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Re: Parse bad xml file

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David Jobes wrote:

> I was given a badly or poor formatted xml file that i need to convert to
> csv file:

There are no "badly formatted" XML files, only valid and invalid ones.
Fortunately  following looks like the beginning of a valid one.

> <?xml version="1.0"?>
> <resultset xmlns:dyn="http://exslt.org/dynamic">
> <table name="SIGNATURE">
> <column name="ID" type="String">	</column>
> <column name="NUM" type="Integer">	</column>
> <column name="SEVERITY_ID" type="Integer">	</column>
> <column name="NAME" type="String">	</column>
> <column name="CLASS" type="String">	</column>
> <column name="PRODUCT_CATEGORY_ID" type="Integer">	</column>
> <column name="PROTOCOL" type="String">	</column>
> <column name="TAXONOMY" type="String">	</column>
> <column name="CVE_ID" type="String">	</column>
> <column name="BUGTRAQ_ID" type="String">	</column>
> <column name="DESCRIPTION" type="String">	</column>
> <column name="MESSAGE" type="String">	</column>
> <column name="FILTERTYPE" type="String">	</column>
> <data>
> <r>
> <c>00000001-0001-0001-0001-000000000027</c>
> <c>27</c>
> <c>2</c>
> <c>0027: IP Options: Record Route (RR)</c>
> <c>Network_equip</c>
> <c>10</c>
> <c>ip</c>
> <c>100741885</c>
> <c>2001-0752,1999-1339,1999-0986</c>
> <c>870</c>
> <c></c>
> <c></c>
> <c></c>
> </r>
> 
> 
> I have been able to load and read the file line by line, 

XML doesn't have an idea of lines, so don't do that. Instead let a parser 
make sense of the document structure.

> but once i get to
> the r line and try to process each c(column) that is where it blows up. I
> need to be able to split the lines and place each one or the r (row) on a
> single line for the csv.
> 
> i have a list set for each one of the headers based on the col name field,
> i just have been able to format properly.

Here's a simple script using ElementTree, to introduce you to basic xml 
handling with Python's stdlib. If you are lucky it might even work ;)

import csv
import sys
from xml.etree import ElementTree

SOURCEFILE = "xml_to_csv.xml"

tree = ElementTree.parse(SOURCEFILE)
table = tree.find("table")
column_names = [c.attrib["name"] for c in table.findall("column")]
writer = csv.writer(sys.stdout)
writer.writerow(column_names)
for row in table.find("data").findall("r"):
    writer.writerow([field.text for field in row.findall("c")])

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Thread

Parse bad xml file David Jobes <david.jobes.sony@gmail.com> - 2014-10-10 04:46 -0700
  Re: Parse bad xml file Peter Otten <__peter__@web.de> - 2014-10-10 14:20 +0200
    Re: Parse bad xml file David Jobes <david.jobes.sony@gmail.com> - 2014-10-10 05:33 -0700
      Re: Parse bad xml file Rustom Mody <rustompmody@gmail.com> - 2014-10-12 01:37 -0700

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