排序对象数组的单一键与日期值

我有一个具有几个键值对的对象数组，我需要根据'updated_at'对它们进行排序:

[
    {
        "updated_at" : "2012-01-01T06:25:24Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-09T11:25:13Z",
        "foo" : "bar"
    },
    {
        "updated_at" : "2012-01-05T04:13:24Z",
        "foo" : "bar"
    }
]

最有效的方法是什么?

当前回答

你可以使用Lodash实用程序库来解决这个问题(它是一个非常有效的库):

Const data = [{ “updated_at”:“2012 - 01 - 01 t06:25:24z”, “foo”:“酒吧” }，｛ “updated_at”:“2012 - 01 - 09年t11:25:13z”, “foo”:“酒吧” }，｛ “updated_at”:“2012 - 01 - 05 t04:13:24z”, “foo”:“酒吧” ｝］ const ordered = _.orderBy( 数据, 函数(项){ 返回item.updated_at; ｝）; console.log(命令) < script src = " https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.15/lodash.min.js " > < /脚本>

你可以在这里找到文档:https://lodash.com/docs/4.17.15#orderBy

2019-10-24 09:30:20

其他回答

数据导入

[
    {
        "gameStatus": "1",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 11:32:04"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:08:24"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:35:40"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 10:42:53"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 10:54:09"
    },
    {
        "gameStatus": "0",
        "userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
        "created_at": "2018-12-19 18:46:22"
    },
    {
        "gameStatus": "1",
        "userId": "7118ed61-d8d9-4098-a81b-484158806d21",
        "created_at": "2018-12-20 10:50:48"
    }
]

由高到低

arr.sort(function(a, b){
    var keyA = new Date(a.updated_at),
        keyB = new Date(b.updated_at);
    // Compare the 2 dates
    if(keyA < keyB) return -1;
    if(keyA > keyB) return 1;
    return 0;
});

例如Asc Order

[
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 10:42:53"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:08:24"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:35:40"
    },
    {
        "gameStatus": "0",
        "userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
        "created_at": "2018-12-19 18:46:22"
    },
    {
        "gameStatus": "1",
        "userId": "7118ed61-d8d9-4098-a81b-484158806d21",
        "created_at": "2018-12-20 10:50:48"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 10:54:09"
    },
    {
        "gameStatus": "1",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 11:32:04"
    }
]

降序为

arr.sort(function(a, b){
    var keyA = new Date(a.updated_at),
        keyB = new Date(b.updated_at);
    // Compare the 2 dates
    if(keyA > keyB) return -1;
    if(keyA < keyB) return 1;
    return 0;
});

描述顺序示例

[
    {
        "gameStatus": "1",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 11:32:04"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-20 10:54:09"
    },
    {
        "gameStatus": "1",
        "userId": "7118ed61-d8d9-4098-a81b-484158806d21",
        "created_at": "2018-12-20 10:50:48"
    },
    {
        "gameStatus": "0",
        "userId": "1a2fefb0-5ae2-47eb-82ff-d1b2cc27875a",
        "created_at": "2018-12-19 18:46:22"
    },
    {
        "gameStatus": "2",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:35:40"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 18:08:24"
    },
    {
        "gameStatus": "0",
        "userId": "c02cfb18-ae66-430b-9524-67d9dd8f6a50",
        "created_at": "2018-12-19 10:42:53"
    }
]

2018-12-20 09:24:48

你可以使用array。sort。

这里有一个例子: Var arr = [{ “updated_at”:“2012 - 01 - 01 t06:25:24z”, “foo”:“酒吧” }，｛ “updated_at”:“2012 - 01 - 09年t11:25:13z”, “foo”:“酒吧” }，｛ “updated_at”:“2012 - 01 - 05 t04:13:24z”, “foo”:“酒吧” ｝］加勒比海盗。排序(函数(a, b) { var keyA = new Date(a.updated_at)， keyB = new Date(b.updated_at); //比较两个日期 if (keyA < keyB)返回-1; if (keyA > keyB)返回1; 返回0; })； console.log (arr);

2012-01-12 15:23:30

array. sort()方法对数组中的元素进行排序并返回数组。小心使用Array.sort()，因为它不是不可变的。对于不可变排序使用immutable-sort。

This method is to sort the array using your current updated_at in ISO format. We use new Data(iso_string).getTime() to convert ISO time to Unix timestamp. A Unix timestamp is a number that we can do simple math on. We subtract the first and second timestamp the result is; if the first timestamp is bigger than the second the return number will be positive. If the second number is bigger than the first the return value will be negative. If the two are the same the return will be zero. This alines perfectly with the required return values for the inline function.

