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    <p>I noticed you have tagged your question as pertaining to Django. If you are going to build a Django application based on this problem, then the <a href="https://docs.djangoproject.com/en/dev/ref/contrib/syndication/">syndication framework</a> is what you want to use. However, this isn't worth it unless you are planning to use other components of Django, such as the database stuff and / or the template language.</p> <p>However, you asked for the "simplest and easiest" solution. I love Django and it only takes me minutes to set up an app so for me the easiest way to do your whole project would probably be to make a quick Django app. The simplest solution is probably to create the feed manually, shouldn't be that hard; something like:</p> <pre><code>inp="""System MA user id = 2084 username = XYZ90 selection = pnq decimal = 6.000 Percentage = 19.1176470588 % System NA user id = 2086 username = pron1 selection = abc decimal = 13.000 Percentage = 13.1147540984 % System M user id = 1664 username = Chrisod selection = pleader decimal = 15.000 Percentage = 16.091954023 %""" inp=inp.split('\n\n') rss_start="""&lt;?xml version="1.0" encoding="ISO-8859-1" ?&gt; &lt;rss version="2.0"&gt; &lt;channel&gt; &lt;title&gt;Your title&lt;/title&gt; &lt;link&gt;http://yoursite.com&lt;/link&gt; &lt;description&gt;Your discription&lt;/description&gt; """ rss_end="""&lt;/channel&gt; &lt;/rss&gt; """ def description(item): return item def title(item): return item.split('\n')[0] def link(item): return 'http://mysite.com/' + item.split('\n')[0] rss_items=[] for counter, item in enumerate(inp): rss_items.append(""" &lt;item&gt; &lt;title&gt;%s&lt;/title&gt; &lt;link&gt;%s&lt;/link&gt; &lt;description&gt;%s&lt;/description&gt; &lt;guid&gt;counter&lt;/guid&gt; &lt;/item&gt;""" % (title(item),description(item),link(item))) rss_output=rss_start+''.join(rss_items)+rss_end </code></pre> <p>You probably also want to add <code>&lt;pubDate&gt;</code> tags. And make sure your <code>&lt;guid&gt;</code>s are unique.</p> <p>Note: rss template copied from w3schools.com</p>
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    1. POGenerating simple RSS feed from parsed data
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    1. POGenerating simple RSS feed from parsed data
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    1. COHello foobarbecue I have made some edits to my question, mainly because the comments don't allow code to be formatted in an aesthetically pleasing way. How would I get the filtered printing section to yield the results in such a way that would be compatible with the supplied `inp=` prefix? Kind regards AEA
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    2. COInstead of printing, you should add your filtered results to some kind of variable, such as a list or a dictionary. I've started with a string variable called "inp" which was just my shorthand for "input." Then I split up the string into a list of strings each time there are two newline characters in a row using .split('\n\n') . You could just build up inp as a list, using inp=[] and then inp.append() in your if statement. If you like, I can edit my answer to add this code, but what you are asking for is pretty basic python.
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    3. COI have to get my code outputting as a variable before I can test this code properly, I will of course remember to upvote and accept your answer if it works :)
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