I'm trying to parse some data from a log file. The premises are that i need a date that is 2 lines above the the line with the pattern that i want.
I'm able to reach this with grep:
> grep -B2 "rc_transaction result=" xml.log
And the output is 3 lines for each pattern find:
03 Apr 12:48:42.503 [6086-16592] DEBUG c.q.q.d.server.XmlServlet - <?xml version="1.0"?>
<stream id="18561">
<rc_transaction result="ok" vst_time="2018-04-03 10:48:42.431" transaction_time="2018-04-03 10:48:42.497" sequence_number="117749" code="0">
--
03 Apr 12:49:21.936 [6086-16592] DEBUG c.q.q.d.server.XmlServlet - <?xml version="1.0"?>
<stream id="18566">
<rc_transaction result="ok" vst_time="2018-04-03 10:49:21.839" transaction_time="2018-04-03 10:49:21.930" sequence_number="117750" code="0">
--
03 Apr 12:49:39.654 [6086-16592] DEBUG c.q.q.d.server.XmlServlet - <?xml version="1.0"?>
<stream id="18569">
<rc_transaction result="ok" vst_time="2018-04-03 10:49:39.582" transaction_time="2018-04-03 10:49:39.648" sequence_number="117751" code="0">
Now, i need to parse this results an get the date on the first line and some properties from the third line, something like this:
03 Apr 12:48:42.503 result="ok" sequence_number="117749"
03 Apr 12:49:21.936 result="ok" sequence_number="117750"
03 Apr 12:49:39.654 result="ok" sequence_number="117751"
Whats the best method to achieve this?