Grep – How to Read a Log File Between Two Dates

You could use some evil trickery:

date -d "Feb 12 10:02:10" +%s

will return the number of seconds since the epoch for that date. Combining this with the same command (for the other date) and running through bc:

echo "$(date -d 'Feb 12 10:53:15' +%s) - $(date -d 'Feb 12 10:02:10' +%s)" | bc

or the shell's arithmetic expansion:

echo "$(( $(date -d 'Feb 12 10:53:15' +%s) - $(date -d 'Feb 12 10:02:10' +%s) ))"

will give you

3065

the number of seconds between the times. You could probably parse this out with awk somehow if you needed to run it more than once.

How about using the date utility to iterate through the range of dates you're interested in? Here is what that might look like for your example:

# Set the date counter to the start date
d=2017-09-03

# Iterate until we reach the end date (i.e. the date after the last date we want)
while [ "$d" != 2017-10-09 ]; do

    # cat each file
    cat "localhost_log_file.${d}.txt";

    # Increment the date counter
    d="$(date -I -d "$d + 1 day")";

done

See this for more information:

• Bash: Looping through dates

Alternatively, you can pass the results of the loop to the cat command instead of invoking cat in the body of the loop.

Here is what that could look like using command-substitution:

d=2017-09-03
cat $(while [ "$d" != 2017-10-09 ]; do
    echo "localhost_log_file.${d}.txt";
    d="$(date -I -d "$d + 1 day")";
done)

And here is the same thing using a pipe and xargs:

d=2017-09-03
while [ "$d" != 2017-10-09 ]; do
    echo "localhost_log_file.${d}.txt";
    d="$(date -I -d "$d + 1 day")";
done | xargs cat