Parsing XML attributes with bash

Question

I need to grab attributes from the XML file with pure bash script.

So I have the following XML file with a root element Group and lots of Person elements, every of them has id and username attributes. id is unique value for each element:

<?xml version="1.0" encoding="UTF-8"?> <Group id="D_8" main="false"> <Person id="P_0001" email="[email protected]" username="person_0001" password="pass_0001" active="true"/> <Person id="P_0002" email="[email protected]" username="person_0002" password="pass_0002" active="true"/> <!-- ...and hundreds of other Person elements ... --> </Group> 

And I need to use bash script to extract the id and username attributes into some key-value structure:

P_0001=person_0001 P_0002=person_0002 

Checked other related answers, but most of them suggest to use some XML parsers like xmllint. But unfortunately I do not have them on the target machine.

Please suggest how I can achieve this.

Answer 1

As long as the username attribute does not come before id attribute, this is a bash script to give the result:

#/usr/bin/env bash id='\bid="([^"]+)"' username='\busername="([^"]+)"' while IFS= read -r line; do [[ $line =~ $id ]] && idv="${BASH_REMATCH[1]}" [[ $line =~ $username ]] && echo "$idv=${BASH_REMATCH[1]}" done < data.xml exit 0 

It works even when username attribute and id attribute are on the same line.

Answer 2

Assumptions:

• the xml file is 'nicely' formatted as the presented example (hence no need for a XML parser)
• id is the first attribute in a Person block (ie, we'll always be able to match on the literal string Person id), otherwise additional parsing will be necessary
• Person id and username only show up in the Group block

One awk solution:

awk -F'"' ' /<Person id=/ { pid=$2 ; next } /[[:space:]]username=/ { printf "%s=%s\n", pid, $2 ; next } ' test.xml 

Running this awk solution against the sample data file generates:

P_0001=person_0001 P_0002=person_0002 

All bets are off if my assumptions are invalid.