按即将到来的日期对日期为字符串的数组进行排序

Posted 2023-02-22

【中文标题】按即将到来的日期对日期为字符串的数组进行排序

【英文标题】：Sort array with date as string by upcoming dates

【发布时间】：2019-06-17 07:10:56

【问题描述】：

所以我正在尝试对数组进行排序，以便第一项与当前日期和月份或最近的条目相同。

我的数组如下所示：

[
    [
       "Firstname Lastname",
        "1979-01-03",
        "40"
    ],
    [
        "Firstname Lastname",
        "1996-01-23",
        "23"
    ],
    [
        "Firstname Lastname",
        "1977-01-28",
        "41"
    ],
    [
        "Firstname Lastname",
        "1983-03-11",
        "35"
    ],
    [
       "Firstname Lastname",
       "1977-03-30",
        "41"
    ],
    [
       "Firstname Lastname",
        "1975-05-08",
        "43"
    ]
]

我确实想出了如何根据月份的日期对数组进行排序，但随后它忽略了月份本身

relativeYearDay(date) 
    let differenceDay = new Date(date).getDate() - new Date().getDate();

     if (differenceDay < 0) 
         differenceDay += 365;
     

     return differenceDay;


getUpcomingBirthdays() 
     return this.birthdays.slice(0).sort((a, b) => 
         return this.relativeYearDay(a[1]) - this.relativeYearDay(b[1]);
    );
,

就像我提到的，这会返回一个基于月份日期的排序数组。

我将如何处理一天和一个月？

【问题讨论】：

所以你试图按日期对数组进行排序，忽略年份？如果您的格式一致，只需剪掉字符串的前 5 个字符 ("yyyy-") 并按剩余内容排序。

将日期转换为时间戳并使用它进行排序。

'Math.round(new Date("2013/09/05 15:34:00").getTime()/1000)'

“最近的条目”是否向后看？ IE，1 月 22 日是距离 1 月 23 日最近的还是最远的值？

@TylerRoper 如果您在创建新的 Date 对象时使用固定年份...

【参考方案1】：

实际上，由于要按下一个生日排序，因此在与current date 进行比较时，可以将当前年份设置为所有dates。当它们之间的差异为负时（即生日已经发生），您可以从现在开始添加 1 年的偏移量。

const birthdays = [
    ["Firstname Lastname", "1979-01-03", "40"],
    ["Firstname Lastname", "1996-01-23", "23"],
    ["Firstname Lastname", "1977-01-28", "41"],
    ["Firstname Lastname", "1983-03-11", "35"],
    ["Firstname Lastname", "1977-03-30", "41"],
    ["Firstname Lastname", "1975-05-08", "43"]
];

function distanceToBirthday(date)

    let currDate = new Date();
    currDate.setHours(0, 0, 0, 0);
    let currYear = currDate.getFullYear();

    let offset = new Date();
    offset.setHours(0, 0, 0, 0);
    offset.setFullYear(currYear + 1);

    date = new Date(date + " 00:00");
    date.setFullYear(currYear);

    let diff = date - currDate;
    return (diff < 0) ? diff + offset.getTime() : diff;


function getUpcomingBirthdays(bdays)

    return bdays.slice(0).sort(
        (a, b) => distanceToBirthday(a[1]) - distanceToBirthday(b[1])
    );


console.log(getUpcomingBirthdays(birthdays));

【讨论】：

谢谢！这和@michael-lynch 的答案几乎一样！这是有道理的，就像一个魅力。

【参考方案2】：

您的原始答案非常接近。排序时，您只需要弄清楚用户下一个即将到来的生日的日期。

const birthdays = [
    [ "Firstname Lastname", "1979-01-03", "40" ],
    [ "Firstname Lastname", "1996-01-23", "23" ],
    [ "Firstname Lastname", "1977-01-28", "41" ],
    [ "Firstname Lastname", "1983-03-11", "35" ],
    [ "Firstname Lastname", "1977-03-30", "41" ],
    [ "Firstname Lastname", "1975-05-08", "43" ]
];


function getNextBirthday(date) 
    // Current Date
    let currentDate = new Date();

    // Set the users birthday to this year (originally from thier birth year)
    let birthday = new Date(date);
    birthday.setFullYear(currentDate.getFullYear());

    // If the birthday has already occured this year.  Then thier next birthday is next year.
    if (birthday - currentDate < 0) 
        birthday.setFullYear(currentDate.getFullYear() + 1);
    

    // Return the users next birthday as a date.
    return birthday;


function getUpcomingBirthdays() 
     return birthdays.slice(0).sort((a, b) => 
         return getNextBirthday(a[1]) - getNextBirthday(b[1]);
    );

编辑：添加 cmets 并修复代码中的小错误。

【参考方案3】：

我误读了这个问题，并认为您只想要最接近当前日期的日期。

在简单的编码中......并且可能会给你一些想法来把它变成一个排序例程

const data = [
    [
       "Firstname Lastname",
        "1979-01-03",
        "40"
    ],
    [
        "Firstname Lastname",
        "1996-01-23",
        "23"
    ],
    [
        "Firstname Lastname",
        "1977-01-28",
        "41"
    ],
    [
        "Firstname Lastname",
        "1983-03-11",
        "35"
    ],
    [
       "Firstname Lastname",
       "1977-03-30",
        "41"
    ],
    [
       "Firstname Lastname",
        "1975-05-08",
        "43"
    ]
];
const now = new Date().getTime();
let nearestIndex = -1;
let nearest = 0;

data.forEach( (item, index) => 
  if(nearest ==0 || now-new Date(item[1]).getTime() < nearest) 
    nearest = now-new Date(item[1]).getTime();
    nearestIndex = index;
    
);

console.log(`Nearest date is at index $nearestIndex`, data[nearestIndex], nearest, nearestIndex);

【参考方案4】：

您可以通过带有部分的排序算法对数组进行排序，其中您将日期作为分隔符并将较小的值移动到数组的末尾，并按日期对所有其他值进行排序。

这种方法使用a部分字符串，因为ISO 8601日期可以按字符串排序。

var array = [["Firstname Lastname", "1979-01-03", "40"], ["Firstname Lastname", "1996-01-23", "23"], ["Firstname Lastname", "1977-01-28", "41"], ["Firstname Lastname", "1983-03-11", "35"], [ "Firstname Lastname","1977-03-30", "41"], [ "Firstname Lastname", "1975-05-08", "43"]],
    day = (new Date).toISOString().slice(5, 10)

array.sort(( 1: a ,  1: b ) =>
    (a.slice(5) < day) - (b.slice(5) < day) || a.slice(5).localeCompare(b.slice(5)));

console.log(array.map(a => a.join(' | ')));