Generating standard xml and making request in Python/Django

Question

I need to generate a standard XML as follows using python:

<?xml version="1.0" encoding="UTF-8"?>
<req:ShipmentValidateRequestAP xmlns:req="http://www.example.com" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.example.com/schema">

<root>
    <doc>
        <field1>some value1</field1>
        <field2>some vlaue2</field2>
    </doc>
</root>

</req:ShipmentValidateRequestAP>

And this needed to be requested to a server URL:

https://sampletest-ea.example.com/XMLShippingServlet

Can anybody please help me to implement this ?

What do you need help with specifically. This is not the place to ask people to do your work for you, this is to ask for help with an attempt that you already made that didn't work. — Burhan Khalid
– Burhan Khalid, Commented Sep 24, 2013 at 9:07
I tried it with urllib2 but with that i can generate only the '<root>,/root>' tags and the contents within that, actually i need help to create an xml file that starts with '<?xml version="1.0" encoding="UTF-8"?>' and '<req:ShipmentValidateRequestAP>' tags as specified above. — Akhil Sundar
– Akhil Sundar, Commented Sep 24, 2013 at 9:28

YaleCheung · Accepted Answer · 2013-09-24 10:45:06Z

1

you can finish your work by the help of http://docs.python.org/2/library/xml.dom.minidom.html. It is a simple libary

answered Sep 24, 2013 at 10:45

YaleCheung

6403 silver badges9 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Generating standard xml and making request in Python/Django

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related