Python将XML输出与列表进行比较

Question

I have an XML that looks something like this:

<Import>
  <spId>1234</spId>
  <GroupFlag>false</GroupFlag>
</Import>

I want to extract the value of spId and compare it with a list and I have the following script:

import xml.etree.ElementTree as ET
xml_file = "c:/somefile.xml"

sp_id_list = ['1234']
tree = ET.parse(xml_file)
root = tree.getroot()

for sp_id in root.findall('./spId'):
  if sp_id.text in sp_id_list:
    print sp_id.text

This doesn't work for spId (numeric) but works for comparing GroupFlag (string) with a list. Why is this happening and how can I rectify this problem?

Sorry for the stupid question, I am a noob to this.

Answer 1

Your code example works correctly if your XML sample posted here is given as input XML file.

However you want to find all elements. So, I assume that your real document has many <Import> items. If a list of items is not wrapped by some parent tag it is not a valid XML. In that case you would have xml.etree.ElementTree.ParseError .

So, I assume that in your real document <Import> is not a root element and <Import> elements are somewhere deeper in the document, for example

<Parent>
  <Import>
    <spId>1234</spId>
    <GroupFlag>false</GroupFlag>
  </Import>
  <Import>
    <spId>1234</spId>
    <GroupFlag>false</GroupFlag>
  </Import>
</Parent>

In that case the search pattern './spId' cannot find those tags, since that pattern matches only direct children of the root element. So, you can use XPath matching tags all levels beneath or even better pointing direct path from the root to the level where spId is located:

# all subelements, on all levels beneath the current element
root.findall('.//spId')

# all spId elements directly in Import tags that are directly
# beneath the root element (as in the above XML example)
root.findall('./Import/spId'):