我有这个功能：
function getInfoSchoolTime() {
var date = new Date();
var schoolBellTime = ["8:10","9:02","9:54","9:59","10:51","11:43","11:58","12:48","13:35","13:40","14:10","15:02","15:54"];
var remainingTime, currentHour;
for (var i = 0; i < schoolBellTime.length-1; i++) {
var startTime = schoolBellTime[i].split(":");
var endTime = schoolBellTime[i+1].split(":");
if (parseInt(startTime[0]) >= date.getHours() && parseInt(startTime[1]) >= date.getMinutes())
if (parseInt(endTime[0]) <= date.getHours() && parseInt(endTime[1]) <= date.getMinutes()) {
currentHour = i;
remainingTime=(parseInt(endTime[1])-date.getMinutes()+60)%60;
break;
}
}
if (currentHour == undefined)
return {current: -1, remaining: "not available"};
return {current: currentHour, remaining: remainingTime};
}
var info = getInfoSchoolTime();
console.log(info.current, info.remaining);
我有schoolBellTime数组，其中包含校钟的时间戳（我知道，我的学校有很多钟时间，这些时间戳包括游戏时间和午餐时间），此功能是为了返回第1小时/第2小时小时/第3小时......以及下一小时/休息时间的剩余时间。
我检查了所有代码但找不到错误，它不断返回{current: -1, remaining: "not available"}
答案 0 :(得分：1)
顶部的函数：setDateTime()占用日期和时间，并构造该时间的日期对象。
然后我更新了您的功能，我将start和end转换为当天的时间，然后检查它们之间是否发生date.getTime()。然后，我只需从date.getTime()中减去end，然后将结果转换为毫秒级的分钟数。
var setDateTime = function(date, str) {
var sp = str.split(':');
date.setHours(parseInt(sp[0], 10));
date.setMinutes(parseInt(sp[1], 10));
return date;
}
function getInfoSchoolTime() {
var date = new Date();
var schoolBellTime = ["8:10", "9:02", "9:54", "9:59", "10:51", "11:43", "11:58", "12:48", "13:35", "13:40", "14:10", "14:10", "15:02", "15:54"];
var remainingTime, currentHour, currentPeriod;
for (var i = 0; i < schoolBellTime.length - 1; i++) {
start = setDateTime(new Date(), schoolBellTime[i])
end = setDateTime(new Date(), schoolBellTime[i + 1])
if (date.getTime() > start.getTime() && date.getTime() < end.getTime()) {
currentHour = i
remainingTime = end.getTime() - date.getTime()
currentPeriod = ([schoolBellTime[i], schoolBellTime[i+1]]).join('-')
}
}
return {current: currentHour, currentPeriod: currentPeriod, remaining: Math.round(remainingTime * 0.0000166667)}
}
console.log(getInfoSchoolTime())
答案 1 :(得分：1)
对于代码和API，这是一种稍微不同的方法。它使用两个辅助函数。每个应该是明显的，只有一个例子：pad(7) //=> "07"和pairs(['foo', 'bar', 'baz', 'qux']) //=> [['foo', 'bar'], ['bar', 'baz'], ['baz', 'qux']]。
main函数获取一个响铃时间列表并返回一个函数，该函数本身接受一个日期对象并返回您正在查找的输出类型（句点，剩余时间段）。此API使测试更容易。
const pad = nbr => ('00' + nbr).slice(-2)
const pairs = vals => vals.reduce((res, val, idx) => idx < 1 ? res : res.concat([[vals[idx - 1], val]]), [])
const schoolPeriods = (schoolBellTime) => {
const subtractTimes = (t1, t2) => 60 * t1.hour + t1.minute - (60 * t2.hour + t2.minute)
const periods = pairs(schoolBellTime.map(time => ({hour: time.split(':')[0], minute: +time.split(':')[1]})))
return date => {
const current = {hour: date.getHours(), minute: date.getMinutes()}
if (subtractTimes(current, periods[0][0]) < 0) {
return {message: 'before school day'}
}
if (subtractTimes(current, periods[periods.length - 1][1]) > 0) {
return {message: 'after school day'}
}
const idx = periods.findIndex(period => subtractTimes(current, period[0]) >= 0 && subtractTimes(period[1], current) > 0)
const period = periods[idx]
return {
current: idx + 1,
currentPeriod: `${period[0].hour}:${pad(period[0].minute)} - ${period[1].hour}:${pad(period[1].minute)}`,
remaining: subtractTimes(period[1], current)
}
}
}
const getPeriod = schoolPeriods(["8:10","9:02","9:54","9:59","10:51","11:43","11:58","12:48","13:35","13:40","14:10","14:10","15:02","15:54"])
console.log("Using current time")
console.log(getPeriod(new Date()))
console.log("Using a fixed time")
console.log(getPeriod(new Date(2017, 11, 22, 14, 27))) // will Christmas break ever come?!
