Develop Reference

Python - multiple logs/structures in the same xml file using ElementTree

I wrote an xml file with ElementTree. The problem i'm facing is that I want to write multiple logs in the same xml file.
The code:
import xml.etree.cElementTree as ET
#XML
root = ET.Element('LOG')
DATE = ET.SubElement(root, 'DATE')
DATE.text = "child_1"
TIME = ET.SubElement(root, 'TIME')
TIME.text = "child_2"
CC = ET.SubElement(root, 'CC')
CC.text = "child_3"
AMOUNT = ET.SubElement(root, 'AMOUNT')
AMOUNT.text = "child_4"
tree = ET.ElementTree(root)
#Generating XML
tree.write("file_name.xlm")
#Print
print(open("file_name.xlm").read())
Current output example:
<LOG><DATE>child_1</DATE><TIME>child_2</TIME><CC>child_3</CC><AMOUNT>child_4</AMOUNT></LOG>
Output that I want:
<LOG><DATE>child_1</DATE><TIME>child_2</TIME><CC>child_3</CC><AMOUNT>child_4</AMOUNT></LOG><LOG><DATE>child_1</DATE><TIME>child_2</TIME><CC>child_3</CC><AMOUNT>child_4</AMOUNT></LOG><LOG><DATE>child_1</DATE><TIME>child_2</TIME><CC>child_3</CC><AMOUNT>child_4</AMOUNT></LOG>...

What you want is just not how xml works. xml documents have one root node, and just one. The closest you can get is by wrapping your <LOG> entries in a container:
<ALL_LOGS>
<LOG>....</LOG>
<LOG>....</LOG>
</ALL_LOGS>

adding elements in xml using python

how do i add an element to my attrib ?
i know how to add with text:
ET.subElement(root[2],'resource')
for x in root.iter('resource'):
but what i'm looking for is:
<resources>
<resource identifier="presentation_3" type="webcontent" href="questions.html" adlcp:scormtype="sco">
<file href="questions.html"/>
</resource>
</resources>
below is my code:
import xml.etree.ElementTree as ET
xmlfile = "imsmanifest.xml"
tree = ET.parse(xmlfile)
root = tree.getroot()
ET.SubElement(root[2],'file').set('href','index.html')
tree.write('new.xml')

case insensitive xml and python

I got this piece of code and I am trying to read all the 'ref' 'href' tags. I am not sure how to make this to be case insensitive as some of my xml files have REF or Ref or ref.
Any suggestions?
f = urllib.urlopen(url)
tree = ET.parse(f)
root = tree.getroot()
for child in root.iter('ref'):
t = child.get('href')
if t not in self.href:
self.href.append(t)
print self.href[-1]

You can normalize tags and attributes by converting them to lowercase using the functions below as a step of preprocessing:
import xml.etree.ElementTree as ET
f = urllib.urlopen(url)
tree = ET.parse(f)
root = tree.getroot()
def normalize_tags(root):
root.tag = root.tag.lower()
for child in root:
normalize_tags(child)
def normalize_attr(root):
for attr,value in root.attrib.items():
norm_attr = attr.lower()
if norm_attr != attr:
root.set(norm_attr,value)
root.attrib.pop(attr)
for child in root:
normalize_attr(child)
normalize_tags(root)
normalize_attr(root)
print(ET.tostring(root))

The following should help
f = urllib.urlopen(url)
tree = ET.parse(f)
root = tree.getroot()
for child in root:
if child.tag.lower() == 'ref':
t = child.attribute.get('href')
if t not in self.href:
self.href.append(t)
print self.href[-1]

If you are using lxml then one option is to use XPath with regular expressions through XSLT extensions (https://stackoverflow.com/a/2756994/2997179):
root.xpath("./*[re:test(local-name(), '(?i)href')]",
namespaces={"re": "http://exslt.org/regular-expressions"})

