Update an XML document using Python or a shell script

Question

I have an XML file that is used as index for a dynamic panel with this structure:

<?xml version="1.0"?>
<Addons>
  <line>
    <text>My collection 1</text>
    <level>3</level>
    <comment/>
    <file>Collection1.zip</file>
    <size>120</size>
    <parent>Collection</parent>
    <directory>picture</directory>
    <type>jpeg</type>
    <filedate>22/05/2014</filedate>
  </line>
  <line>
    <text>My collection 2</text>
    <level>3</level>
    <comment/>
    <file>Collection2.zip</file>
    <size>880</size>
    <parent>Collection</parent>
    <directory>picture</directory>
    <type>jpeg</type>
    <filedate>01/04/2013</filedate>
  </line>
</Addons>

My panel uses this file as building index. When I update the files on the server, I need to update the filedate element by hand, ...but the file has 80 lines and it's terrible.

Is there a way by a script to do the job? The sequence is:

parse the file line by line
read the file name from the <file> element
get the last-modification date of the file
update the <filedate> element if present
go to next line

Shell scripts and Python are available on the server.

Thanks!!

The proper tool would be an XML processor. Do you have xsltproc or similar? — tripleee
– tripleee, Commented May 28, 2014 at 20:55

glenn jackman · Accepted Answer · 2014-05-29 00:11:13Z

2

Using the xmlstarlet command line tool:

xmlstarlet sel -t -v '//file' -n file.xml |
while IFS= read -r filename; do
    filedate=$(date -d "@$(stat -c %Y "$filename")" +%d/%m/%Y)
    xmlstarlet ed --inplace -u "//filedate[../file = '$filename']" -v "$filedate" file.xml
done

edited May 29, 2014 at 0:11

answered May 28, 2014 at 21:16

glenn jackman

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2 Comments

mklement0 Over a year ago

+1; delightfully concise; however, you probably meant file.xml instead of xml.xml in the 2nd xmlstarlet invocation; also, to get the xmlstarlet sel command (1st invocation) to terminate its output with \n, I had to insert option -n after '//file' - without it, read wouldn't pick up the last filename. Finally, a heads-up: implies use of GNU date and stat (probably a fair assumption on a server). xmlstarlet can be found at xmlstar.sourceforge.net

glenn jackman Over a year ago

@user3685289, I would highly recommend using the date format %Y-%m-%d -- anything else is ambiguous.

miindlek · Accepted Answer · 2014-05-29 18:32:04Z

1

Here is a little python script that does that for you:

#!/usr/bin/python

import re
from datetime import date

input_file = open("input.xml", "r")
output_file = open("output.xml", "w")

today = date.today().strftime("%d/%m/%Y")
replacement = '<filedate>' + today + '</filedate>' 

for line in input_file:
    updated_line = re.sub(r'<filedate>.*?</filedate>', replacement, line)
    output_file.write(updated_line)

input_file.close()
output_file.close()

edited May 29, 2014 at 18:32

answered May 28, 2014 at 21:34

miindlek

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3 Comments

mklement0 Over a year ago

This works for the sample input file as formatted, but what if the filedate element is spread across multiple lines? Use of an XML processor avoids such issues.

miindlek Over a year ago

Shure, this is not a general solution for xml parsing. But I think it does the job for the author. And why would you split such short information like a date into multiple lines?

mklement0 Over a year ago

SO is as much about the general as it is about the specific. If you're parsing a format that is not line-oriented line by line, it makes your solution fragile (less general). If you decide to go with this approach (for a simpler, quicker solution), its limitations are worth noting - which is what I did in my comment, but doing so in your answer is preferable. Also note that your solution is not retrieving the last-modified time stamp from the files referenced (another thing worth mentioning, if you decide not to implement that part).

mklement0 · Accepted Answer · 2014-05-29 06:44:41Z

A Python solution that uses an XML parser (xml.etree) to robustly process the input:

#!/usr/bin/env python

import os
import datetime
from xml.etree import ElementTree

inFile = "in.xml"
outFile = "out.xml"

# Parse the entire document.
tree = ElementTree.parse(inFile)

# Loop over the <line> elements
for elem in tree.iter('line'):

  # Get the child element [values] of interest.
  fname = elem.find('file').text.strip()
  elemFDate = elem.find('filedate')

  # Determine the new modification date.
  # Note: As @glenn jackman notes, %Y-%m-%d would be a better date format to use.
  newDate = datetime.datetime.fromtimestamp(os.path.getmtime(fname)) \
    .strftime("%d/%m/%y")

  # Update the date element.
  elemFDate.text = newDate


# Write the output file.
tree.write(outFile)

Note:

The entire document is read and written at once, which is only appropriate for smaller XML files.
There is no error handling - all relevant elements and the files referenced are assumed to exist.

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