Reputation: 25
I need to parse a directory of xml files into one large csv file, I need certain attributes under the element 'Param' (attributes are 'Name' and 'PNum'). There is another XML file in the directory called Content.xml which I can get all the names of the other XML files and set them as the FileName. The issue is that I cannot figure out how to get these attributes in each XML file as each XML file has a different organisation and some don't seem to have these attributes in the first place.
I have written code that works for one of the XML files in the directory that outputs a CSV file with all the relevant information.
import xml.etree.ElementTree as ET
import csv
import os
FileName = '------.xml'
tree = ET.parse(FileName)
root = tree.getroot()[4]
csv_out = open('CsvOut', 'w')
csvwriter = csv.writer(csv_out)
count = 0
for child in root:
generation = []
parameters = []
if count == 0:
csv_head = ['Generation', 'Parameter Name', 'Parameter Number']
csvwriter.writerow(csv_head)
count = count + 1
gen = FileName[:-4]
generation.append(gen)
parameters.append(generation)
name = child.get('Name')
parameters.append(name)
num = child.get('PNum')
parameters.append(num)
csvwriter.writerow(parameters)
csv_out.close()
Upvotes: 2
Views: 583
Reputation: 12992
I rather simple and you can do it in two steps:
import xml.etree.ElementTree as ET
import csv
import os
from glob import glob
# create csv writer
csv_out = open('CsvOut', 'w')
csvwriter = csv.writer(csv_out)
# write the header
csv_head = ['Generation', 'Parameter Name', 'Parameter Number']
csvwriter.writerow(csv_head)
# iterate over the xml files in the current directory
for FileName in glob("*.xml"):
tree = ET.parse(FileName)
root = tree.getroot()[4]
for child in root:
generation = []
parameters = []
gen = FileName[:-4]
generation.append(gen)
parameters.append(generation)
name = child.get('Name')
parameters.append(name)
num = child.get('PNum')
parameters.append(num)
csvwriter.writerow(parameters)
# after iterating, close the csv file
csv_out.close()
Upvotes: 1