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Extract XML message from a log file using awk

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# 1  
Old 06-21-2011
Extract XML message from a log file using awk
Dear all

I have a log file and the content like this

file name: temp.log
Code:
<?xml version="1.0" encoding="cp850"?>

<!DOCTYPE aaabbb SYSTEM '/dtdpath'>

<aaabbb>
  <tranDtl>
    <msgId>000001</msgId>
  </tranDtl>
.....

</aaabbb>
...
...
...
<?xml version="1.0" encoding="cp850"?>

<!DOCTYPE aaabbb SYSTEM '/dtdpath'>

<aaabbb>
  <tranDtl>
    <msgId>000002</msgId>
  </tranDtl>
.....

</aaabbb>

What I want to do is to extract the xml message and redirect it to a file called 000001.xml, 000002.xml etc...
start string = <?xml version="1.0" encoding="cp850"?>
end string = </aaabbb>

My idea is to get the start string (ss) and end string (es) line number first to a txt file, then use sed -n ss,eep > msgId.xml....
Code:
grep -n "xml version" temp.log | awk -F":" '{print $1}' | cat -n > start.txt
grep -n "</gcrLis>" temp.log | awk -F":" '{print $1}' | cat -n > end.txt
grep "<msgId>" temp.log | awk -F"[>,<]" '{print $3}' | cat -n > msgid.txt

Then loop the *.txt file and use sed to do so.

Although this approach is work but it is not pretty enough, how can I achieve this with awk. Please kindly advise.. Thank you so much.
Last edited by pludi; 06-22-2011 at 06:25 AM..
# 2  
Old 06-22-2011
With gnu coreutils you can:
Code:
% csplit -z testfile '/^<?xml/' '{*}'

You get your chunks in files xx00, xx01, etc.
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