How to execute a date function inside an awk command in an fast way?

I am reading a file using awk and parsing the date and converting it into seconds. I wrote a working code that achieves this but it runs relatively slow and I have large log files that I intend to read. I need to know if there is another way that can achieve this in less time. Here is my code:

awk '
 {
  currentDateTime=$1 " " $2
  seconds ="date \"+%s\" -d \""currentDateTime" \""
  print "The time in seconds is: [ " seconds "]"
  seconds | getline result
  print result 
 }'  out.log

Here is the output:

The time in seconds is: [ date "+%s" -d "2014-04-01 10:20:22,357 "]
1396336822

Note that in the first output, the variable currentDateTime is outputted as a command and not as a value. I am wondering if I could execute this command without using getline since the getline seems slow when trying it out on larger files.

Solution

The reason getline is slow is that in this case you are running a shell command.

That's what is slow.

You are spawning a new shell and running date in it for each entry.

Stop doing that.

Use the GNU awk mktime function if you are using GNU awk.