Hello I am a programming in python to make a xml parse. And While I can get it to parse out the first xml tag. I want to be able to do it all in the file. but I dont know how to do it. Here is the code I am sure I am missing something I know I gotta replace the 0 with something that can be counted but can't seem to figure out what I am missing. Any help would be appreciated. Here is the code I can't really write it here without it looking funny. Hope that helps.
Here is my code:
from xml.dom import minidom
def xml_data ():
f = open('C:/opidea.xml','r')
data = f.read()
f.close()
dom = minidom.parseString(data)
list(data) = ic
xmlTag = dom.getElementsByTagName('author')[0].toxml()
xmlData=xmlTag.replace('<author>','Changes by: ').replace('</author>','')
xmlDate = dom.getElementsByTagName('date')[0].firstChild.nodeValue
xmlPath = dom.getElementsByTagName('path')[0].toxml()
xmlPathm =xmlPath.replace('<kind>',' What Changed: ').replace('</path>','')
xmlMsg = dom.getElementsByTagName('msg')[0].toxml()
xmlMsgm =xmlMsg.replace('<msg>','Comments: ').replace('</msg>','')
content = ''
content += xmlData + '\n'
content += xmlDate + '\n'
content += xmlPathm + '\n'
content += xmlMsgm + '\n'
send_email(content)
def send_email(content):
print content
And here is what the xml data would look like.
<log>
<logentry
revision="33185">
<author>glv</author>
<date>2012-08-06T21:01:52.494219Z</date>
<paths>
<path
action="M"
kind="dir">/branches/somefolder</path>
</paths>
<msg>PATCH_BRANCH:N/A
BUG_NUMBER:N/A
FEATURE_AFFECTED:N/A
OVERVIEW:N/A
Adding the SVN log size requirement to the branch
</msg>
</logentry>
</log>
Related
After reading from an existing file with 'ugly' XML and doing some modifications, pretty printing doesn't work. I've tried etree.write(FILE_NAME, pretty_print=True).
I have the following XML:
<testsuites tests="14" failures="0" disabled="0" errors="0" time="0.306" name="AllTests">
<testsuite name="AIR" tests="14" failures="0" disabled="0" errors="0" time="0.306">
....
And I use it like this:
tree = etree.parse('original.xml')
root = tree.getroot()
...
# modifications
...
with open(FILE_NAME, "w") as f:
tree.write(f, pretty_print=True)
For me, this issue was not solved until I noticed this little tidbit here:
http://lxml.de/FAQ.html#why-doesn-t-the-pretty-print-option-reformat-my-xml-output
Short version:
Read in the file with this command:
>>> parser = etree.XMLParser(remove_blank_text=True)
>>> tree = etree.parse(filename, parser)
That will "reset" the already existing indentation, allowing the output to generate it's own indentation correctly. Then pretty_print as normal:
>>> tree.write(<output_file_name>, pretty_print=True)
Well, according to the API docs, there is no method "write" in the lxml etree module. You've got a couple of options in regards to getting a pretty printed xml string into a file. You can use the tostring method like so:
f = open('doc.xml', 'w')
f.write(etree.tostring(root, pretty_print=True))
f.close()
Or, if your input source is less than perfect and/or you want more knobs and buttons to configure your out put you could use one of the python wrappers for the tidy lib.
http://utidylib.berlios.de/
import tidy
f.write(tidy.parseString(your_xml_str, **{'output_xml':1, 'indent':1, 'input_xml':1}))
http://countergram.com/open-source/pytidylib
from tidylib import tidy_document
document, errors = tidy_document(your_xml_str, options={'output_xml':1, 'indent':1, 'input_xml':1})
f.write(document)
fp = file('out.txt', 'w')
print(e.tree.tostring(...), file=fp)
fp.close()
Here is an answer that is fixed to work with Python 3:
from lxml import etree
from sys import stdout
from io import BytesIO
parser = etree.XMLParser(remove_blank_text = True)
file_obj = BytesIO(text)
tree = etree.parse(file_obj, parser)
tree.write(stdout.buffer, pretty_print = True)
where text is the xml code as a sequence of bytes.
