Question

我正在设置一些JavaScript来比较当前时间与场地的开放和关闭时间（每天可能不同，因此每天的变量不同）。

这是我到目前为止所拥有的：

// 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();
}

显然非常简单 - 有没有办法优化这个？

Answer 1

首先，我们可以合并这些background-color:transparent语句的负载：

if

那仍然很难看，对吧？但那里有一些逻辑......我们可以在开/关时使用数组：

// 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();
}

或者，使用ternary condition时更短：

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();

Answer 2

以下是另一种可以缩短代码的方法：

// 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();   
}

有没有办法进一步优化这个JavaScript（比较当前时间与开放时间）？

2 个答案: