Bash command to get the last occurence of a tag in a XML structure

Question

I have below a sample XML structure:

<root>
<id>1</id>
<name>abc</name>
<id>2</id>
<name>def</name>
<id>3</id>
<name>ghi</name>
</root>

I am trying to achieve the one single following job:

Retrieve the text value of the last occurrence of an id tag in the structure. (Should output 3 in this case).

Can you please help?

I have tried using sed but could not get it to output 3. :D

[Don't Parse XML/HTML With Regex.](https://stackoverflow.com/a/1732454/3776858) I suggest to use an XML/HTML parser (xmlstarlet, xmllint ...). — Cyrus, Aug 13 '19 at 15:52
If you're parsing XML in a shell script, use the `xmlstarlet` utility. — Barmar, Aug 13 '19 at 16:12
If the format is always like that, `grep '' filename | tail -1` will find the last `` line, then you can extract the number from it. — Barmar, Aug 13 '19 at 16:13

user192234 · Answer 1 · 2019-08-13T20:08:32.713

You can use xpath queries that traverse an XML file or string.

For example:

xpath -q -e '//root/id[last()]/text()' file.xml

xpath

is an executable/command that takes a query and an XML.

-q stands for quiet, meaning you only get what you asked for in the stdout.

-e comes right before a query string.

// stands for a root to start searching.

last() is what you'd expect -1 to do :).

text() takes only the value and doesn't print it surrounded by tags.

1 Answers1