我想计算我的工作时间。输入
时效果很好08:00 - 09:00 = 01:00
但是当我这次输入
23:30 - 01:30 = 10:00
它应该返回02:00
function pad(num) {
return ("0" + num).slice(-2);
}
function diffTime(start, end) {
var s = start.split(":"),
sMin = +s[1] + s[0] * 60,
e = end.split(":"),
eMin = +e[1] + e[0] * 60,
diff = eMin - sMin;
if (diff < 0) {
sMin -= 12 * 60;
diff = eMin - sMin
}
var h = Math.floor(diff / 60),
m = diff % 60;
return "" + pad(h) + ":" + pad(m);
}
document.getElementById('button').onclick = function() {
document.getElementById('delay').value = diffTime(
document.getElementById('timeOfCall').value,
document.getElementById('timeOfResponse').value
);
}<input type="time" id="timeOfCall">
<input type="time" id="timeOfResponse">
<button type="button" id="button">CLICK</button>
<input type="time" id="delay">
答案 0 :(得分：1)
所需算法中最重要的部分是确定结束日期是否是明天。
根据您的代码，这里是我建议的working example。
<!DOCTYPE html>
<html>
<head>
<style>
</style>
</head>
<body>
<input type="time" id="timeOfCall">
<input type="time" id="timeOfResponse">
<button type="button" id="button" onclick="diffTime()">CLICK
</button>
<input type="time" id="delay">
<script>
function pad(num) {
return ("0" + num).slice(-2);
}
function diffTime() {
var start = document.getElementById("timeOfCall").value;
var end = document.getElementById("timeOfResponse").value;
// start date will be today
var d1 = new Date();
var s = start.split(":")
var date1 = new Date(d1.getFullYear(),d1.getMonth(),d1.getDate(),s[0],s[1],0,0);
var s2 = end.split(":")
// end date
if(s2[0] < s[0])
{
// its tommorow...
var ms = new Date().getTime() + 86400000;
var tomorrow = new Date(ms);
d1=tomorrow;
}
var date2 = new Date(d1.getFullYear(),d1.getMonth(),d1.getDate(),s2[0],s2[1],0,0);
var diff = date2.getTime() - date1.getTime();
var msec = diff;
var hh = Math.floor(msec / 1000 / 60 / 60);
msec -= hh * 1000 * 60 * 60;
var mm = Math.floor(msec / 1000 / 60);
msec -= mm * 1000 * 60;
var ss = Math.floor(msec / 1000);
msec -= ss * 1000;
alert(hh + ":" + mm + ":" + ss);
}
document.getElementById("timeOfCall").defaultValue = "23:30";
document.getElementById("timeOfResponse").defaultValue = "01:30";
</script>
</body>
</html>
答案 1 :(得分：0)
我将使用Date对象来计算时间差。由于您只对时间感兴趣，因此可以使用任何日期来构造有效的日期字符串。您获得10个小时的原因是因为没有日期显示第二天是凌晨1点（这是根据我对您问题的理解）。
您可以执行以下操作来完成工作。
const pad = num => (num < 10) ? `0${num}` : `${num}`;
const addADay = (start, end) => {
const sHour = parseInt(start.split(':')[0], 10);
const eHour = parseInt(end.split(':')[0], 10);
return (eHour < sHour);
};
const diffTime = (start, end) => {
const startDate = new Date(`2019/01/01 ${start}:00`);
const endDate = addADay(start, end)
? new Date(`2019/01/02 ${end}:00`)
: new Date(`2019/01/01 ${end}:00`);
const diff = endDate.getTime() - startDate.getTime();
const hours = Math.floor(diff / 3600000);
const min = (diff - (hours * 3600000)) / 60000;
return `${pad(hours)}:${pad(min)}`;
}
console.log(diffTime('08:00','09:00')); // returns 01:00
console.log(diffTime('23:00','01:30')); // returns 02:30
答案 2 :(得分：0)
