How do you find the difference between two dates in bash?

How do you find the difference between two dates in bash?

1. Provide valid time string in A and B.
2. Use date -d to handle time strings.
3. Use date %s to convert time strings to seconds since 1970 (unix epoche)
4. Use bash parameter expansion to subtract seconds.
5. divide by seconds per day (86400=60*60*24) to get difference as days.
6. ! DST is not taken into account ! See this answer at unix.

How do you compare two date variables?

For comparing the two dates, we have used the compareTo() method. If both dates are equal it prints Both dates are equal. If date1 is greater than date2, it prints Date 1 comes after Date 2. If date1 is smaller than date2, it prints Date 1 comes after Date 2.

How to do a date comparison in Bash?

The ( ( )) syntax is used to create an arithmetic context as we are comparing two numbers. You could also use the more general [ $ {x} -lt $ {y} ] style syntax if portability is a concern. date -d “next year” +%s. Furthermore, date understands timezones, so it can correctly handle comparisons between pairs of dates where the timezone is different.

What’s the best way to compare points of time?

You can use the timestamp format date +%s and the -r option. The best way to compare points of time (dates) is in seconds (since Epoch). Thanks for contributing an answer to Unix & Linux Stack Exchange!

Why do I get the wrong value when comparing two dates?

I am trying to compare two dates in a shell script but I always get a wrong value. This is the code I am using: Even if it is smaller. I always get peak time is bigger than current. So the if statement always prints the first sentence. This is because you are using < in a [ ], and that doesn’t do arithmetic comparison, it’s input redirection.

How to compare dates to days in MySQL?

dates are strings, not integers; you can’t compare them with standard arithmetic operators. This works as far back as bash 2.05b.0 (1)-release. You can also use mysql’s builtin function to compare the dates. It gives the result in ‘days’. Just insert the values of $DBUSER and $DBPASSWORD variables according to yours.