How to parse an XML file using Python?

Question

I am trying to print all the elements and attributes in an xml file. The xml file's contents are:

<topology>
<switch id="10">
    <port no="1">h1</port>
    <port no="2">h2</port>
</switch>

<tunnel id="91">
<port no="1">s1</port>
<port no="8">s8</port>
</tunnel>
</topology>

How do I do it? Also, how do I search for an element like switch inside topology?

Step 1 search. Python has several XML parsers already in the standard library. docs.python.org/library/markup.html What part of the standard library documentation is hard to read? Please choose one of the existing XML parsers, and follow the examples. When you get stuck, please ask a specific question about the parser your chose. — S.Lott
– S.Lott, Commented Jan 24, 2012 at 10:45
That's a stilly question. There are several. They have different purposes. I've used them all. Please Read the documentation first before asking silly questions. — S.Lott
– S.Lott, Commented Jan 24, 2012 at 10:48
Please have a look at this link. It should be simple and clear. — Frankline
– Frankline, Commented Jan 24, 2012 at 10:49
@Bruce: Clearly. And I resent that. There's a lot to learn, and you need to actually learn it yourself. — S.Lott
– S.Lott, Commented Jan 24, 2012 at 11:24
possible duplicate of Really simple way to deal with XML in Python? — phihag
– phihag, Commented Jan 24, 2012 at 11:25

Joao Figueiredo · Accepted Answer · 2012-01-24 11:35:48Z

4

Like S.Lott expressed, you have way too many ways to skin this cat,

here is an example using lxml,

from lxml import etree

xml_snippet = '''<topology>
 <switch id="10">
     <port no="1">h1</port>
     <port no="2">h2</port>
 </switch>

 <tunnel dpid="91">
 <port no="1">s1</port>
 <port no="8">s8</port>
 </tunnel>
 </topology>'''

root = etree.fromstring(xml_snippet)

for element in root.iter("*"):
  print element.tag, element.items()

output:

topology []
switch [('id', '10')]
port [('no', '1')]
port [('no', '2')]
tunnel [('dpid', '91')]
port [('no', '1')]
port [('no', '8')]

Using XPath to find an attribute

attribute = '10'
element = root.find('.//switch[@id="%s"]' % attribute)
element.items()

output:

[('id', '10')]

answered Jan 24, 2012 at 11:35

Joao Figueiredo

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Comments

Shriram · Accepted Answer · 2012-07-01 17:00:20Z

4

Here is my working code:

import xml.etree.ElementTree as ET

doc = ET.parse("nm.xml")
s = doc.find("switch")
print s.attrib["id"]
for item in s:
  print item.attrib["no"]
  print item.text

t = doc.find("tunnel")
print t.attrib["dpid"]
for item in t:
  print item.attrib["no"]
  print item.text

P.S: You can replace ET.parse with ET.fromstring and change input argument to a string type It works

edited Jul 1, 2012 at 17:00

Shriram

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answered Jan 24, 2012 at 11:24

Bruce

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