Need to cut a specific pattern from a line

Shell Programming and Scripting
1

Hello,
I need to cut a specific pattern from a line irrespecitve of knowing field place.

I am aware to cut field if you know the place of the field, but for me The sting place varies.

1468129514436,0,something_error,Non HTTP response code: java.net.URISyntaxException,Non HTTP response message: Illegal character in query at index 81: http://xxx:000/abcd/jsp/soh/fixedinquiries/InquiryFrameSet.jsp?accountSuffix=${SUFFIX_ID}&accountNumber=Number Not Found&domainName=%20&corpUserName=%20&accountId=${ACCOUNT_ID}&pgCode=M&pCode=${PCODE}&imsiNumber=${IMSI_ID}&evisionSTBSerialNo=%20&eccNumber12=%20&arabicDomainName=%20&operation=3&searchPage=SOHACCOUNTIDINQ,Thread Group 1-1,text,false,1226,1,1,0

out of above line I am able to get

cat log.jtl | egrep 'false'| egrep  'http://' | cut -d ',' -f3

False and http will always be there , by above command I get error line and i also get 3rd filed. but i also want to print URL in the error

something_error   http://xxx:000/abcd
2

Not too clear what you really want. Based on what I infer, try

awk -F, '/false/ && match ($0, /http:\/\/[^\/]*\/[^\/]*/) {print $3, substr ($0, RSTART, RLENGTH)}' file3
something_error http://xxx:000/abcd
3

I got below error.
I want to print the third field and the the url .
url is not at specific position every time, it can be at filed 10, 12 13 and so on.
i want script to pick where ever it is in the line

awk: syntax error near line 1
awk: bailing out near line 1
4

When using awk on Solaris/SunOS systems, change awk to /usr/xpg4/bin/awk or nawk .

5

Thanks DON, it worked.

I will consider to use [CODE] instead of [ICODE]

6

Again, you're not too eloquent with your request nor error description. Reading between your lines, I guess I recognize the problem. Let me cite