Datetime to epoch conversion

Question

I have a bash question (when using awk). I\'m extracting every single instance of the first and fifth column in a textfile and piping it to a new file with the following cod

Answer 1

awk -F '[.[:blank:]]+' '
   # use separator for dot and space (to avoid trailing time info)

   {
   # for line other than header
   if( NR>1) {
      # time is set for format "YYYY MM DD HH MM SS [DST]"
      #  prepare with valuable info
      T = "20"$1 " " $2
      # use correct separator
      gsub( /[\/:]/, " ", T)
      # convert to epoch
      E = mktime( T)

      # print result, adding fractionnal as mentionned later
      printf("%d.%d %s\n", E, $3, $7)
      }
     else {
        # print header (line 1)
        print $1 " "$7
        }
    }
   ' test170201.rawtxt \
   > Redirected.file

• self commented, code is longer for understanding purpose
• use of gnu awk for the mktime function not available in posix or older version

Oneliner a bit optimized here after

awk -F '[.[:blank:]]+' '{if(NR>1){T="20"$1" "$2;gsub(/[\/:]/," ", T);$1=mktime(T)}print $1" "$7}' test170201.rawtxt