我有一个日期，格式是2014年5月11日太阳。如何使用JavaScript将其转换为2014-05-11 ?
函数taskDate(dateMilli) { var d = (new Date(dateMilli) + ")。分割(' '); D [2] = D [2] + '，'; 返回[d[0]， d[1]， d[2]， d[3]]。加入(' '); ｝ var datemilli =日期。解析(' 2014年5月11日'); console.log (taskDate (datemilli));
上面的代码给了我相同的日期格式，2014年5月11日。我该如何解决这个问题?
当前回答
Date.js很适合这个。
require("datejs")
(new Date()).toString("yyyy-MM-dd")
其他回答
new Date('Tue Nov 01 2022 22:14:53 GMT-0300').toLocaleDateString('en-CA');
new Date().toLocaleDateString('pt-br').split( '/' ).reverse( ).join( '-' );
or
new Date().toISOString().split('T')[0]
new Date('23/03/2020'.split('/').reverse().join('-')).toISOString()
new Date('23/03/2020'.split('/').reverse().join('-')).toISOString().split('T')[0]
试试这个!
之前的一些答案是可以的，但是不是很灵活。我想要一些能够真正处理更多边缘情况的东西，所以我采用了@orangleliu的答案并扩展了它。https://jsfiddle.net/8904cmLd/1/
function DateToString(inDate, formatString) {
// Written by m1m1k 2018-04-05
// Validate that we're working with a date
if(!isValidDate(inDate))
{
inDate = new Date(inDate);
}
// See the jsFiddle for extra code to be able to use DateToString('Sun May 11,2014', 'USA');
//formatString = CountryCodeToDateFormat(formatString);
var dateObject = {
M: inDate.getMonth() + 1,
d: inDate.getDate(),
D: inDate.getDate(),
h: inDate.getHours(),
m: inDate.getMinutes(),
s: inDate.getSeconds(),
y: inDate.getFullYear(),
Y: inDate.getFullYear()
};
// Build Regex Dynamically based on the list above.
// It should end up with something like this: "/([Yy]+|M+|[Dd]+|h+|m+|s+)/g"
var dateMatchRegex = joinObj(dateObject, "+|") + "+";
var regEx = new RegExp(dateMatchRegex,"g");
formatString = formatString.replace(regEx, function(formatToken) {
var datePartValue = dateObject[formatToken.slice(-1)];
var tokenLength = formatToken.length;
// A conflict exists between specifying 'd' for no zero pad -> expand
// to '10' and specifying yy for just two year digits '01' instead
// of '2001'. One expands, the other contracts.
//
// So Constrict Years but Expand All Else
if (formatToken.indexOf('y') < 0 && formatToken.indexOf('Y') < 0)
{
// Expand single digit format token 'd' to
// multi digit value '10' when needed
var tokenLength = Math.max(formatToken.length, datePartValue.toString().length);
}
var zeroPad = (datePartValue.toString().length < formatToken.length ? "0".repeat(tokenLength) : "");
return (zeroPad + datePartValue).slice(-tokenLength);
});
return formatString;
}
使用示例:
DateToString('Sun May 11,2014', 'MM/DD/yy');
DateToString('Sun May 11,2014', 'yyyy.MM.dd');
DateToString(new Date('Sun Dec 11,2014'),'yy-M-d');
const today = new Date(); // or whatever const yearFirstFormater = (date): string => { const modifiedDate = new Date(date).toISOString().slice(0, 10); return `${modifiedDate.split('-')[0]}/${modifiedDate.split('-')[1]}/${modifiedDate.split('-')[2]}`; } const monthFirstFormater = (date): string => { const modifiedDate = new Date(date).toISOString().slice(0, 10); return `${modifiedDate.split('-')[1]}/${modifiedDate.split('-')[2]}/${modifiedDate.split('-')[0]}`; } const dayFirstFormater = (date): string => { const modifiedDate = new Date(date).toISOString().slice(0, 10); return `${modifiedDate.split('-')[2]}/${modifiedDate.split('-')[1]}/${modifiedDate.split('-')[0]}`; } console.log(yearFirstFormater(today)); console.log(monthFirstFormater(today)); console.log(dayFirstFormater(today));
你可以试试这个:https://www.npmjs.com/package/timesolver
npm i timesolver
在你的代码中使用它:
const timeSolver = require('timeSolver');
const date = new Date();
const dateString = timeSolver.getString(date, "YYYY-MM-DD");
你可以使用下面的方法获取日期字符串:
getString
以下是一些答案的组合:
var d = new Date(date);
date = [
d.getFullYear(),
('0' + (d.getMonth() + 1)).slice(-2),
('0' + d.getDate()).slice(-2)
].join('-');
