I want to make code to execute a task 3 seconds before a given time.
Something like this:
var Time = new Date('December 25, 1995 08:00:00:000');
Time.getSeconds() - 3 ;
This should give me 57, instead of -3
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VLAZ
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11. Modify the date to be 3 seconds earlier 2. Get the seconds. 3. Done – VLAZ Apr 15 '21 at 11:44
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If you get the seconds of `08:00:00`, it's `00`. If you subtract 3, the result is `-3`. Makes sense, doesn't it? Why should `0 - 3` give `57`? – Jeremy Thille Apr 15 '21 at 11:46
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Because if its 08:00:00am, -3 sec = 07:59:57 :) – Fábio Vilela Apr 15 '21 at 11:50
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1You are not subtracting 3 from `08:00:00am`. You are subtracting 3 from `Time.getSeconds()`, which is `0`. – Jeremy Thille Apr 15 '21 at 11:53
1 Answers
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You need to set the seconds via the the dedicated method first - the logic to handle such “rollovers” is already implemented in there - and then you read it again:
// var Time = new Date('December 25, 1995 08:00:00:000');
// careful with the above date format, not all browsers will parse this correctly
// modified version, for this example:
var Time = new Date('1995-12-25T08:00:00');
Time.setSeconds( Time.getSeconds() - 3 );
console.log(Time.getSeconds());
CBroe
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1Should work [better than this](https://i.imgur.com/lajk92i.png) if the date was valid. [Demo](https://jsbin.com/vatonay/1/edit?js,console) – VLAZ Apr 15 '21 at 11:49
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@VLAZ yes, good point. Fabio, the date format used in the initial example does not work in all browsers - see https://stackoverflow.com/questions/2587345/why-does-date-parse-give-incorrect-results – CBroe Apr 15 '21 at 11:51