Bash Script to take the timestamp of a logged files and subtract it from the timestamp of another log file

问题

I am trying to write a script that will have the following information:

   4824597 11:26 /home/customer1/ITAM.xml
   .
   .
   .
   .
   4824597 14:08 /home/customer46/ecds/dropoff/ITAM.xml

I have another file that will log the same information

   4824597 11:28 /home/customer1/ITAM.xml
   .
   .
   .
   .
   4824597 14:11 /home/customer46/ecds/dropoff/ITAM.xml

I want see how long it takes to receive files. To do this I want to subtract the timestamp of all the files logged in the log, and subtract them from their initial timestamp in the first log file.

I am new to scripting and am struggling to make this work. I am trying:

   #!/bin/bash
   time=$11:48:30
   day=$2012-10-12
   time2=$13:13:48
   t=$(date -d "day time" +%s)
   t1=$(date -d "day time2" +%s)
   diff=$(expr $t1 - $t)
   echo $diff

So far no luck. Any help would be greatly appreciated.

回答1:

Try something like this:

awk '{
  sub(/:/," ",$2);
  t1=mktime(strftime("%Y %m %d")" "$2" 00");
  getline < "input2.txt";
  sub(/:/," ",$2);
  t2=mktime(strftime("%Y %m %d")" "$2" 00");
  print $3": "t2-t1" s"
}' input1.txt

来源：https://stackoverflow.com/questions/17241982/bash-script-to-take-the-timestamp-of-a-logged-files-and-subtract-it-from-the-tim