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    <p>Your problem is that the forloop.counter is an integer and you are using the <code>add</code> template filter which will behave properly if you pass it all strings or all integers, but not a mix.</p> <p>One way to work around this is:</p> <pre><code>{% for x in some_list %} {% with y=forloop.counter|stringformat:"s" %} {% with template="mod"|add:y|add:".html" %} &lt;p&gt;{{ template }}&lt;/p&gt; {% endwith %} {% endwith %} {% endfor %} </code></pre> <p>which results in:</p> <pre><code>&lt;p&gt;mod1.html&lt;/p&gt; &lt;p&gt;mod2.html&lt;/p&gt; &lt;p&gt;mod3.html&lt;/p&gt; &lt;p&gt;mod4.html&lt;/p&gt; &lt;p&gt;mod5.html&lt;/p&gt; &lt;p&gt;mod6.html&lt;/p&gt; ... </code></pre> <p>The second with tag is required because stringformat tag is implemented with an automatically prepended <code>%</code>. To get around this you can create a custom filter. I use something similar to this:</p> <p><a href="http://djangosnippets.org/snippets/393/" rel="noreferrer">http://djangosnippets.org/snippets/393/</a></p> <p>save the snipped as some_app/templatetags/some_name.py</p> <pre><code>from django import template register = template.Library() def format(value, arg): """ Alters default filter "stringformat" to not add the % at the front, so the variable can be placed anywhere in the string. """ try: if value: return (unicode(arg)) % value else: return u'' except (ValueError, TypeError): return u'' register.filter('format', format) </code></pre> <p>in template:</p> <pre><code>{% load some_name.py %} {% for x in some_list %} {% with template=forloop.counter|format:"mod%s.html" %} &lt;p&gt;{{ template }}&lt;/p&gt; {% endwith %} {% endfor %} </code></pre>
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    1. POHow can I concatenate forloop.counter to a string in my django template
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    1. COWhy is this not marked as the answer?!
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