第二天,跳过周末
Get next day, skip weekends
我想使用 JavaScript 生成下一个工作日。
这是我目前的代码
var today = new Date();
today.setDate(today.getDate());
var tdd = today.getDate();
var tmm = today.getMonth()+1;
var tyyyy = today.getYear();
var date = new Date();
date.setDate(date.getDate()+3);
问题是,星期五 returns 星期六的日期,而我希望它是星期一
这将选择下一个工作日。
var today = new Date(2016, 7, 26,12,0,0,0,0); // Friday at noon
console.log("today, Monday",today,"day #"+today.getDay());
var next = new Date(today.getTime());
next.setDate(next.getDate()+1); // tomorrow
while (next.getDay() == 6 || next.getDay() == 0) next.setDate(next.getDate() + 1);
console.log("no change ",next,"day #"+next.getDay());
console.log("-------");
// or without a loop:
function getNextWork(d) {
d.setDate(d.getDate()+1); // tomorrow
if (d.getDay()==0) d.setDate(d.getDate()+1);
else if (d.getDay()==6) d.setDate(d.getDate()+2);
return d;
}
next = getNextWork(today); // Friday
console.log("today, Friday",today);
console.log("next, Monday ",next);
console.log("-------");
today = new Date(2016, 7, 29,12,0,0,0); // Monday at noon
next = getNextWork(today); // Still Monday at noon
console.log("today, Monday",today);
console.log("no change ",next);
console.log("-------");
// Implementing Rob's comment
function getNextWork1(d) {
var day = d.getDay(),add=1;
if (day===5) add=3;
else if (day===6) add=2;
d.setDate(d.getDate()+add);
return d;
}
today = new Date(2016, 7, 26,12,0,0,0,0); // Friday at noon
next = getNextWork1(today); // Friday
console.log("today, Friday",today);
console.log("next, Monday ",next);
console.log("-------");
today = new Date(2016, 7, 26,12,0,0,0,0); // Monday at noon
next = getNextWork1(today); // Monday
console.log("today, Monday",today);
console.log("no change ",next);
您可以一次添加 1 天,直到到达不是星期六或星期日的那一天:
function getNextBusinessDay(date) {
// Copy date so don't affect original
date = new Date(+date);
// Add days until get not Sat or Sun
do {
date.setDate(date.getDate() + 1);
} while (!(date.getDay() % 6))
return date;
}
// today, Friday 26 Aug 2016
[new Date(), new Date(2016,7,26)].forEach(function(d) {
console.log(d.toLocaleString() + ' : ' + getNextBusinessDay(d).toLocaleString());
});
您也可以测试当天并添加额外的内容以度过周末:
// Classic Mon to Fri
function getNextWorkDay(date) {
let d = new Date(+date);
let day = d.getDay() || 7;
d.setDate(d.getDate() + (day > 4? 8 - day : 1));
return d;
}
for (let i=0, d=new Date(); i<7; i++) {
console.log(`${d.toDateString()} -> ${getNextWorkDay(d).toDateString()}`);
d.setDate(d.getDate() + 1);
}
这是另一种方法,可以使用 ECMAScript 工作日编号(周日 = 0、周一 = 1 等)指定工作周。超出范围的日期将移至下一个工作周的开始。
这在一周不是典型的周一到周五的情况下很有用,例如在中东,周六到周三很常见,或者对于一些可能在周五到周一(或其他时间)工作的人。
function getNext(start, end, date) {
let d = new Date(+date);
d.setDate(d.getDate() + 1);
let day = d.getDay();
// Adjust end and day if necessary
// The order of tests and adjustment is important
if (end < start) {
if (day <= end) {
day += 7;
}
end += 7;
}
// If day is before start, shift to start
if (day < start) {
d.setDate(d.getDate() + start - day);
// If day is after end, shift to next start (treat Sunday as 7)
} else if (day > end) {
d.setDate(d.getDate() + 8 - (day || 7));
}
return d;
}
// Examples
let f = new Intl.DateTimeFormat('en-GB', {
weekday:'short',day:'2-digit', month:'short'});
let d = new Date();
[{c:'Work days Mon to Fri',s:1,e:5},
{c:'Work days Sat to Wed',s:6,e:3},
{c:'Work days Fri to Mon',s:5,e:1}
].forEach(({c,s,e}) => {
for (let i = 0; i < 7; i++) {
!i? console.log(`\n${c}`) : null;
console.log(`${f.format(d)} => ${f.format(getNext(s, e, d))}`);
d.setDate(d.getDate() + 1);
}
});
看看这个:https://jsfiddle.net/e9a4066r/
function get_next_weekday (date) {
var tomorrow = new Date(date.setDate(date.getDate() + 1))
return tomorrow.getDay() % 6
? tomorrow
: get_next_weekday(tomorrow)
}
接受的答案将一次跳过一天,这回答了 OP 问题,但对于希望添加可变天数同时仍跳过周末的任何人来说,以下功能可能会有所帮助:
function addWorkDays(date, days) {
while (days > 0) {
date.setDate(date.getDate() + 1);
if (date.getDay() != 0 && date.getDay() != 6) {
days -= 1;
}
}
return date;
}
我想我会在这里把我的帽子扔进戒指:
function getNextBusinessDate(date) {
// Create date array [S, M, T, W, T, F, S]
const days = new Array(7);
let nextDate = date;
for(let i = 0; i < 7; i++) {
days[nextDate.getDay()] = new Date(nextDate);
nextDate.setDate(nextDate.getDate() + 1);
}
// Shift indices to index as though array was [M, T, W, T, F, S, S]
// Then truncate with min to make F, S, S all yield M for next date
return days[Math.min((date.getDay() + 6) % 7 + 1, 5) % 5 + 1];
