如果可能的话，不使用JavaScript库或大量笨重的代码，我正在寻找格式化距今两周的日期的最简单方法，格式如下:
2013年3月13日
我正在使用以下代码:
var newdate = new Date(+new Date + 12096e5);
document.body.innerHTML = newdate;
该函数返回两周后的日期和时间，格式如下：
Wed Mar 27 2013 21:50:29 GMT+0000 (GMT Standard Time)这是在jsFiddle上的代码。
希望能得到任何帮助！
const nth = (d) => {
if (d > 3 && d < 21) return 'th';
switch (d % 10) {
case 1: return "st";
case 2: return "nd";
case 3: return "rd";
default: return "th";
}
};
// test code
const fortnightAway = new Date(+new Date + 12096e5);
const date = fortnightAway.getDate();
const month = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"][fortnightAway.getMonth()];
document.getElementById("date").innerHTML = `In two weeks it will be the ${date}<sup>${nth(date)}</sup> of ${month} ${fortnightAway.getFullYear()}`;
// test
const dates = [...Array(32).keys()].slice(1).map(i => `${i}${nth(i)}`)
console.log(dates.join(", "))sup {
font-size: x-small
}<span id="date"></span>const nth = (d) => {
const dString = String(d);
const last = +dString.slice(-2);
if (last > 3 && last < 21) return 'th';
switch (last % 10) {
case 1: return "st";
case 2: return "nd";
case 3: return "rd";
default: return "th";
}
};
// test
const numbers = [...Array(1225).keys()].map(i => `${i}${nth(i)}`)
console.log(numbers.join(", "))const nth = n => n>3&&n<21?"th":n%10==1?"st":n%10==2?"nd":n%10==3?"rd":"th";
// to call: nth(i)
// test
document.getElementById("nums").innerHTML = [...Array(1225).keys()].map(i => `<pre>${String(i).padStart(3," ")}<sup>${nth(i)}</sup></pre>`).join(" ");pre { display:inline-block; margin:2px; }<div id="nums"></div>如果你不喜欢开关，这个同样有效
const nth = (d) => {
const last = +String(d).slice(-2);
if (last > 3 && last < 21) return 'th';
const remainder = last % 10;
if (remainder === 1) return "st";
if (remainder === 2) return "nd";
if (remainder === 3) return "rd";
return "th";
};
// test
const numbers = [...Array(1225).keys()].map(i => `${i}${nth(i)}`)
console.log(numbers.join(", "))d>3 && d<21 - Temüjinfunction nth(n){return n>3&&n<21?"th":n%10==2?"nd":n%10==2?"nd":n%10==3?"rd":"th"} - ashleedawgfunction nth(n){return n>3&&n<21?"th":n%10==1?"st":n%10==2?"nd":n%10==3?"rd":"th"} - Sam Tolton以下是受其他答案启发的一行代码。经过测试，它可以接受0和负数。
function getOrdinalNum(n) {
return n + (n > 0 ? ['th', 'st', 'nd', 'rd'][(n > 3 && n < 21) || n % 10 > 3 ? 0 : n % 10] : '');
}
2020年6月23日更新。以下是上述函数更易读的答案：
const getOrdinalNum = (number) => {
let selector;
if (number <= 0) {
selector = 4;
} else if ((number > 3 && number < 21) || number % 10 > 3) {
selector = 0;
} else {
selector = number % 10;
}
return number + ['th', 'st', 'nd', 'rd', ''][selector];
};
["th", "st", "nd", "rd"][n % 100 > 10 && n % 100 < 14) || n % 10 > 3 ? 0 : n % 10]但是，这段代码并不适用于日期的情况。如果有人需要将其用于任何正整数，则可以使用该代码。 - Oğuz K我也用这种方法来处理日期，但因为每月的日期只能在1到31之间，所以最终采用了一种简化的解决方案。
function dateOrdinal(dom) {
if (dom == 31 || dom == 21 || dom == 1) return dom + "st";
else if (dom == 22 || dom == 2) return dom + "nd";
else if (dom == 23 || dom == 3) return dom + "rd";
else return dom + "th";
};
或使用条件运算符的简洁版本
function dateOrdinal(d) {
return d+(31==d||21==d||1==d?"st":22==d||2==d?"nd":23==d||3==d?"rd":"th")
};
var newdate = new Date();
moment(newdate).format("Do MMMM YYYY")
//Returns 1st January 2020
moment("01/01/2020", "MM/DD/YYYY").format("Do")
//Returns 1st
moment("01/01/2020", "MM/DD/YYYY").format("Do MMM YYYY")
//Returns 1st Jan 2020
这是一个适用于任何数字的简单函数：
function getOrdinal(n) {
let ord = ["st", "nd", "rd"]
let exceptions = [11, 12, 13]
let nth =
ord[(n % 10) - 1] == undefined || exceptions.includes(n % 100) ? "th" : ord[(n % 10) - 1]
return n + nth
}
getOrdinal(28) //Outputs: 28th
getOrdinal('108') //Outputs: 108th
