Re: Rendering XML using XSLT

On Sat, Sep 27, 2014 at 9:02 AM, John <john-django@martinhome.org.uk> wrote:

Do you have a URL conf that matches the /get_xsd/ccd-description.xsl stylesheet? You have to match and serve that as well.

 

Taking a step back though. It looks like you are serving static files that would be best served from a webserver, perhaps with some symlinks, so that the overhead of Django is not required at all.

John

​This gave me the hint in a solution. The overhead isn't a big deal in this use case so using Django is easier and more portable (I think). ​

​I created a URL:  ​    url(r'^get_xsd/ccd-description.xsl', "ccdgen.views.get_xsl")

and the associated view:

def get_xsl(request):

    CWD = os.getcwd()

    if CWD == '/':    # Live site

        xslfile = "/home/ccdgen/ccdlib/ccd-description.xsl"

    else:             # development

        xslfile = "ccdlib/ccd-description.xsl"

    xsl = open(xslfile,encoding='utf-8')

    response = HttpResponse(mimetype="text/xsl")

    for line in xsl:

        response.write(line)

    xsl.close()

    return response

​ ​

​Thanks for the nudge in the right direction. ​

 

On 27/09/14 12:37, Timothy W. Cook wrote:

Some details I forgot are that this is Django 1.6.4, Python 3.4.1.

The 404 error is on http://192.168.1.9:8080/get_xsd/ccd-description.xsl 

On Sat, Sep 27, 2014 at 8:10 AM, Timothy W. Cook <tim@mlhim.org> wrote:

I have a view that opens an XML file (it happens to be a schema but that doesn't matter) that specifies a stylesheet.

It works as expected when opening the XML file locally.

​The stylesheet isn't applied when viewing the file via Django and I see in Firebug that I am getting a 404 on the stylesheet. The stylesheet is in the same directory as the XML file.

So how do I setup a view or add this file to my existing view so it is found? 

The view:

​def get_ccddetails(request, id):

    ccd = CCD.objects.get(pk=id)

    ccdid= ccd.ct_id

    CWD = os.getcwd()

    if CWD == '/':    # Live site

        frm_dir = "/home/ccdgen/ccdlib/CCD_"+ ccdid

    else:             # development

        frm_dir = "ccdlib/CCD_"+ ccdid

    filename = frm_dir+"/ccd-"+ccdid+".xsd"

    if os.path.isfile(filename):

        response = HttpResponse(mimetype="application/xml; charset=utf-8")

        f = open(filename, encoding="utf-8")

        for line in f:

            response.write(line)

        f.close()

    else:

        response = HttpResponseNotFound("<h1>XSD: "+filename+" was not found.</h1>")

    return response

​The URL:

​    url(r'^get_xsd/(?P<id>\d+)$', "ccdgen.views.get_ccddetails", name='getxsd'),

​ Thanks,

Tim ​

 

============================================
Timothy Cook
LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook

MLHIM http://www.mlhim.org

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Timothy Cook
LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook

MLHIM http://www.mlhim.org

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Timothy Cook
LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook

MLHIM http://www.mlhim.org

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