I am not sure why other answers did not mention this. If you want to obtain the root of the xml there is a method called getroot(). I hope I answered your question (though a little late).
tree = et.parse(xmlFile)
root = tree.getroot()
Of course - pretty print of lxml.etree is possible.
In my case, the old trick with remove_blank_text=True and pretty_print=True was not working as I expected (was too delicate), so I decided to write it by myself.
Here is it - a modern, forcible, native pythonic way to correct lxml.etee.Element tree indentation.
This gives a nicely prettified XML string:
from typing import Optional
import lxml.etree
def indent_lxml(element: lxml.etree.Element, level: int = 0, is_last_child: bool = True) -> None:
space = " "
indent_str = "\n" + level * space
element.text = strip_or_null(element.text)
if element.text:
element.text = f"{indent_str}{space}{element.text}"
num_children = len(element)
if num_children:
element.text = f"{element.text or ''}{indent_str}{space}"
for index, child in enumerate(element.iterchildren()):
is_last = index == num_children - 1
indent_lxml(child, level + 1, is_last)
elif element.text:
element.text += indent_str
tail_level = max(0, level - 1) if is_last_child else level
tail_indent = "\n" + tail_level * space
tail = strip_or_null(element.tail)
element.tail = f"{indent_str}{tail}{tail_indent}" if tail else tail_indent
def strip_or_null(text: Optional[str]) -> Optional[str]:
if text is not None:
return text.strip() or None
It's decent fast, because it doesn't allocate any additional structures in memory and also traversing the tree - it visits each node only once, giving the best possible - O x N computational complexity.
It rearranges all the existing indentation "in place" in the tree (the DOM) by correcting contents of Element.text and Element.tail attributes (affects white-spaces only).
Naturally, it also can be used with HTML parsed by lxml.
In order to use it, do something like that:
root = lxml.etree.parse("path/to/the_file.xml").getroot()
# or
root = lxml.etree.fromstring("<xml><body><leaf1/><leaf2/></body></xml>")
indent_lxml(root) # corrects indentation "in place"
result = lxml.etree.tostring(root, encoding="unicode")
print(result)
Which prints:
<xml>
<body>
<leaf1/>
<leaf2/>
</body>
</xml>
After reading from an existing file with 'ugly' XML and doing some modifications, pretty printing doesn't work. I've tried etree.write(FILE_NAME, pretty_print=True).
I have the following XML:
<testsuites tests="14" failures="0" disabled="0" errors="0" time="0.306" name="AllTests">
<testsuite name="AIR" tests="14" failures="0" disabled="0" errors="0" time="0.306">
....
And I use it like this:
tree = etree.parse('original.xml')
root = tree.getroot()
...
# modifications
...
with open(FILE_NAME, "w") as f:
tree.write(f, pretty_print=True)
For me, this issue was not solved until I noticed this little tidbit here:
http://lxml.de/FAQ.html#why-doesn-t-the-pretty-print-option-reformat-my-xml-output
Short version:
Read in the file with this command:
>>> parser = etree.XMLParser(remove_blank_text=True)
>>> tree = etree.parse(filename, parser)
That will "reset" the already existing indentation, allowing the output to generate it's own indentation correctly. Then pretty_print as normal:
>>> tree.write(<output_file_name>, pretty_print=True)
Well, according to the API docs, there is no method "write" in the lxml etree module. You've got a couple of options in regards to getting a pretty printed xml string into a file. You can use the tostring method like so:
f = open('doc.xml', 'w')
f.write(etree.tostring(root, pretty_print=True))
f.close()
Or, if your input source is less than perfect and/or you want more knobs and buttons to configure your out put you could use one of the python wrappers for the tidy lib.
http://utidylib.berlios.de/
import tidy
f.write(tidy.parseString(your_xml_str, **{'output_xml':1, 'indent':1, 'input_xml':1}))
http://countergram.com/open-source/pytidylib
from tidylib import tidy_document
document, errors = tidy_document(your_xml_str, options={'output_xml':1, 'indent':1, 'input_xml':1})
f.write(document)
fp = file('out.txt', 'w')
print(e.tree.tostring(...), file=fp)
fp.close()
Here is an answer that is fixed to work with Python 3:
from lxml import etree
from sys import stdout
from io import BytesIO
parser = etree.XMLParser(remove_blank_text = True)
file_obj = BytesIO(text)
tree = etree.parse(file_obj, parser)
tree.write(stdout.buffer, pretty_print = True)
where text is the xml code as a sequence of bytes.
