Python用Etree替换XML内容

Question

I'd like to parse and compare 2 XML files with the Python Etree parser as follows:

I have 2 XML files with loads of data. One is in English (the source file), the other one the corresponding French translation (the target file). Eg:

source file:

<AB>
  <CD/>
  <EF>

    <GH>
      <id>123</id>
      <IJ>xyz</IJ>
      <KL>DOG</KL>
      <MN>dogs/dog</MN>
      some more tags and info on same level
      <metadata>
        <entry>
           <cl>Translation</cl>
           <cl>English:dog/dogs</cl>
        </entry>
        <entry>
           <string>blabla</string>
           <string>blabla</string>
        </entry>
            some more strings and entries
      </metadata>
    </GH>

  </EF>
  <stuff/>
  <morestuff/>
  <otherstuff/>
  <stuffstuff/>
  <blubb/>
  <bla/>
  <blubbbla>8</blubbla>
</AB>

The target file looks exactly the same, but has no text at some places:

<MN>chiens/chien</MN>
some more tags and info on same level
<metadata>
  <entry>
    <cl>Translation</cl>
    <cl></cl>
  </entry>

The French target file has an empty cross-language reference where I'd like to put in the information from the English source file whenever the 2 macros have the same ID. I already wrote some code in which I replaced the string tag name with a unique tag name in order to identify the cross-language reference. Now I want to compare the 2 files and if 2 macros have the same ID, exchange the empty reference in the French file with the info from the English file. I was trying out the minidom parser before but got stuck and would like to try Etree now. I have hardly any knowledge about programming and find this very hard. Here is the code I have so far:

    macros = ElementTree.parse(english)

    for tag in macros.getchildren('macro'):
        id_ = tag.find('id')
        data = tag.find('cl')
        id_dict[id_.text] = data.text

    macros = ElementTree.parse(french)

    for tag in macros.getchildren('macro'):
        id_ = tag.find('id')
        target = tag.find('cl')
        if target.text.strip() == '':
        target.text = id_dict[id_.text]

    print (ElementTree.tostring(macros))

I am more than clueless and reading other posts on this confuses me even more. I'd appreciate it very much if someone could enlighten me :-)

Answer 1

There is probably more details to be clarified. Here is the sample with some debug prints that shows the idea. It assumes that both files have exactly the same structure, and that you want to go only one level below the root:

import xml.etree.ElementTree as etree

english_tree = etree.parse('en.xml')
french_tree = etree.parse('fr.xml')

# Get the root elements, as they support iteration
# through their children (direct descendants)
english_root = english_tree.getroot()
french_root = french_tree.getroot()

# Iterate through the direct descendants of the root
# elements in both trees in parallel.
for en, fr in zip(english_root, french_root):
   assert en.tag == fr.tag # check for the same structure
   if en.tag == 'id':
       assert en.text == fr.text # check for the same id

   elif en.tag == 'string':
       if fr.text is None:
           fr.text = en.text
           print en.text      # displaying what was replaced

etree.dump(french_tree)

For more complex structures of the file, the loop through the direct children of the node can be replaced by iteration through all the elements of the tree. If the structures of the files are exactly the same, the following code will work:

import xml.etree.ElementTree as etree

english_tree = etree.parse('en.xml')
french_tree = etree.parse('fr.xml')

for en, fr in zip(english_tree.iter(), french_tree.iter()):
   assert en.tag == fr.tag        # check if the structure is the same
   if en.tag == 'id':
       assert en.text == fr.text  # identification must be the same
   elif en.tag == 'string':
       if fr.text is None:
           fr.text = en.text
           print en.text          # display the inserted text

# Write the result to the output file.
with open('fr2.xml', 'w') as fout:
    fout.write(etree.tostring(french_tree.getroot()))

However, it works only in cases when both files have exactly the same structure. Let's follow the algorithm that would be used when the task is to be done manually. Firstly, we need to find the French translation that is empty. Then it should be replaced by the English translation from the GH element with the same identification. A subset of XPath expressions is used in the case when searching for the elements:

import xml.etree.ElementTree as etree

def find_translation(tree, id_):
    # Search fot the GH element with the given identification, and return
    # its translation if found. Otherwise None is returned implicitly.
    for gh in tree.iter('GH'):
       id_elem = gh.find('./id')
       if id_ == id_elem.text:
           # The related GH element found.
           # Find metadata entry, extract the translation.
           # Warning! This is simplification for the fixed position 
           # of the Translation entry.
           me = gh.find('./metadata/entry')
           assert len(me) == 2     # metadata/entry has two elements
           cl1 = me[0]
           assert cl1.text == 'Translation'
           cl2 = me[1]

           return cl2.text


# Body of the program. --------------------------------------------------

english_tree = etree.parse('en.xml')
french_tree = etree.parse('fr.xml')

for gh in french_tree.iter('GH'): # iterate through the GH elements only 
   # Get the identification of the GH section
   id_elem = gh.find('./id')      
   id_ = id_elem.text

   # Find and check the metadata entry, extract the French translation.
   # Warning! This is simplification for the fixed position of the Translation 
   # entry.
   me = gh.find('./metadata/entry')
   assert len(me) == 2     # metadata/entry has two elements
   cl1 = me[0]
   assert cl1.text == 'Translation'
   cl2 = me[1]
   fr_translation = cl2.text

   # If the French translation is empty, put there the English translation
   # from the related element.
   if cl2.text is None:
       cl2.text = find_translation(english_tree, id_)


with open('fr2.xml', 'w') as fout:
   fout.write(etree.tostring(french_tree.getroot()).decode('utf-8'))