我有根据铁路时间往返的时间,现在我将上午AM转换为正常时间。但是之后,我需要增加30分钟的间隔。我对迭代器方法有点困惑。请提前帮助谢谢
我的密码
let fromtime = '09:00:00'
let totime = '21:00:00'
let getGenTime = (timeString) => {
let H = +timeString.substr(0, 2);
let h = (H % 12) || 12;
let ampm = H < 12 ? " AM" : " PM";
return timeString = h + timeString.substr(2, 3) + ampm;
}
现在我有上午和下午两种格式的往返时间,但我需要在往返时间之间增加30分钟的间隔
fromtime = getGenTime(fromtime) // 09:00 AM
totime = getGenTime(totime) // 09:00 PM
控制台中的预期结果:
09:00 AM
09:30 AM
10:00 AM
10:30 AM
11:00 AM
....
.....
08:30 PM
09:00 PM
最佳答案
如果使用Date
对象而不是strings
,或者使用moment.js
为此提供了许多有用的方法,那会更好。
但是无论如何,它都只需要一个带有一些检查的循环即可实现您想要的,我做了一个函数,该函数将返回一个在fromtime
和totime
之间所有时间的数组:
function returnTimesInBetween(start, end) {
var timesInBetween = [];
var startH = parseInt(start.split(":")[0]);
var startM = parseInt(start.split(":")[1]);
var endH = parseInt(end.split(":")[0]);
var endM = parseInt(end.split(":")[1]);
if (startM == 30)
startH++;
for (var i = startH; i < endH; i++) {
timesInBetween.push(i < 10 ? "0" + i + ":00" : i + ":00");
timesInBetween.push(i < 10 ? "0" + i + ":30" : i + ":30");
}
timesInBetween.push(endH + ":00");
if (endM == 30)
timesInBetween.push(endH + ":30")
return timesInBetween.map(getGenTime);
}
演示:
let fromtime = '09:00:00'
let totime = '21:00:00'
let getGenTime = (timeString) => {
let H = +timeString.substr(0, 2);
let h = (H % 12) || 12;
let ampm = H < 12 ? " AM" : " PM";
return timeString = h + timeString.substr(2, 3) + ampm;
}
function returnTimesInBetween(start, end) {
var timesInBetween = [];
var startH = parseInt(start.split(":")[0]);
var startM = parseInt(start.split(":")[1]);
var endH = parseInt(end.split(":")[0]);
var endM = parseInt(end.split(":")[1]);
if (startM == 30)
startH++;
for (var i = startH; i < endH; i++) {
timesInBetween.push(i < 10 ? "0" + i + ":00" : i + ":00");
timesInBetween.push(i < 10 ? "0" + i + ":30" : i + ":30");
}
timesInBetween.push(endH + ":00");
if (endM == 30)
timesInBetween.push(endH + ":30")
return timesInBetween.map(getGenTime);
}
console.log(returnTimesInBetween(fromtime, totime));
编辑:
要从
Date
和fromtime
字符串获取totime
对象,可以使用Date
构造函数:let fromtime = '09:00:00';
var d = new Date(Date.UTC(2017, 10, 10, parseInt(fromtime.split(":")[0]), parseInt(fromtime.split(":")[1])));
请注意Locale和TimeZones,结果可能会有所不同。
演示:
let fromtime = '09:00:00'
let totime = '21:00:00'
let getGenTime = (timeString) => {
let H = +timeString.substr(0, 2);
let h = (H % 12) || 12;
let ampm = H < 12 ? " AM" : " PM";
return timeString = h + timeString.substr(2, 3) + ampm;
}
function returnTimesInBetween(start, end) {
var timesInBetween = [];
var startH = parseInt(start.split(":")[0]);
var startM = parseInt(start.split(":")[1]);
var endH = parseInt(end.split(":")[0]);
var endM = parseInt(end.split(":")[1]);
if (startM == 30)
startH++;
for (var i = startH; i < endH; i++) {
timesInBetween.push(i < 10 ? "0" + i + ":00" : i + ":00");
timesInBetween.push(i < 10 ? "0" + i + ":30" : i + ":30");
}
timesInBetween.push(endH + ":00");
if (endM == 30)
timesInBetween.push(endH + ":30")
timesInBetween.map(getGenTime);
return timesInBetween.map(time => new Date(Date.UTC(2017, 10, 10, parseInt(time.split(":")[0]), parseInt(time.split(":")[1]))));
}
console.log(returnTimesInBetween(fromtime, totime));