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Getting all instances of child node using xml.etree.ElementTree

I have the following XML file as input:

<Test>
  <callEvents>
    <moc>
      <causeForTermination>0</causeForTermination>
      <serviceCode>
        <teleServiceCode>11</teleServiceCode>
      </serviceCode>
      <dialledDigits>5555555</dialledDigits>
      <connectedNumber>77777</connectedNumber>
    </moc>

    <moc>
      <causeForTermination>0</causeForTermination>
      <serviceCode>
        <teleServiceCode>11</teleServiceCode>
      </serviceCode>
      <dialledDigits>2222222</dialledDigits>
    </moc>
  </callEvents>
  <callEventsCount>100</callEventsCount>
</Test> 

I want to output all the values for dialledDigits. However, my code only displays the first instance of dialledDigits.

dialledDigits {} 5555555

My desired output should contain both instances.

dialledDigits {} 5555555
dialledDigits {} 2222222

Here is my code

import xml.etree.ElementTree as ET
tree = ET.parse('as.xml')
root = tree.getroot()
callevent=root.find('callEvents')

Moc1=callevent.find('moc')

for node in Moc1.getiterator():
    if node.tag=='dialledDigits':
        print node.tag, node.attrib, node.text
like image 316
Ash Avatar asked Mar 25 '15 21:03

Ash


2 Answers

You can also write an XPath expression. Just 2 lines instead of 5 and a single loop:

for node in tree.findall('.//callEvents/moc/dialledDigits'):
    print node.tag, node.attrib, node.text 

Demo:

>>> import xml.etree.ElementTree as ET
>>> 
>>> 
>>> tree = ET.parse('as.xml')
>>> root = tree.getroot()
>>> 
>>> for node in tree.findall('.//callEvents/moc/dialledDigits'):
...     print node.tag, node.attrib, node.text
... 
dialledDigits {} 5555555
dialledDigits {} 2222222
like image 189
alecxe Avatar answered Oct 14 '22 06:10

alecxe


Use findall:

moc1 = callevent.findall('moc')

for moc in moc1:
    for node in moc.getiterator():
        if node.tag=='dialledDigits':
            print node.tag, node.attrib, node.text

Output:

dialledDigits {} 5555555
dialledDigits {} 2222222
like image 37
Celeo Avatar answered Oct 14 '22 04:10

Celeo