Bash command to get start time and endtime from log file-CodePudding

I need to get the start time and end time of from a log file which looks like the below format:

REQUEST     test 1 1651474077633    
.
.
.
.
REQUEST     test n  1651474676561

I want to print start and end time. The time is in Unix format and I want to convert it into standard date and time format.

code which I tried:

cat ${FILENAME} | awk 'BEGIN {
   print strftime("Time = %m/%d/%Y %H:%M:%S", NF)
}'

But it prints all time.

I just want time for 1st entry and last entry.

CodePudding user response：

I think your code uses BEGIN and NF incorrectly.

Try the code below.

cat ${FILENAME} | awk '$1 == "REQUEST" { print strftime("Time = %m/%d/%Y %H:%M:%S", $4) }'

If you want to output only the first and last outputs:

cat ${FILENAME} | awk '
BEGIN { first = 1 }
$1 == "REQUEST" {
  if (first) {
    firsttime = $4
    first = 0
  }
  lasttime = $4
}
END {
  print strftime("First Time = %m/%d/%Y %H:%M:%S", firsttime)
  print strftime("Last  Time = %m/%d/%Y %H:%M:%S", lasttime)
}'

The numbers in the text in question do not seem to be in UNIX time, but you can check there.

CodePudding user response：

They're indeed valid posix epochs time, in milliseconds

just divide it by a 1000, and you'll get a reasonable timestamp - May 3rd, 2022 :

echo "${ccccc}" \
\
| gawk -be '
  function printtime(_,__,___,____) { 
      print \
          substr((____=RS)(RS="\n"),
          (__=sprintf("gdate  \47%5s Time = "(___\
                 )"\47",__,_*1e-3)) | getline _,
               RS*close(__)^(RS=____))_ 
 } 1<NF {
          __[  _]=$!(NF=NF) 
 }  END { 
          printtime(__[_^!_], "First", ___)
          printtime(__[( _)],  "Last", ___)

 }' ___='%D %T.%6N'  FS='^REQUEST[^t] test[ ] ([n]|[0-9] )[ ] ' | ecp
 
First Time = 05/03/22 00:55:57.458000
 Last Time = 05/03/22 00:55:57.462939