Just a reference post for how to determine the number of seconds between two date strings in awk. If you have a normally formatted date string in Linux, such as “Sun Mar 19 00:50:33 EDT 2017”, you can convert this to a time object with the following…<\/p>\n
\r\nbash-4.1$ grep 2017$ list2.out | awk 'BEGIN{split(\"Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec\",l,\"|\")} {for (i in l) {if ($2 == l[i]) {month=i}};split($4,t,\":\");t1=mktime($6\" \"month\" \"$3\" \"t[1]\" \"t[2]\" \"t[3]);if (last > 0 && t1-last > 35) {print $0,t1-last};last=t1}'\r\nSat Mar 18 17:14:18 EDT 2017 96\r\nSun Mar 19 00:50:33 EDT 2017 95\r\n<\/pre>\nThe mktime() command expects a date formatted as “year month day hour minute second”. If you can parse that out of string, you can convert it to a date object and perform math operations on it.<\/p>\n
What is above prints only those lines that have a timestamp greater than 35 seconds after the previous sample. It also prints the difference in seconds between the two samples.<\/p>\n","protected":false},"excerpt":{"rendered":"
Just a reference post for how to determine the number of seconds between two date strings in awk. If you have a normally formatted date string in Linux, such as “Sun Mar 19 00:50:33 EDT 2017”, you can convert this…<\/p>\n