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Comment by cldellow | Hacker News Reader

No, for example, the datediff in years for New Year's Eve and New Year's Day should be 1 (because it spans a year boundary), but EXTRACT on the difference would give you 0:

    select extract(year from '2021-01-01'::timestamptz - '2020-12-31'::timestamptz);

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Am I missing something here, or is the subtraction operator exactly right?

    > select ('2021-01-01'::timestamptz - '2020-12-31'::timestamptz) as diff;
    1 day

I think they want a function that output "1 year" given those two dates + "years" as arguments.

Correct.

But the difference between 2021-01-01 and 2020-12-31 is only one day, not a year.

It gets even strange for e.g. select DATEDIFF(year, '2021-12-01', '2022-01-01') which still returns 1 even though that's a whole month. I don't see a use case for this kind of result

Yes, but that's not what datediff does. The article doesn't really make it clear, but Microsoft explains Datediff as "This function returns the count (as a signed integer value) of the specified datepart boundaries crossed between the specified startdate and enddate." That's very different from "the difference between two date" that is what the subtract operator calculates

Oh I see! What a weird function. I've used datediff in mysql and it's ... different: https://dev.mysql.com/doc/refman/8.0/en/date-and-time-functi...

So this is correct for years, right?

    > select floor(extract(epoch from date_trunc('year', '2021-01-01'::timestamptz) - date_trunc('year', '2020-12-31'::timestamptz)) / 3600 / 24 / 365) as diff;
    1

Years aren't 365 days long, though. Technically, there are about 365.24 days in a year, but human dating doesn't take that into account well. The 3600 * 24 * 365 approach is tricky because there's a 1/4 chance that it'll work for you when you test it...

I don't think there's a good, mathematical expression for determining something as arbitrary as dates. If there was, I don't think MS would implement a special function like DateDiff for it.

So, it is this:

> # SELECT date_part('year', '2021-01-01'::timestamptz) - date_part('year', '2020-12-31'::timestamptz) as years;

Or what am I missing?

[edit: formatting]

So isn't that `age ( timestamp, timestamp ) → interval`?

The question is: what do you do with the result?

If you e.g. want to check if two timestamps are more than a year apart you just compare the difference to an interval of 1 year.

isn't that what the AGE function do?

> No, for example, the datediff in years for New Year's Eve and New Year's Day should be 1

I don't understand the purpose of such a calculation. If you want to check if two dates (or timestamps) fall into the same year, you compare the year part of them. If you want to check how far they are apart, you compare the difference between the two to an interval.

Yep. That's exactly why this function isn't trivial to write. Boundaries are not obvious and native pg functions (insofar as I'm aware of them) don't do these kinds of diffs.

Semantically the code has to diff days (for example) but be aware that you crossed a year boundary (or any other).

alternative - for 1y results:

  select date_part('year','2021-01-01'::date) - date_part('year','2020-12-31'::date) as yeardiff;

  yeardiff 
  ----------
      1
  (1 row)

That doesn't make sense to me. Why wouldn't datediff be 0 in this case, as it is less than 365 days? Else one should simply:

    extract(year from d1) - extract(year from d2)

and call it a day?

Try it again, except looking for months. You'll get -11 instead of 1.

    select extract(month from justify_interval(timestamp '2022-01-01' - timestamp '2021-11-01'))

returns 2

I'm pretty sure this will provide the desired semantics, though I am away from a usable postgres and not bothering to test it.

CREATE FUNCTION date_diff(granularity text, t1 timestamptz, t2 timestamptz) RETURNS int8 AS $$ SELECT date_part(granularity, date_trunc(granularity, t2) - date_trunc(granularity, t1))::int8 $$ LANGUAGE SQL;

Why would you want this to round up instead of down or to nearest?

It's not even rounded up, datediff('years', 2020-01-01, 2021-12-31) returns 1.

I'm not sure what it's used for though

extract(year from '2021-01-01'::timestamptz) - extract(year from '2020-12-31'::timestamptz)

For months and days it would be a bit bigger

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