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Home/ Questions/Q 715861
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T05:14:04+00:00 2026-05-14T05:14:04+00:00

I want to build an API service using Django. A basic workflow goes like

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I want to build an API service using Django. A basic workflow goes like this:

First, an http request goes to http://mycompany.com/create?id=001&callback=http://callback.com. It will create a folder on the server with name 001.

Second, if the folder does not exist, it will be created. You get response immediately in XML format. It will look like:

<?xml version="1.0" encoding="UTF-8"?>
<response>
    <status>
        <statusCode>0</statusCode>
        <message>Success</message>
    </status>
    <group id="001"/>
</response>

Finally, the server will do its job (i.e. creating the folder). After it is done, the server does a callback to the URL provided.

Currently, I use

return render_to_response('create.xml', {'statusCode': statusCode,
                                                   'statusMessage': statusMessage,
                                                   'groupId': groupId,
                                                   }, mimetype = 'text/xml')

to send the XML response back. I have an XML template which has statusCode, statusMessage, groupId placeholders.

<?xml version="1.0" encoding="UTF-8"?> 
<response>
    <status>
        <statusCode>{{ statusCode }}</statusCode>
        <message>{{ statusMessage }}</message>
    </status>
    {% if not statusCode %}
        <group id="{{ groupId }}"/>
    {% endif %} 
</response>

But in this way I have to put step 3 before step 2, because otherwise step 3 will not be executed if it is after return statement.

Can somebody give me some suggestions how to do this? Thanks.

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  1. Editorial Team
    Editorial Team
    2026-05-14T05:14:04+00:00Added an answer on May 14, 2026 at 5:14 am

    I have a feeling that you might be missing some Django fundamentals here.

    Why is create.py inside of your url?

    If you’re using Django’s url routing and views, the render_to_response should work fine. You might be jumping to an incorrect conclusion regarding why your response isn’t getting returned.

    I’m not sure I understand the statement:

    But in this way I have to put step 3
    before step 2, because otherwise step
    3 will not be executed if it is after
    return statement.

    Step 3 is not after the return statement. It is part of the return statement.

    You could always do something like this to split up the process:

    # Code that creates folder, statusCode, statusMessage, groupId
    response = render_to_response('create.xml', {'statusCode': statusCode,
                                                       'statusMessage': statusMessage,
                                                       'groupId': groupId,
                                                       }, mimetype = 'text/xml')
    # Some other code, maybe an import pdb; pdb.set_trace() 
    # So that you can inspect the response inside of a python shell.
    return response
    
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