[SOLVED] Shell script to find the string and insert after 2 lines in xml

Issue

Am unable to get the exact output for a XML

<String="12345">
<Bin>
<Value1 />
<Value2 />
</Bin>
</String>

Here need to find <String=12345>, then insert a line after <Bin> something similar to below

<String="12345">
 <Bin>
 <ValueXxxx />
 <Value1 />
 <Value2 />
 </Bin>
 </String>

Thanks in advance for ur help.

Solution

Although I don't intend to encourage people to parse or edit structured data as xml, the following would work for this specific case:

nl=$'\n'
nl="\\$nl"

sed '
:l
N
$!b l
s#\(<String="12345">\n[^\n]*\n\)#\1<ValueXxxx />'"$nl"'#
' input.xml

which yields:

<String="12345">
<Bin>
<ValueXxxx />
<Value1 />
<Value2 />
</Bin>
</String>

• At first it slurps all lines in pattern space to enable matching with multiple lines.
• the regex <String="12345">\n[^\n]*\n\ matches with the specified line followed by an arbitrary line.
• Then the s command inserts the string <ValueXxxx /> and a newline just after the matched lines.

Answered By - tshiono