}
我想使用 JavaScript 生成下一个工作日。
这是我目前的代码
var today = new Date();
today.setDate(today.getDate());
var tdd = today.getDate();
var tmm = today.getMonth()+1;
var tyyyy = today.getYear();
var date = new Date();
date.setDate(date.getDate()+3);
问题是,星期五 returns 星期六的日期,而我希望它是星期一
这将选择下一个工作日。
var today = new Date(2016, 7, 26,12,0,0,0,0); // Friday at noon
console.log("today, Monday",today,"day #"+today.getDay());
var next = new Date(today.getTime());
next.setDate(next.getDate()+1); // tomorrow
while (next.getDay() == 6 || next.getDay() == 0) next.setDate(next.getDate() + 1);
console.log("no change ",next,"day #"+next.getDay());
console.log("-------");
// or without a loop:
function getNextWork(d) {
d.setDate(d.getDate()+1); // tomorrow
if (d.getDay()==0) d.setDate(d.getDate()+1);
else if (d.getDay()==6) d.setDate(d.getDate()+2);
return d;
}
next = getNextWork(today); // Friday
console.log("today, Friday",today);
console.log("next, Monday ",next);
console.log("-------");
today = new Date(2016, 7, 29,12,0,0,0); // Monday at noon
next = getNextWork(today); // Still Monday at noon
console.log("today, Monday",today);
console.log("no change ",next);
console.log("-------");
// Implementing Rob's comment
function getNextWork1(d) {
var day = d.getDay(),add=1;
if (day===5) add=3;
else if (day===6) add=2;
d.setDate(d.getDate()+add);
return d;
}
today = new Date(2016, 7, 26,12,0,0,0,0); // Friday at noon
next = getNextWork1(today); // Friday
console.log("today, Friday",today);
console.log("next, Monday ",next);
console.log("-------");
today = new Date(2016, 7, 26,12,0,0,0,0); // Monday at noon
next = getNextWork1(today); // Monday
console.log("today, Monday",today);
console.log("no change ",next);
您可以一次添加 1 天,直到到达不是星期六或星期日的那一天:
function getNextBusinessDay(date) {
// Copy date so don't affect original
date = new Date(+date);
// Add days until get not Sat or Sun
do {
date.setDate(date.getDate() + 1);
} while (!(date.getDay() % 6))
return date;
}
// today, Friday 26 Aug 2016
[new Date(), new Date(2016,7,26)].forEach(function(d) {
console.log(d.toLocaleString() + ' : ' + getNextBusinessDay(d).toLocaleString());
});
您也可以测试当天并添加额外的内容以度过周末:
// Classic Mon to Fri
function getNextWorkDay(date) {
let d = new Date(+date);
let day = d.getDay() || 7;
d.setDate(d.getDate() + (day > 4? 8 - day : 1));
return d;
}
for (let i=0, d=new Date(); i<7; i++) {
console.log(`${d.toDateString()} -> ${getNextWorkDay(d).toDateString()}`);
d.setDate(d.getDate() + 1);
}
这是另一种方法,可以使用 ECMAScript 工作日编号(周日 = 0、周一 = 1 等)指定工作周。超出范围的日期将移至下一个工作周的开始。
这在一周不是典型的周一到周五的情况下很有用,例如在中东,周六到周三很常见,或者对于一些可能在周五到周一(或其他时间)工作的人。
function getNext(start, end, date) {
let d = new Date(+date);
d.setDate(d.getDate() + 1);
let day = d.getDay();
// Adjust end and day if necessary
// The order of tests and adjustment is important
if (end < start) {
if (day <= end) {
day += 7;
}
end += 7;
}
// If day is before start, shift to start
if (day < start) {
d.setDate(d.getDate() + start - day);
// If day is after end, shift to next start (treat Sunday as 7)
} else if (day > end) {
d.setDate(d.getDate() + 8 - (day || 7));
}
return d;
}
// Examples
let f = new Intl.DateTimeFormat('en-GB', {
weekday:'short',day:'2-digit', month:'short'});
let d = new Date();
[{c:'Work days Mon to Fri',s:1,e:5},
{c:'Work days Sat to Wed',s:6,e:3},
{c:'Work days Fri to Mon',s:5,e:1}
].forEach(({c,s,e}) => {
for (let i = 0; i < 7; i++) {
!i? console.log(`\n${c}`) : null;
console.log(`${f.format(d)} => ${f.format(getNext(s, e, d))}`);
d.setDate(d.getDate() + 1);
}
});
看看这个:https://jsfiddle.net/e9a4066r/
function get_next_weekday (date) {
var tomorrow = new Date(date.setDate(date.getDate() + 1))
return tomorrow.getDay() % 6
? tomorrow
: get_next_weekday(tomorrow)
}
接受的答案将一次跳过一天,这回答了 OP 问题,但对于希望添加可变天数同时仍跳过周末的任何人来说,以下功能可能会有所帮助:
function addWorkDays(date, days) {
while (days > 0) {
date.setDate(date.getDate() + 1);
if (date.getDay() != 0 && date.getDay() != 6) {
days -= 1;
}
}
return date;
}
我想我会在这里把我的帽子扔进戒指:
function getNextBusinessDate(date) {
// Create date array [S, M, T, W, T, F, S]
const days = new Array(7);
let nextDate = date;
for(let i = 0; i < 7; i++) {
days[nextDate.getDay()] = new Date(nextDate);
nextDate.setDate(nextDate.getDate() + 1);
}
// Shift indices to index as though array was [M, T, W, T, F, S, S]
// Then truncate with min to make F, S, S all yield M for next date
return days[Math.min((date.getDay() + 6) % 7 + 1, 5) % 5 + 1];
}