python xml pretty print not working

I am changing an some xml by adding some nodes and values from a list. I can successfully create all the new tags and values, I am creating them between the contributors tags, but when I save the xml out to a new file, the tags I create are all on one line. Here is a sample of my code:
templateXml = """<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<package>
<delivery_type>new</delivery_type>
<feature>
<feature_type>Movie</feature_type>
<contributors>
</contributors>
</package>"""
from lxml import etree
tree = etree.fromstring(templateXml)
node_video = tree.xpath('//feature/contributors')[0]
for cast in castList:
pageElement = etree.SubElement(node_video, 'contributor')
node_video1 = tree.xpath('//feature/contributors/contributor')[0]
pageElement.attrib['type'] = 'cast'
pageElement1 = etree.SubElement(pageElement, 'name')
pageElement1.text = cast.text
pageElement2 = etree.SubElement(pageElement, 'role')
pageElement2.text = "actor"
xmlFileOut = '/Users/User1/Desktop/Python/Done.xml'
with open(xmlFileOut, "w") as f:
f.write(etree.tostring(tree, pretty_print = True, xml_declaration = True, encoding='UTF-8', standalone="yes"))
Here is saved xml file:
<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<package>
<delivery_type>new</delivery_type>
<feature>
<feature_type>Movie</feature_type>
<contributors>
<contributor type="cast"><name>John Doe</name><role>actor</role></contributor><contributor type="cast"><name>Another Actors name</name><role>actor</role></contributor><contributor type="cast"><name>Jane Doe</name><role>actor</role></contributor><contributor type="cast"><name>John Smith</name><role>actor</role></contributor></contributors>
</package>
I have solved this issue when opening an xml file to work on using the below code:
from lxml import etree
parser = etree.XMLParser(remove_blank_text=True) # makes pretty print work
path3 = 'path_to_xml_file'
open(path3)
tree = etree.parse(path3, parser)
root = tree.getroot()
tree.write(xmlFileOut, pretty_print = True, xml_declaration = True, encoding = 'UTF-8')
This works, but how do I get it to work with a string xml?

Taken from http://ruslanspivak.com/2014/05/12/how-to-pretty-print-xml-with-lxml/
import StringIO
import lxml.etree as etree
def prettify(xml_text):
"""Pretty prints xml."""
parser = etree.XMLParser(remove_blank_text=True)
file_obj = StringIO.StringIO(xml_text)
tree = etree.parse(file_obj, parser)
return etree.tostring(tree, pretty_print=True)

A simple solution might be to use StringIO:
from StringIO import StringIO
from lxml import etree
parser = etree.XMLParser(remove_blank_text=True)
tree = etree.parse(StringIO(templateXml), parser)

Changing a specific xml element using Python 3 ElementTree

I have a set of metadata files in xml which are updated regularly and I'm trying to automate.
I've worked out how to itteratively find and then replace text in the desired element of the xml but thought there must be a direct way to access and change the element. I just can't work it out.
The metadata xml is formatted:
<?xml version="1.0" ?>
<metadata xml:lang="en">
<Esri>
<CreaDate>20120405</CreaDate>
<CreaTime>13113000</CreaTime>
<ArcGISFormat>1.0</ArcGISFormat>
<SyncOnce>TRUE</SyncOnce>
<ModDate>20121129</ModDate>
<ModTime>11433300</ModTime>
<ArcGISProfile>ItemDescription</ArcGISProfile>
</Esri>
<dataIdInfo>
<idPurp>Updated :: 121129_114038</idPurp>
</dataIdInfo>
</metadata>
My iterative approach was:
for child in root:
for xel in child.iter('idPurp'):
download_new_datetime = strftime('%y%m%d_%H%M%S')
download_new_text = 'Downloaded :: '
xel.text = download_new_text + download_new_datetime
tree.write(xmlfile)
Ideas appreciated on a better way.

I would write to the file only once I'm done with the loop:
import xml.etree.ElementTree as ET
from time import strftime
xmlfile = '/tmp/file'
tree = ET.parse(xmlfile)
root = tree.getroot()
for child in root:
for xel in child.iter('idPurp'):
download_new_datetime = strftime('%y%m%d_%H%M%S')
download_new_text = 'Downloaded :: '
xel.text = download_new_text + download_new_datetime
tree.write(xmlfile)
I would even simplify that loop further to:
for child in root:
for xel in child.iter('idPurp'):
xel.text = 'Downloaded :: ' + time.strftime('%y%m%d_%H%M%S')

Two simpler ways, both work, tested.
First:
import xml.etree.ElementTree as ET
from time import strftime
xmlfile = 'metadata.xml'
tree = ET.parse(xmlfile)
root = tree.getroot()
xel = root.find('./dataIdInfo/idPurp')
xel.text = 'Downloaded :: ' + strftime('%y%m%d_%H%M%S')
tree.write(xmlfile)
Second:
import xml.etree.ElementTree as ET
from time import strftime
xmlfile = 'metadata.xml'
tree = ET.parse(xmlfile)
root = tree.getroot()
xel = root[1][0]
xel.text = 'Downloaded :: ' + strftime('%y%m%d_%H%M%S')
tree.write(xmlfile)
I prefer the first one, it's more readable in my opinion.