Find a pattern in XML using unix command

Question

I have XML like below

<a>
</a>
<a>
<b></b>
<c></c>
</a>

I need to find line numbers of the empty tags which are in two consecutive lines without any tags in between them like below.

    <a>
    </a>

Output: 1 -> line number

I'm using PCGREP but my unix admins have restricted or not supporting perl commands hence can this be achieved using SED or AWK? Note: xmlstarlet is not supported too.

Answer 1

Sed solution

#!/bin/bash

sed -nr '
    /^<[^<]*>$/ {
        N
        /^<([^<]*)>\n<\/\1>$/=
    }
' "$1" | awk '{print "Output: "$NF - 1" -> line number"}'

Explanation:

1. sed
   • /^<[^<]*>$/ if we are have one open tag in the line
   • N - append the next line of input into the pattern space.
   • /^<([^<]*)>\n<\/\1>$/ and check, does the next line have the equivalent closed tag.
   • if so, print this line number by = command. Bear in mind, that it is the closed tag line number. We should decrease it by one in further.
2. awk - decreases the line number and print it in the message string.

Testing:

Input

<a>
</a>
<a>
<b></b>
<c></c>
<c>
</c>
</a>

Output

./empty_tag.sh input.txt 
Output: 1 -> line number
Output: 6 -> line number

---

AWK solution

Usage: ./empty_tag.sh input.txt

#!/bin/bash

awk -F'[>/]' '
    line_num {
        if(NF == 3) {print "Output: " line_num " -> line number";}
        line_num = 0;
    }
    NF == 2 {line_num = NR;}
' "$1"