这里已经有很多好的答案了，不过有一个使用Intl.PluralRules的答案可以标准化各种语言环境下序数词的分类，可能仍然有用。
以下是一些适用于en-GB的实现。
a one liner:
console.log({one:'st',two:'nd',few:'rd',other:'th'}[new Intl.PluralRules('en-GB', { type: 'ordinal' }).select(new Date().getDate())])
a verbose example:
const suffixMap = {
one: 'st',
two: 'nd',
few: 'rd',
other: 'th',
};
const locale = 'en-GB';
const moment = new Date();
const dayOfMonth = moment.getDate();
const pluralRuleOptions = {
type: 'ordinal',
};
const pluralRule = new Intl.PluralRules(locale, pluralRuleOptions);
const ordinal = pluralRule.select(dayOfMonth);
console.log(suffixMap[ordinal])
const suffix = {
one: 'st',
two: 'nd',
few: 'rd',
other: 'th',
};
document.getElementById("tomorrow-month").innerHTML = new Intl.DateTimeFormat('en-GB', { month: 'long' }).format(new Date(Date.now() + 86400000));
document.getElementById("tomorrow-day").innerHTML = new Intl.DateTimeFormat('en-GB', { day: 'numeric' }).format(new Date(Date.now() + 86400000));
document.getElementById("tomorrow-ordinal").innerHTML = suffix[new Intl.PluralRules('en-GB', { type: 'ordinal' }).select(new Date(Date.now() + 86400000).getDate())];
document.getElementById("yesterday-month").innerHTML = new Intl.DateTimeFormat('en-GB', { month: 'long' }).format(new Date(Date.now() - 86400000));
document.getElementById("yesterday-day").innerHTML = new Intl.DateTimeFormat('en-GB', { day: 'numeric' }).format(new Date(Date.now() - 86400000));
document.getElementById("yesterday-ordinal").innerHTML = suffix[new Intl.PluralRules('en-GB', { type: 'ordinal' }).select(new Date(Date.now() - 86400000).getDate())];Tomorrow, on <span id="tomorrow-month"></span> <span id="tomorrow-day"></span><sup id="tomorrow-ordinal"></sup>, I ordered a time machine.
It arrived yesterday on <span id="yesterday-month"></span> <span id="yesterday-day"></span><sup id="yesterday-ordinal"></sup>.在众多解决方案中又多了一个选择。
let suffix = (day >= 4 && day <= 20) || (day >= 24 && day <= 30)
? "th"
: ["st", "nd", "rd"][day % 10 - 1];
dayjs(date).format('Do') 就可以了，但我会提供一个示例来展示如何使用任何你想要的格式。
请注意，Dayjs 高级格式提供了日期序数。var advancedFormat = require('dayjs/plugin/advancedFormat')
dayjs.extend(advancedFormat)
// date variable - your date to format eg:- 2022-04-01T21:27:00
dayjs(date).format('dddd, MMMM Do [at] h:mma')
这里有很多格式化的答案，我只关注于任意整数的第n个-
Number.prototype.nth= function(){
if(this%1) return this;
var s= this%100;
if(s>3 && s<21) return this+'th';
switch(s%10){
case 1: return this+'st';
case 2: return this+'nd';
case 3: return this+'rd';
default: return this+'th';
}
}
this%1？因为模1的结果总是0，所以该语句将始终失败，不是吗？ - Foragefunction addOrd(n) {
var ords = [, 'st', 'nd', 'rd'];
var ord, m = n % 100;
return n + ((m > 10 && m < 14) ? 'th' : ords[m % 10] || 'th');
}
// Return date string two weeks from now (14 days) in
// format 13th March 2013
function formatDatePlusTwoWeeks(d) {
var months = ['January', 'February', 'March', 'April', 'May', 'June',
'July', 'August', 'September', 'October', 'November', 'December'
];
// Copy date object so don't modify original
var e = new Date(d);
// Add two weeks (14 days)
e.setDate(e.getDate() + 14);
return addOrd(e.getDate()) + ' ' + months[e.getMonth()] + ' ' + e.getFullYear();
}
console.log(formatDatePlusTwoWeeks(new Date()));
// Alternative using Intl.DateTimeFormat
function datePlusTwoWeeks(date = new Date()) {
let d = new Date(date.getFullYear(), date.getMonth(), date.getDate() + 14);
let parts = new Intl.DateTimeFormat('en',{
year: 'numeric',
month: 'long',
day: 'numeric'
}).formatToParts(d).reduce((acc, part) => {
acc[part.type] = part.value;
return acc;
}, Object.create(null));
return `${addOrd(parts.day)} ${parts.month} ${parts.year}`;
}
console.log(datePlusTwoWeeks())