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Date: Wed, 27 Apr 2011 14:41:49 -0400
Subject: Re: ElementTree XML parsing problem
From: Benjamin Kaplan <benjamin.kaplan@case.edu>
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On Wed, Apr 27, 2011 at 2:26 PM, Mike <Mike@invalid.invalid> wrote:
> I'm using ElementTree to parse an XML file, but it stops at the second
> record (id =3D 002), which contains a non-standard ascii character, =E4. =
Here's
> the XML:
>
> <?xml version=3D"1.0"?>
> <snapshot time=3D"Mon Apr 25 08:47:23 PDT 2011">
> <records>
> <record id=3D"001" education=3D"High School" employment=3D"7 yrs" />
> <record id=3D"002" education=3D"Universit=E4t Bremen" employment=3D"3 yea=
rs" />
> <record id=3D"003" education=3D"River College" employment=3D"5 yrs" />
> </records>
> </snapshot>
>
> The complaint offered up by the parser is
>
> Unexpected error opening simple_fail.xml: not well-formed (invalid token)=
:
> line 5, column 40
>
> and if I change the line to eliminate the =E4, everything is wonderful. T=
he
> parser is perfectly happy with this modification:
>
> <record id=3D"002" education=3D"University Bremen" employment=3D"3 yrs" /=
>
>
> I can't find anything in the ElementTree docs about allowing additional t=
ext
> characters or coercing strange ascii to Unicode.
>
> Is there a way to coerce the text so it doesn't cause the parser to raise=
 an
> exception?
>

Have you tried specifying the file encoding? =E4 is not "strange ascii".
It's outside the ASCII range so if the parser expects ASCII, it will
get confused.

> Here's my test script (simple_fail contains the offending line, and
> simple_pass contains the line that passes).
>
> import sys
> import xml.etree.ElementTree as ET
>
> def main():
>
> =A0 =A0xml_files =3D ['simple_fail.xml', 'simple_pass.xml']
> =A0 =A0for xml_file in xml_files:
>
> =A0 =A0 =A0 =A0print
> =A0 =A0 =A0 =A0print 'XML file: %s' % (xml_file)
>
> =A0 =A0 =A0 =A0try:
> =A0 =A0 =A0 =A0 =A0 =A0tree =3D ET.parse(xml_file)
> =A0 =A0 =A0 =A0except Exception, inst:
> =A0 =A0 =A0 =A0 =A0 =A0print "Unexpected error opening %s: %s" % (xml_fil=
e, inst)
> =A0 =A0 =A0 =A0 =A0 =A0continue
>
> =A0 =A0 =A0 =A0root =3D tree.getroot()
> =A0 =A0 =A0 =A0records =3D root.find('records')
> =A0 =A0 =A0 =A0for record in records:
> =A0 =A0 =A0 =A0 =A0 =A0print record.attrib['id'], record.attrib['educatio=
n']
>
> if __name__ =3D=3D "__main__":
> =A0 =A0 =A0 =A0main()
>
>
> Thanks,
>
> -- Mike --
>
> --
> http://mail.python.org/mailman/listinfo/python-list
>