Simplest way to do basic xml parsing from unix command line

Question

I'm searching for xml files that have certain properties. For example, files that contain the following pattern:

<param-value>
  <name>Hosts</name>
  <description>some description</description>
  <value></value>
</param-value>

For such files, I'd like to parse the value of another tag, such as:

<param-value>
  <name>Roles</name>
  <description>some description</description>
  <value>asdf</value>
</param-value>

And print out the file name along with "asdf". What's the simplest way to accomplish this from the command line?

One approach I was thinking of was just using grep with the -l option to filter the matching files out, and then using xargs grep to extract the value of Roles. However, grep doesn't work well with multi-line regexes. I saw another question that showed it could be done with the -Pzo options, but didn't have any luck getting it to work in my case. Is there a simpler approach?

Is there any particular reason you don't want to use a scripting language such as perl? — Tom
– Tom, Commented Feb 8, 2012 at 20:07
The simplest for me is to use Saxon from the command line. Here's an example of using XPath on the command line. This, combined with a shell script, would do exactly what you're asking. — Daniel Haley
– Daniel Haley, Commented Feb 8, 2012 at 20:10
According to the answer to this question, XMLStarlet seems to be very good for this kind of thing. — Lars Kotthoff
– Lars Kotthoff, Commented Feb 8, 2012 at 20:12
No, a perl solution would be great, preferably a compact one-liner, but I don't know the best way to go about writing it. — jonderry
– jonderry, Commented Feb 8, 2012 at 20:28

Mark O'Connor · Accepted Answer · 2012-02-09 01:19:34Z

13

The following linux command uses XPath to access specified values within the XML file

for xml in `find . -name "*.xml"`
do  
echo $xml `xmllint --xpath "/param-value/value/text()" $xml`| awk 'NF>1'
done

Example output for matching XML files:

./test1.xml asdf
./test4.xml 1234

edited Feb 9, 2012 at 1:19

answered Feb 9, 2012 at 1:12

Mark O'Connor

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1 Comment

th3penguinwhisperer Over a year ago

Didn't knew xmllint could be used to parse xml. To me this is the best answer because it's always installed as it's a system dependency (at least on CentOS/Redhat/...)

jonderry · Accepted Answer · 2012-02-09 03:51:36Z

I worked out a couple of solutions using basic perl/awk functionality (basically a poor man's parsing of the tags). If you see any improvements using only basic perl/awk functionality, let me know. I avoided dealing with multiline regular expressions by setting a flag with I see a particular tag. Kind of clumsy but it works.

perl:

perl -ne '$h = 1 if m/Host/; $r = 1 if m/Role/; if ($h && m/<value>/) { $h = 0; print "hosts: ", $_ =~ /<value>(.*)</, "\n"}; if ($r && m/<value>/) { $r = 0; print "\nrole: ", $_ =~ /<value>(.*)</, "\n" }'

awk:

awk '/Host/ {h = 1} /Role/ {r = 1} h && /<value>/ {h = 0; match($0, "<value>(.*)<", a); print "hosts: " a[1]} r && /<value>/ {r = 0; match($0, "<value>(.*)<", a); print "\nrole: " a[1]}'

kev · Accepted Answer · 2012-02-09 04:32:33Z

1

$ xmlstarlet ed -u /param-value/name -v Roles -u /param-value/value -v asdf data.xml

<?xml version="1.0"?>
<param-value>
  <name>Roles</name>
  <description>some description</description>
  <value>asdf</value>
</param-value>

edited Feb 9, 2012 at 4:32

answered Feb 9, 2012 at 3:58

kev

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Comments

choroba · Accepted Answer · 2012-02-09 00:01:18Z

0

I usually use Perl's XML::XSH2. You can process XML files interactively in it, or script it. The script would be something like (untested):

for my $file in { glob "*.xml" } {
    open $file ;
    my $param_value = //param-value[name="Hosts"] ;
    if $param_value echo $file $value/value ;
}

answered Feb 9, 2012 at 0:01

choroba

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