_to_response('thanks.html', {'res':result, 'res1':command}, context_instance=RequestContext(request))

And then use the join filter in your template to separate each element in the list with the HTML line-break tag <br />:

<td>{{ res|join:"<br/>" }}</td>