Simple timer to measure seconds an operation took to complete

I run my own script to dump databases into files on a nightly basis.

I wanted to count time (in seconds) it takes to dump each database, so I was trying to write some functions to help me achieve it, but I'm running into problems.

I am no expert in scripting in bash, so if I'm doing it plain wrong, just say so and ideally suggest alternative, please.

Here's the script:

#!/bin/bash

declare -i time_start

function get_timestamp {
        declare -i time_curr=`date -j -f "%a %b %d %T %Z %Y" "\`date\`" "+%s"`
        echo "get_timestamp:" $time_curr
        return $time_curr
}

function timer_start {
        get_timestamp
        time_start=$?
        echo "timer_start:" $time_start
}

function timer_stop {
        get_timestamp
        declare -i time_curr=$?
        echo "timer_stop:" $time_curr
        declare -i time_diff=$time_curr-$time_start

        return $time_diff
}

timer_start
sleep 3
timer_stop
echo $?

The code should really be quite self-explanatory. echo commands are only for debugging.

I expect the output to be something like this:

$ bash timer.sh
get_timestamp: 1285945972
timer_start: 1285945972
get_timestamp: 1285945975
timer_stop: 1285945975
3

Now this is not the case unfortunately. What I get is:

开发者_如何学编程$ bash timer.sh
get_timestamp: 1285945972
timer_start: 116
get_timestamp: 1285945975
timer_stop: 119
3

As you can see, the value that local var time_curr gets from the command is a valid timestamp, but returning this value causes it to be changed to an integer between 0 and 255.

Can someone please explain to me why this is happening?

PS. This obviously is just my timer test script without any other logic.

---

UPDATE Just to be perfectly clear, I want this to be part of a bash script very similar to this one, where I want to measure each loop cycle.

Unless of course I can do it with time, then please suggest a solution.

---

You don't need to do all this. Just run time <yourscript> in the shell.

---

$? is used to hold the exit status of a command and can only hold a value between 0 and 255. If you pass an exit code outside this range (say, in a C program calling exit(-1)), the shell will still receive a value in that range and set $? accordingly.

As a workaround, you could just set a different value in your bash function:

function get_timestamp {
        declare -i time_curr=`date -j -f "%a %b %d %T %Z %Y" "\`date\`" "+%s"`
        echo "get_timestamp:" $time_curr
        get_timestamp_return_value=$time_curr
}

function timer_start {
        get_timestamp
        #time_start=$?
        time_start=$get_timestamp_return_value
        echo "timer_start:" $time_start
}

...

---

I believe you should be able to use the existing "time" function.

After Update to the question: This was the bit of script from your link which was doing a for loop.

# dump each database in turn
for db in $databases; do
    echo $db
    $MYSQLDUMP --force --opt --user=$USER --password=$PASSWORD
    --databases $db > "$OUTPUTDIR/$db.bak"
done

You could extract the inner portion of the loop into a new script (call it dump_one_db.sh) and do this inside the loop:

  # dump each database in turn
    for db in $databases; do
        time dump_one_db.sh $db
    done

Make sure to write the output of the time against the db name into some file.

---

This is happening because return codes need to be between 0-255. You can't return an arbitrary number. If you continue to refuse to use the builtin time function and roll your own, change your functions to echo their stamp and use a process expansion ($()) to grab the value.