Echo each shell script command with a timestamp

Question

set -x or set -v prints every executed command.

How do I get the command printed with the time when the command started executing?

Answer 1

In Bash, you can use PS4 - which is used by set -x to prefix tracing output.

PS4='+ $(date "+%s.%N")\011 '
set -x
...

\011 is the horizontal tab character

Answer 2

in Bash, execution of code after each command can be achieved using the "trap" builtin and the "DEBUG" level.

in bash do: help trap or refer to bash manual page, and look for the trap section

example:

trap 'date' DEBUG

This will execute the command "date" juste after each command.

Of course you can format the timestamp as you want, refer to the "date" manual page.

If you want to remove the newline, so that timestamp appears in front of next command, you can do:

trap 'echo -n $(date)' DEBUG

And to terminate this:

trap '' DEBUG

You can combine with "set -x" / "set +x"

Answer 3

Not saying this is a perfect solution by any means, but I think it basically accomplishes what you need. Just wrap the main script in a function and have a separate function to call the time. I'm sure this could be greatly improved upon to get the exact desired result, but here's what I came up with:

#!/bin/bash

tstamp() { date +[%T:%N]; }

yourscript() {
  set -x
  echo hello
  sleep 1
  echo this
  sleep 2
  echo is
  sleep 3
  echo an
  echo example
}

yourscript | while read line; do echo -e "$(tstamp)\t$line"; done

The above script outputs the following:

+ echo hello
+ sleep 1
[16:17:33:344851100]    hello
+ echo this
+ sleep 2
[16:17:34:352269700]    this
+ echo is
+ sleep 3
[16:17:36:380810900]    is
+ echo an
+ echo example
[16:17:39:392033300]    an
[16:17:39:414361400]    example