Is there any way to optimise this JavaScript further (comparing current time to opening hours)?

Question

I am setting up some JavaScript to compare the current time to a venue's opening and closing hours (potentially different each day, so different var for each day).

Here's what I have so far:

// Compare current time to today's hours
if (day === 1 && time > monOpen && time < monClose) {
    venueIsOpen();
} else if (day === 2 && time > tuesOpen && time < tuesClose) {
    venueIsOpen();
} else if (day === 3 && time > wedOpen && time < wedClose) {
    venueIsOpen();
} else if (day === 4 && time > thursOpen && time < thursClose) {
    venueIsOpen();
} else if (day === 5 && time > friOpen && time < friClose) {
    venueIsOpen();
} else if (day === 6 && time > satOpen && time < satClose) {
    venueIsOpen();
} else if (day === 0 && time > sunOpen && time < sunClose) {
    venueIsOpen();
} else {
    venueIsClosed();
}

Obviously very straightforward - is there any way to optimise this?

Why do you have so many variables? Shouldn't you put the open and close times in an array of size 7? An "optimisation" would be quite straightforward then. — Bergi, Nov 20 '15 at 14:59
Review/refactoring requests are offtopic here. Go to [CodeReview](http://codereview.stackexchange.com/). — hindmost, Nov 20 '15 at 15:10

Cerbrus · Accepted Answer · 2015-11-20T15:14:26.087

First of all, we can combine a load of those if statements:

// Compare current time to today's hours
if (day === 1 && time > monOpen && time < monClose ||
    day === 2 && time > tuesOpen && time < tuesClose ||
    day === 3 && time > wedOpen && time < wedClose ||
    day === 4 && time > thursOpen && time < thursClose ||
    day === 5 && time > friOpen && time < friClose ||
    day === 6 && time > satOpen && time < satClose ||
    day === 0 && time > sunOpen && time < sunClose) {
        venueIsOpen();
} else {
    venueIsClosed();
}

That's still quite ugly, right? But there's some logic in there... We could use arrays for the open / close times:

var open = [sunOpen, monOpen, tuesOpen, wedOpen, thursOpen, friOpen, satOpen],
    close = [sunClose, monClose, tuesClose, wedClose, thursClose, friClose, satClose];

if(time > open[day] && time < close[day])
    venueIsOpen();
else
    venueIsClosed();

Or, even shorter using a ternary condition:

(time > open[day] && time < close[day] ? venueIsOpen : venueIsClosed)();

That ternary is ugly. But if you want to go for it, then it should be `(time > open[day] && time < close[day] ? venueIsOpen : venueIsClosed)()` — Bergi, Nov 20 '15 at 15:13

Jonas Grumann · Answer 2 · 2015-11-20T15:09:25.790

0

Here's another way that could shorten your code:

// These are the 7 days of the week
var openings = [
{opening: 1,close: 2},
{opening: 2,close: 4},
{opening: 5,close: 7},
{opening: 1,close: 5},
{opening: 1,close: 3},
{opening: 2,close: 9},
{opening: 1,close: 2}
];

// loop through all of them
for(var i=0, l=openings.length; i<l; i++) {
    // if the current time is bigger than the current day opening time and smaller than the current closing time call venueIsOpen(), otherwise call venueIsClosed()
    (openings[day].opening < time && openings[day].close > time) ? venueIsOpen() : venueIsClosed();   
}

edited Nov 20 '15 at 15:09

answered Nov 20 '15 at 15:06

Jonas Grumann

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`var openings = { open: [], close: [] }` would be shorter, but this is also a good option. – Cerbrus Nov 20 '15 at 15:08
True, I just think the other way around makes it more readable but since the question is about optimization I guess you're right. – Jonas Grumann Nov 20 '15 at 15:10

Is there any way to optimise this JavaScript further (comparing current time to opening hours)?

2 Answers2