For ES6:

arr.sort((a,b) => new Date(a.updated_at).getTime() - new Date(b.updated_at).getTime());

一对ES5的:

arr.sort(function(a,b){ 
 return new Date(a.updated_at).getTime() - new Date(b.updated_at).getTime();
});

如果你把updated_at改为unix时间戳，你可以这样做:

For ES6:

arr.sort((a,b) => a.updated_at - b.updated_at);

一对ES5的:

arr.sort(function(a,b){ 
 return a.updated_at - b.updated_at;
});

在撰写本文时，现代浏览器还不支持ES6。要在现代浏览器中使用ES6，请使用babel将代码转译为ES5。期待在不久的将来浏览器对ES6的支持。

Array.sort()应该接收三种可能结果之一的返回值:

正数(第一项>第二项) 负数(第一项<第二项) 如果两项相等，则为0

注意，内联函数的返回值可以是任何值正数或负数。Array.Sort()并不关心返回编号为。它只关心返回值是否为正，负或者零。

对于不可变排序:(ES6中的例子)

const sort = require('immutable-sort');
const array = [1, 5, 2, 4, 3];
const sortedArray = sort(array);

你也可以这样写:

import sort from 'immutable-sort';
const array = [1, 5, 2, 4, 3];
const sortedArray = sort(array);

import-from是在ES6中包含javascript的一种新方法，它使你的代码看起来非常干净。我个人的最爱。

不可变排序不会改变源数组，而是返回一个新数组。建议在不可变数据上使用const。

2017-03-15 16:50:35

我已经在Typescript中创建了一个排序函数，我们可以使用它来搜索字符串，日期和对象数组中的数字。它还可以对多个字段进行排序。

export type SortType = 'string' | 'number' | 'date';
export type SortingOrder = 'asc' | 'desc';

export interface SortOptions {
  sortByKey: string;
  sortType?: SortType;
  sortingOrder?: SortingOrder;
}


class CustomSorting {
    static sortArrayOfObjects(fields: SortOptions[] = [{sortByKey: 'value', sortType: 'string', sortingOrder: 'desc'}]) {
        return (a, b) => fields
          .map((field) => {
            if (!a[field.sortByKey] || !b[field.sortByKey]) {
              return 0;
            }

            const direction = field.sortingOrder === 'asc' ? 1 : -1;

            let firstValue;
            let secondValue;

            if (field.sortType === 'string') {
              firstValue = a[field.sortByKey].toUpperCase();
              secondValue = b[field.sortByKey].toUpperCase();
            } else if (field.sortType === 'number') {
              firstValue = parseInt(a[field.sortByKey], 10);
              secondValue = parseInt(b[field.sortByKey], 10);
            } else if (field.sortType === 'date') {
              firstValue = new Date(a[field.sortByKey]);
              secondValue = new Date(b[field.sortByKey]);
            }
            return firstValue > secondValue ? direction : firstValue < secondValue ? -(direction) : 0;

          })
          .reduce((pos, neg) => pos ? pos : neg, 0);
      }
    }
}

用法:

const sortOptions = [{
      sortByKey: 'anyKey',
      sortType: 'string',
      sortingOrder: 'asc',
    }];

arrayOfObjects.sort(CustomSorting.sortArrayOfObjects(sortOptions));

2019-08-22 08:36:36

你可以使用Lodash实用程序库来解决这个问题(它是一个非常有效的库):

你可以在这里找到文档:https://lodash.com/docs/4.17.15#orderBy

2019-10-24 09:30:20

排序对象数组的单一键与日期值

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