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Finding Leap Year using SQL

Before we start writing a statement to find a Leap Year, let us first understand what is a Leap Year.

According to Gregorian Calendar each year consists of 365 day. In reality Earth takes approximately 365.242375 days to circle once around the Sun.

As the Gregorian calendar has only 365 days in a year, so if we didn't add a day on February 29 nearly every 4 years, we would lose almost six hours off our calendar every year. After only 100 years, our calendar would be off by approximately 24 days!

Lets look at it in more details,

Because the Earth rotates about 365.242375 times a year ...

... but a normal year is 365 days, ...

... so something has to be done to "catch up" the extra 0.242375 days a year.

So every 4th year we add an extra day (the 29th of February), which makes 365.25 days a year. This is fairly close, but is wrong by about 1 day every 100 years.

So every 100 years we don't have a leap year, and that results in 365.24 days per year (1 day less in 100 year = -0.01 days per year). Closer, but still not accurate enough!

So another rule says that every 400 years is a leap year again. This gets us 365.2425 days per year (1 day regained every 400 years = 0.0025 days per year), which is close enough to 365.242375 not to matter much.

So this is why we have Leap Year once every 4 years.

According to Gregorian calendar 3 criteria must be taken into account to identify leap years:

1. The year is evenly divisible by 4;
2. If the year can be evenly divided by 100, it is NOT a leap year, unless;
3. The year is also evenly divisible by 400. Then it is a leap year.

This means that 2000 and 2400 are leap years, while 1800, 1900, 2100, 2200, 2300 and 2500 are NOT leap years.

The year 2000 was somewhat special as it was the first instance when the third criterion was used in most parts of the world since the transition from the Julian to the Gregorian Calendar.

Now its time for an SQL which can check all 3 conditions.

Following example is with CASE conditional Expression,

CASE WHEN (mod(&&a,4) = 0 and mod(&a,100) <> 0) OR
(mod(&a,4) = 0 and mod(&a,100) = 0 and mod(&a,400) = 0)
THEN '&&a is a Leap Year'
ELSE '&a is not a Leap Year' END
FROM dual;

Following example is without CASE,

SELECT '&&a is a Leap Year'
FROM dual
WHERE (mod(&a,4) = 0 and mod(&a,100) <> 0)
(mod(&a,4) = 0 and mod(&a,100) = 0 and mod(&a,400) = 0);


Unknown said…
Hi Mithun,
When we execute this statement for second time it is directly giving the same output of the previous input i.e once the variable given a value or initialized we need to take another variable name and execute the statement for another input; Is there any solution that we can give different input for the same variable.
Mithun said…
Hi Chandra,

Once you execute the query to set the variable again, execute the command below,

undef a
Unknown said…
thank u Mithun ...
Unknown said…
hello mithun..
I need a query to display my current age .... so my birth date is 16th sep 1988...
Thank you..
Unknown said…
This comment has been removed by the author.
Unknown said…
select trunc(months_between(sysdate,'16-sep-1988')/12) years from dual

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