Hi there I am hoping you could help me with a script that would search for a word and extract the block of text containing it. This is a log file containing multiple messages. I would like to first search for a keyword, ("bwSipRequestTimeOutReceived")and then extract the lines of text containing it. Each log is seperated by a blank line.See below as an example of lines of text of the message any help with awk or similar command to acheive this would be greatly appreciated. Thankyou very much 3234049,1197578062818,1,0,false, "bwSipRequestTimeOutReceived", "as1", "Application Server", "SIP", "SIP respond 408 Request Timeout received. Response Message: SIP/2.0 408 Request Timeout Via:SIP/2.0/UDP 218.101.55.1:5060;received=218.101.55.1;branch=z9hG4bK-BroadWorks.218.101.55.1-218.101.55.201V5060-0-714830972-479130644-1197578053878- Call-ID:BW093413878141207-1215374075@218.101.55.1 From:<sip:+xxxxxxx@218.101.55.1;user=phone>;tag=479130644-1197578053878- To:<sip:011691yyyyyyyyy@csps.softswitch.ddddddd.net:5060;user=phone>;tag=2891-587c CSeq:714830972 INVITE Content-Length:0

i think this may work, but will need some manipulation. FILE="path of log file" SEARCH="\"bwSipRequestTimeOutReceived\"" cat $FILE | \ while read line do index1="`awk -v a="$line" -v b="$SEARCH" 'BEGIN{print index(a,b)}'` " if [ index1 -ne 0 ] ; then echo $line fi done Jain Sahab

perl -00 -ne 'print if /bwSipRequestTimeOutReceived/' file.log