How to get only first occurrence in log file using awk - awk

Question

i've a log file like this
some text line
other text line
<a>
<b>1</b>
<c>2</c>
</a>
another text line
<a>
<b>1</b>
<c>2</c>
</a>
yet another text line
I need to get only ther first occurrence of the XML "a":
<a>
<b>1</b>
<c>2</c>
</a>
I know
awk '/<a>/,/<\/a>/' file.log
will find all occurrences, how can I get just the first? (adding |head -n1 obvously doesn't work because it will capture only first line, and I can't know for sure how long "a" is because the awk expression must be generic because I've different log files with different "a" contents)

Answer 1

Another slight variation is to simply use a simple counter variable to indicate when you are in the first <a>...</a> block, outputting that block and then exiting afterwards. In your case using n as the variable to indicate in the first block, e.g.
awk -v n=0 '$1=="</a>" {print $1; exit} $1=="<a>" {n=1}; n==1' f.xml
Example Use/Output
With your input file as f.xml you would get:
$ awk -v n=0 '$1=="</a>" {print $1; exit} $1=="<a>" {n=1}; n==1' f.xml
<a>
<b>1</b>
<c>2</c>
</a>
(note: the {n=1} and n==1 rules rely on the default operation (print) to output the record)

Answer 2

This awk:
awk '
match($0,/<a>/) {
$0=substr($0,RSTART)
flag=1
}
match($0,/<\/a/) {
$0=substr($0,1,RSTART+RLENGTH)
print
exit
}
flag' file
can handle these forms:
The above awk handles this:
<a><b>1</b><c>2</c></a>
and this:
<a>
<b>1</b>
<c>2</c>
</a>
and also <a>
<b>1</b>
<c>2</c>
</a> this
the end
Another for GNU awk:
$ gawk -v RS="</?a>" '
NR==1 { printf RT }
NR==2 { print $0 RT }
' file

Answer 3

First:
$ awk '/<a>/{f=1} f; /<\/a>/{exit}' file
<a>
<b>1</b>
<c>2</c>
</a>
Last:
$ tac file | awk '/<\/a>/{f=1} f; /<a>/{exit}' | tac
<a>
<b>1</b>
<c>2</c>
</a>
Nth:
$ awk -v n=2 '/<a>/{c++} c==n{print; if (/<\/a>/) exit}' file
<a>
<b>1</b>
<c>2</c>
</a>