如何将Date对象格式化为字符串？
当前回答
设置日期格式的其他方式：
function formatDate(dDate,sMode){
var today = dDate;
var dd = today.getDate();
var mm = today.getMonth()+1; //January is 0!
var yyyy = today.getFullYear();
if(dd<10) {
dd = '0'+dd
}
if(mm<10) {
mm = '0'+mm
}
if (sMode+""==""){
sMode = "dd/mm/yyyy";
}
if (sMode == "yyyy-mm-dd"){
return yyyy + "-" + mm + "-" + dd + "";
}
if (sMode == "dd/mm/yyyy"){
return dd + "/" + mm + "/" + yyyy;
}
}
其他回答
这是我刚刚编写的一些代码，用于处理我正在处理的项目的日期格式。它模仿了PHP日期格式功能，以满足我的需要。请随意使用它，它只是扩展了现有的Date（）对象。这可能不是最优雅的解决方案，但它符合我的需求。
var d = new Date();
d_string = d.format("m/d/Y h:i:s");
/**************************************
* Date class extension
*
*/
// Provide month names
Date.prototype.getMonthName = function(){
var month_names = [
'January',
'February',
'March',
'April',
'May',
'June',
'July',
'August',
'September',
'October',
'November',
'December'
];
return month_names[this.getMonth()];
}
// Provide month abbreviation
Date.prototype.getMonthAbbr = function(){
var month_abbrs = [
'Jan',
'Feb',
'Mar',
'Apr',
'May',
'Jun',
'Jul',
'Aug',
'Sep',
'Oct',
'Nov',
'Dec'
];
return month_abbrs[this.getMonth()];
}
// Provide full day of week name
Date.prototype.getDayFull = function(){
var days_full = [
'Sunday',
'Monday',
'Tuesday',
'Wednesday',
'Thursday',
'Friday',
'Saturday'
];
return days_full[this.getDay()];
};
// Provide full day of week name
Date.prototype.getDayAbbr = function(){
var days_abbr = [
'Sun',
'Mon',
'Tue',
'Wed',
'Thur',
'Fri',
'Sat'
];
return days_abbr[this.getDay()];
};
// Provide the day of year 1-365
Date.prototype.getDayOfYear = function() {
var onejan = new Date(this.getFullYear(),0,1);
return Math.ceil((this - onejan) / 86400000);
};
// Provide the day suffix (st,nd,rd,th)
Date.prototype.getDaySuffix = function() {
var d = this.getDate();
var sfx = ["th","st","nd","rd"];
var val = d%100;
return (sfx[(val-20)%10] || sfx[val] || sfx[0]);
};
// Provide Week of Year
Date.prototype.getWeekOfYear = function() {
var onejan = new Date(this.getFullYear(),0,1);
return Math.ceil((((this - onejan) / 86400000) + onejan.getDay()+1)/7);
}
// Provide if it is a leap year or not
Date.prototype.isLeapYear = function(){
var yr = this.getFullYear();
if ((parseInt(yr)%4) == 0){
if (parseInt(yr)%100 == 0){
if (parseInt(yr)%400 != 0){
return false;
}
if (parseInt(yr)%400 == 0){
return true;
}
}
if (parseInt(yr)%100 != 0){
return true;
}
}
if ((parseInt(yr)%4) != 0){
return false;
}
};
// Provide Number of Days in a given month
Date.prototype.getMonthDayCount = function() {
var month_day_counts = [
31,
this.isLeapYear() ? 29 : 28,
31,
30,
31,
30,
31,
31,
30,
31,
30,
31
];
return month_day_counts[this.getMonth()];
}
// format provided date into this.format format
Date.prototype.format = function(dateFormat){
// break apart format string into array of characters
dateFormat = dateFormat.split("");
var date = this.getDate(),
month = this.getMonth(),
hours = this.getHours(),
minutes = this.getMinutes(),
seconds = this.getSeconds();
// get all date properties ( based on PHP date object functionality )
var date_props = {
d: date < 10 ? '0'+date : date,
D: this.getDayAbbr(),
j: this.getDate(),
l: this.getDayFull(),
S: this.getDaySuffix(),
w: this.getDay(),
z: this.getDayOfYear(),
W: this.getWeekOfYear(),
F: this.getMonthName(),
m: month < 10 ? '0'+(month+1) : month+1,
M: this.getMonthAbbr(),
n: month+1,
t: this.getMonthDayCount(),
L: this.isLeapYear() ? '1' : '0',
Y: this.getFullYear(),
y: this.getFullYear()+''.substring(2,4),
a: hours > 12 ? 'pm' : 'am',
A: hours > 12 ? 'PM' : 'AM',
g: hours % 12 > 0 ? hours % 12 : 12,
G: hours > 0 ? hours : "12",
h: hours % 12 > 0 ? hours % 12 : 12,
H: hours,
i: minutes < 10 ? '0' + minutes : minutes,
s: seconds < 10 ? '0' + seconds : seconds
};
// loop through format array of characters and add matching data else add the format character (:,/, etc.)
var date_string = "";
for(var i=0;i<dateFormat.length;i++){
var f = dateFormat[i];
if(f.match(/[a-zA-Z]/g)){
date_string += date_props[f] ? date_props[f] : '';
} else {
date_string += f;
}
}
return date_string;
};
/*
*
* END - Date class extension
*
************************************/
字符串转换
// date
const dateConvert = {
dasher: dt => {
let m = (dt.getMonth() + 1) === 13 ? 1 : (dt.getMonth() + 1);
m = m < 10 ? `0${m}` : m.toString();
let d = dt.getDate();
d = d < 10 ? `0${d}` : d.toString();
return `${dt.getFullYear()}-${m}-${d}`;
},
slasher: dt => {
return dateConvert.slash(dateConvert.dasher(dt));
},
dash: str => {
// 03/11/2022 -> 2022-03-11
let [d, m, y] = str.split('/');
return `${y}-${m}-${d}`;
},
slash: str => {
// 2022-03-11 -> 03/11/2022
let [y, m, d] = str.split('-');
return `${d}/${m}/${y}`
}
}
// console.log(dateConvert.dasher(new Date('01/31/2001')));
如果在代码中使用jQueryUI，则有一个内置函数，名为formatDate（）。我用这种方式设置今天的日期：
var testdate = Date();
testdate = $.datepicker.formatDate( "d-M-yy",new Date(testdate));
alert(testdate);
在jQueryUI文档中可以看到许多其他格式化日期的示例。
设置日期格式的其他方式：
function formatDate(dDate,sMode){
var today = dDate;
var dd = today.getDate();
var mm = today.getMonth()+1; //January is 0!
var yyyy = today.getFullYear();
if(dd<10) {
dd = '0'+dd
}
if(mm<10) {
mm = '0'+mm
}
if (sMode+""==""){
sMode = "dd/mm/yyyy";
}
if (sMode == "yyyy-mm-dd"){
return yyyy + "-" + mm + "-" + dd + "";
}
if (sMode == "dd/mm/yyyy"){
return dd + "/" + mm + "/" + yyyy;
}
}
在我的情况下，我已将日期表“2022年7月1日”格式化为“2022-07-01”
常量格式日期=日期=>{const d=新日期（日期）let month=（d.getMonth（）+1）.toString（）let day=d.getDate（）.toString（）const year=d.getFullYear（）如果（月长度<2）{月=“0”+月}如果（日长度<2）{天=“0”+天}return[年，月，日]。join（'-'）}console.log（格式日期（'01/07/2022'））