您好，我已经稍微更改了您的代码。解释是，让您的开始时间为10:00，结束时间为09:00。现在顺时针思考。时间必须到24小时的9:00。因此，计算结果是24小时与10小时之间的时差，然后加上其余时间。
D = E +（24-S）
function pad(num) {
return ("0" + num).slice(-2);
}
function diffTime(start, end) {
var s = start.split(":"),
sMin = +s[1] + s[0] * 60,
e = end.split(":"),
eMin = +e[1] + e[0] * 60,
diff = eMin - sMin;
if (diff < 0) {
diff = eMin + (24 * 60 - sMin); /* You had to caculate with 24 hours */
}
var h = Math.floor(diff / 60),
m = diff % 60;
return "" + pad(h) + ":" + pad(m);
}
document.getElementById('button').onclick = function() {
document.getElementById('delay').value = diffTime(
document.getElementById('timeOfCall').value,
document.getElementById('timeOfResponse').value
);
}<input type="time" id="timeOfCall">
<input type="time" id="timeOfResponse">
<button type="button" id="button">CLICK</button>
<input type="time" id="delay">
答案 3 :(得分：0)
这是解决您所有测试用例的问题的另一种更简单的方法，如果有任何失败的情况，请尝试所有测试用例，然后告诉我我将解决它。
您只需要先花几个小时，然后检查是上午还是下午，然后简单地数分钟即可。
function diffTime(start, end) {
var s = start.split(":");
var e = end.split(":");
var dHour;
var dMinute ;
var startHour = parseInt(s[0]);
var endHour = parseInt(e[0]);
var startMinute = parseInt(s[1]);
var endMinute = parseInt(e[1]);
// For counting difference of hours
if((startHour>12 && endHour>12) || (startHour<12 && endHour<12))
{
if(startHour<endHour)
{
dHour = endHour - startHour;
}
else if(startHour>endHour)
{
dHour = 24 - ( startHour - endHour);
}
else
{
dHour = 24;
}
}
else if(startHour>12 && endHour<=12)
{
dHour = (24 - startHour) + endHour;
}
else if(startHour<=12 && endHour > 12)
{
dHour = endHour - startHour;
}
else
{
dHour = 24
}
// For Counting Difference of Minute
if (startMinute>endMinute)
{
dMinute = 60 - (startMinute - endMinute);
dHour = dHour - 1;
}
else if(startMinute<endMinute)
{
dMinute = endMinute - startMinute;
}
else
{
dMinute = 0
}
return dHour + " Hours " + dMinute + " Minutes";
}
document.getElementById('button').onclick = function() {
document.getElementById('delay').value = diffTime(
document.getElementById('timeOfCall').value,
document.getElementById('timeOfResponse').value);
}<input type="time" id="timeOfCall">
<input type="time" id="timeOfResponse">
<button type="button" id="button">CLICK</button>
<input type="text" id="delay">
答案 4 :(得分：0)
谢谢你的朋友，我解决了我的问题，Miraz Chowdhury的代码完成了我的工作
答案 5 :(得分：-1)
function diff(t1, t2) {
const day = 86400000;
function pad(num) {
return ("0" + num).slice(-2);
}
let time1 = t1.split(":").map(el => parseInt(el));
let time2 = t2.split(":").map(el => parseInt(el));
let zero = (new Date(1990, 0, 1, 0, 0)).setMilliseconds(0)
let aaa = (new Date(1990, 0, 1, time1[0], time1[1])).setMilliseconds(0)
let bbb = (new Date(1990, 0, 1, time2[0], time2[1])).setMilliseconds(0)
let diff = day -Math.abs(aaa - bbb)<Math.abs(aaa - bbb)?day -Math.abs(aaa - bbb):Math.abs(aaa - bbb)
return `${pad(Math.round(diff/1000/60/60))}:${pad(Math.abs(Math.round(diff/1000/60%60)))}`;
}
console.log(diff("09:00", "08:00"));
console.log(diff("23:30", "01:30"));
console.log(diff("01:30", "23:30"));