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#8
01-16-2008
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Thanks to this thread, awk: need to extract a line before a pattern I finally managed to extract out the data that I wanted.
Code:
awk '$0~/Start Time/{lineno=NR; text=$0} NR==lineno+6 && $0~/send messages/{print text}' filename
Using matrixmadhan's code : Code:
sed -n '/Start Time/s/^.*CDATA\[\(.*\)]].*$/\1/p' filename Thanks!  
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