如果日期超出期间范围，我会随机猜测你想要的行为。
在内部，它会创建一个类似于
的句点对象列表[{hour:9, minute: 59}, {hour: 10, minute: 51}]
如果不是一个双元素数组，那么它可能会更加清晰，它是一个具有start和end属性的对象。这将是一个很容易的改变。
请注意，为了理所当然，需要按顺序列出铃声。我们可以通过sort调用解决此问题，但我认为没有充分的理由这样做。
答案 2 :(得分：0)
这是一个使用deconstruct（const [a,b]=[1,2]），数组映射，数组缩减，部分应用程序（闭包）和胖箭头函数语法的ES6示例。
这可能不适用于旧浏览器。
//pass date and bellTimes to function so you can test it more easily
// you can partially apply bellTimes
const getInfoSchoolTime = bellTimes => {
//convert hour and minute to a number
const convertedBellTimes = bellTimes
.map(bellTime=>bellTime.split(":"))//split hour and minute
.map(([hour,minute])=>[new Number(hour),new Number(minute)])//convert to number
.map(([hour,minute])=>(hour*60)+minute)//create single number (hour*60)+minutes
.reduce(//zip with next
(ret,item,index,all)=>
(index!==all.length-1)//do not do last one, create [1,2][2,3][3,4]...
? ret.concat([[item,all[index+1]]])
: ret,
[]
);
return date =>{
//convert passed in date to a number (hour*60)+minutes
const passedInTime = (date.getHours()*60)+date.getMinutes();
return convertedBellTimes.reduce(
([ret,goOn],[low,high],index,all)=>
//if goOn is true and passedInTime between current and next bell item
(goOn && passedInTime<high && passedInTime>=low)
? [//found the item, return object and set goOn to false
{
current: index+1,
currentPeriod: bellTimes[index]+"-"+bellTimes[index+1],
remaining: high-passedInTime
},
false//set goOn to false, do not continue checking
]
: [ret,goOn],//continue looking or continue skipping (if goOn is false)
[
{current: 0, currentPeriod: "School is out", remaining: 0},//default value
true//initial value for goOn
]
)[0];//reduced to multiple values (value, go on) only need value
}
};
//some tests
const date = new Date();
//partially apply with some bell times
const schoolTime = getInfoSchoolTime(
[
"8:10", "9:02", "9:54", "9:59", "10:51",
"11:43", "11:58", "12:48", "13:35", "13:40",
"14:10", "14:10", "15:02", "15:54"
]
);
//helper to log time from a date
const formatTime = date =>
("0"+date.getHours()).slice(-2)+":"+("0"+date.getMinutes()).slice(-2);
date.setHours(11);
date.setMinutes(1);
console.log(formatTime(date),schoolTime(date));//11:01
date.setHours(15);
date.setMinutes(53);
console.log(formatTime(date),schoolTime(date));//15:53
date.setHours(23);
date.setMinutes(1);
console.log(formatTime(date),schoolTime(date));//23:01