I am not sure why other answers did not mention this. If you want to obtain the root of the xml there is a method called getroot(). I hope I answered your question (though a little late).
tree = et.parse(xmlFile)
root = tree.getroot()
Of course - pretty print of lxml.etree is possible.
In my case, the old trick with remove_blank_text=True and pretty_print=True was not working as I expected (was too delicate), so I decided to write it by myself.
Here is it - a modern, forcible, native pythonic way to correct lxml.etee.Element tree indentation.
This gives a nicely prettified XML string:
from typing import Optional
import lxml.etree
def indent_lxml(element: lxml.etree.Element, level: int = 0, is_last_child: bool = True) -> None:
space = " "
indent_str = "\n" + level * space
element.text = strip_or_null(element.text)
if element.text:
element.text = f"{indent_str}{space}{element.text}"
num_children = len(element)
if num_children:
element.text = f"{element.text or ''}{indent_str}{space}"
for index, child in enumerate(element.iterchildren()):
is_last = index == num_children - 1
indent_lxml(child, level + 1, is_last)
elif element.text:
element.text += indent_str
tail_level = max(0, level - 1) if is_last_child else level
tail_indent = "\n" + tail_level * space
tail = strip_or_null(element.tail)
element.tail = f"{indent_str}{tail}{tail_indent}" if tail else tail_indent
def strip_or_null(text: Optional[str]) -> Optional[str]:
if text is not None:
return text.strip() or None
It's decent fast, because it doesn't allocate any additional structures in memory and also traversing the tree - it visits each node only once, giving the best possible - O x N computational complexity.
It rearranges all the existing indentation "in place" in the tree (the DOM) by correcting contents of Element.text and Element.tail attributes (affects white-spaces only).
Naturally, it also can be used with HTML parsed by lxml.
In order to use it, do something like that:
root = lxml.etree.parse("path/to/the_file.xml").getroot()
# or
root = lxml.etree.fromstring("<xml><body><leaf1/><leaf2/></body></xml>")
indent_lxml(root) # corrects indentation "in place"
result = lxml.etree.tostring(root, encoding="unicode")
print(result)
Which prints:
<xml>
<body>
<leaf1/>
<leaf2/>
</body>
</xml>
I'm trying to write the list elements to an xml file. I have written the below code. The xml file is created, but the data is repeated. I'm unable to figure out why is the data written twice in the xml file.
users_list = ['Group1User1', 'Group1User2', 'Group2User1', 'Group2User2']
def create_xml(self):
usrconfig = Element("usrconfig")
usrconfig = ET.SubElement(usrconfig,"usrconfig")
for user in range(len( users_list)):
usr = ET.SubElement(usrconfig,"usr")
usr.text = str(users_list[user])
usrconfig.extend(usrconfig)
tree = ET.ElementTree(usrconfig)
tree.write("details.xml",encoding='utf-8', xml_declaration=True)
Output File: details.xml
-
<usr>Group1User1</usr>
<usr>Group1User2</usr>
<usr>Group2User1</usr>
<usr>Group2User2</usr>
<usr>Group1User1</usr>
<usr>Group1User2</usr>
<usr>Group2User1</usr>
<usr>Group2User2</usr>
enter image description here
usrconfig.extend(usrconfig)
This line looks suspicious to me. if userconfig was a list, this line would be equivalent to "duplicate every element in this list". I suspect that something similar happens for Elements, too. Try deleting that line.
import xml.etree.ElementTree as ET
users_list = ["Group1User1", "Group1User2", "Group2User1", "Group2User2"]
def create_xml():
usrconfig = ET.Element("usrconfig")
usrconfig = ET.SubElement(usrconfig,"usrconfig")
for user in range(len( users_list)):
usr = ET.SubElement(usrconfig,"usr")
usr.text = str(users_list[user])
tree = ET.ElementTree(usrconfig)
tree.write("details.xml",encoding='utf-8', xml_declaration=True)
create_xml()
Result:
<?xml version='1.0' encoding='utf-8'?>
<usrconfig>
<usr>Group1User1</usr>
<usr>Group1User2</usr>
<usr>Group2User1</usr>
<usr>Group2User2</usr>
</usrconfig>
For such a simple xml structure, we can directly write out the file. But this technique might also be useful if one is not up to speed with the python xml modules.
import os
users_list = ["Group1User1", "Group1User2", "Group2User1", "Group2User2"]
os.chdir("C:\\Users\\Mike\\Desktop")
xml_out_DD = open("test.xml", 'wb')
xml_out_DD.write(bytes('<usrconfig>', 'utf-8'))
for i in range(0, len(users_list)):
xml_out_DD.write(bytes('<usr>' + users_list[i] + '</usr>', 'utf-8'))
xml_out_DD.write(bytes('</usrconfig>', 'utf-8'))
xml_out_DD.close()
My python (2.7) script is outputting the following XML using lxml library:
<Button android:id="#+id/button1" android:layout_width="wrap_content" android:layout_height="wrap_content" android:layout_marginLeft="17dp" android:layout_marginTop="16dp" android:text="Button"/>
I would like to output it in multiple lines, one per attribute:
<Button
android:id="#+id/button1"
android:layout_width="wrap_content"
android:layout_height="wrap_content"
android:layout_marginLeft="17dp"
android:layout_marginTop="16dp"
android:text="Button" />
I came up with a very naive and inefficient approach.
After generating the xml using lxml library, I process the output.
The following code is only tested to work with lxml output and assumes we have a tag per line.
output = etree.tostring(
tree,
xml_declaration=True,
pretty_print=True,
encoding=tree.docinfo.encoding,
)
output = output.replace("\" ","\"\n")
with open(filename, "w") as f:
parent_tag_line = None
for i, line in enumerate(output.splitlines()):
line_stripped = line.lstrip(" ")
line_ident = len(line) - len(line_stripped)
if parent_tag_line is not None:
if line_ident == 0 and line[:2] != "</":
line = (parent_line_ident+2)*" " + line
else:
parent_tag_line = line
parent_line_ident = line_ident
line_stripped = line.lstrip()
if line_stripped[:4] != "<!--" and line_stripped[:2] != "</":
line="\n"+line
else:
parent_tag_line = line
parent_line_ident = line_ident
print >>f, line
Although this gets the job done, it is far from being an optimal approach.
I wonder if there's a better and simpler way of doing this.
I have some code that is parsing an xml file and saving it as a csv. I can do this two ways, one by manually downloading the xml file and then parsing it, the other by taking the xml feed directly using ET.fromstring and then parsing. When I go directly I get data errors it appears to be an integrity issue. I am trying to include the xml download in to the code, but I am not quite sure the best way to approach this.
import xml.etree.ElementTree as ET
import csv
import urllib
url = 'http://www.capitalbikeshare.com/data/stations/bikeStations.xml'
connection = urllib.urlopen(url)
data = connection.read()
#I need code here!!!
tree = ET.parse('bikeStations.xml')
root = tree.getroot()
#for child in root:
#print child.tag, child.attrib
locations = []
for station in root.findall('station'):
name = station.find('name').text
bikes = station.find('nbBikes').text
docks = station.find('nbEmptyDocks').text
time = station.find('latestUpdateTime').text
sublist = [name, bikes, docks, time]
locations.append(sublist)
#print 'Station:', name, 'has', bikes, 'bikes and' ,docks, 'docks'
#print locations
s = open('statuslog.csv', 'wb')
w = csv.writer(s)
w.writerows(locations)
s.close()
f = open('filelog.csv', 'ab')
w = csv.writer(f)
w.writerows(locations)
f.close()
What you need is:
root = ET.fromstring(data)
and omit the line of: tree = ET.parse('bikeStations.xml')
As the response from connection.read() returns String, you can directly read the XML string by using fromstring method, you can read more from HERE.