DEMO

Question

假设你有一个像这样的javascript类

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

假设您创建了该类的多个实例并将它们存储在数组中

var objArray = [];
objArray.push(DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

所以我现在将拥有一个由DepartmentFactory创建的对象数组。我如何使用array.sort()方法按每个对象的DepartmentName属性对这个对象数组进行排序？

array.sort()方法在排序字符串数组时工作正常

var myarray=["Bob", "Bully", "Amy"];
myarray.sort(); //Array now becomes ["Amy", "Bob", "Bully"]

但是如何使它与对象列表一起使用？

Answer 1

你必须做这样的事情：

objArray.sort(function(a, b) {
    var textA = a.DepartmentName.toUpperCase();
    var textB = b.DepartmentName.toUpperCase();
    return (textA < textB) ? -1 : (textA > textB) ? 1 : 0;
});

注意：更改大小写（向上或向下）可确保不区分大小写的排序。

Answer 2

支持unicode：

objArray.sort(function(a, b) {
   return a.DepartmentName.localeCompare(b.DepartmentName);
});

Answer 3

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

// use `new DepartmentFactory` as given below. `new` is imporatant

var objArray = [];
objArray.push(new DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(new DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(new DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(new DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

function sortOn(property){
    return function(a, b){
        if(a[property] < b[property]){
            return -1;
        }else if(a[property] > b[property]){
            return 1;
        }else{
            return 0;   
        }
    }
}

//objArray.sort(sortOn("id")); // because `this.id = data.Id;`
objArray.sort(sortOn("name")); // because `this.name = data.DepartmentName;`
console.log(objArray);

演示：http://jsfiddle.net/diode/hdgeH/

Answer 4

objArray.sort((a, b) => a.DepartmentName.localeCompare(b.DepartmentName))

Answer 5

// Sorts an array of objects "in place". (Meaning that the original array will be modified and nothing gets returned.)
function sortOn (arr, prop) {
    arr.sort (
        function (a, b) {
            if (a[prop] < b[prop]){
                return -1;
            } else if (a[prop] > b[prop]){
                return 1;
            } else {
                return 0;   
            }
        }
    );
}

//Usage example:

var cars = [
        {make:"AMC",        model:"Pacer",  year:1978},
        {make:"Koenigsegg", model:"CCGT",   year:2011},
        {make:"Pagani",     model:"Zonda",  year:2006},
        ];

// ------- make -------
sortOn(cars, "make");
console.log(cars);

/* OUTPUT:
AMC         : Pacer : 1978
Koenigsegg  : CCGT  : 2011
Pagani      : Zonda : 2006
*/



// ------- model -------
sortOn(cars, "model");
console.log(cars);

/* OUTPUT:
Koenigsegg  : CCGT  : 2011
AMC         : Pacer : 1978
Pagani      : Zonda : 2006
*/



// ------- year -------
sortOn(cars, "year");
console.log(cars);

/* OUTPUT:
AMC         : Pacer : 1978
Pagani      : Zonda : 2006
Koenigsegg  : CCGT  : 2011
*/

Answer 6

DEMO

var DepartmentFactory = function(data) {
    this.id = data.Id;
    this.name = data.DepartmentName;
    this.active = data.Active;
}

var objArray = [];
objArray.push(new DepartmentFactory({Id: 1, DepartmentName: 'Marketing', Active: true}));
objArray.push(new DepartmentFactory({Id: 2, DepartmentName: 'Sales', Active: true}));
objArray.push(new DepartmentFactory({Id: 3, DepartmentName: 'Development', Active: true}));
objArray.push(new DepartmentFactory({Id: 4, DepartmentName: 'Accounting', Active: true}));

console.log(objArray.sort(function(a, b) { return a.name > b.name}));

Answer 7

ES6中的短代码

objArray.sort((a, b) => a.DepartmentName.toLowerCase().localeCompare(b.DepartmentName.toLowerCase()))

Answer 8

这样做

objArrayy.sort(function(a, b){
 var nameA=a.name.toLowerCase(), nameB=b.name.toLowerCase()
 if (nameA < nameB) //sort string ascending
  return -1
 if (nameA > nameB)
  return 1
 return 0 //default return value (no sorting)
});
console.log(objArray)

Answer 9

objArray.sort( (a, b) => a.id.localeCompare(b.id, 'en', {'sensitivity': 'base'}));

这按字母顺序对它们进行排序，并且不区分大小写。它也很干净而且易于阅读：D

Answer 10

这是一个简单的函数，可用于通过对象的属性对对象数组进行排序；不管该属性是字符串还是整数，都可以使用。

(width, height, nb_channels=3, max_pow+1)

Answer 11

因为这里提供的所有解决方案都没有空/未定义的安全操作，所以我是这样处理的（您可以根据需要处理空值）：

ES5

service.request(request).map { _ in () }.eraseToAnyPublisher()

ES6+

objArray.sort(
  function(a, b) {
    var departmentNameA = a.DepartmentName ? a.DepartmentName : '';
    var departmentNameB = b.DepartmentName ? b.DepartmentName : '';

    departmentNameA.localeCompare(departmentNameB);
  }
);

我还删除了其他人使用的 toLowerCase，因为 localeCompare 不区分大小写。此外，我更喜欢在使用 Typescript 或 ES6+ 时对参数更加明确，以便为未来的开发人员提供更加明确的参数。

Answer 12

你必须传递一个接受两个参数的函数，比较它们并返回一个数字，所以假设你想按ID对它们进行排序，你会写...

objArray.sort(function(a,b) {
    return a.id-b.id;
});
// objArray is now sorted by Id

Answer 13

尝试了一下，并尝试减少循环次数后，我得到了以下解决方案：

Demo on codepen

const items = [
      {
        name: 'One'
      },
      {
        name: 'Maria is here'
      },
      {
        name: 'Another'
      },
      {
        name: 'Z with a z'
      },
      {
        name: '1 number'
      },
      {
        name: 'Two not a number'
      },
      {
        name: 'Third'
      },
      {
        name: 'Giant'
      }
    ];

    const sorted = items.sort((a, b) => {
      return a[name] > b[name];
    });

    let sortedAlphabetically = {};

    for(var item in sorted) {
      const firstLetter = sorted[item].name[0];
      if(sortedAlphabetically[firstLetter]) {
        sortedAlphabetically[firstLetter].push(sorted[item]);
      } else {
        sortedAlphabetically[firstLetter] = [sorted[item]]; 
      }
    }

    console.log('sorted', sortedAlphabetically);

Answer 14

一个简单的答案：

objArray.sort(function(obj1, obj2) {
   return obj1.DepartmentName > obj2.DepartmentName;
});

ES6方式：

objArray.sort((obj1, obj2) => {return obj1.DepartmentName > obj2.DepartmentName};

如果你需要将它设置为小写/大写等，只需这样做并将结果存储在变量中，而不是比较该变量。示例：

objArray.sort((obj1, obj2) => {
   var firstObj = obj1.toLowerCase();
   var secondObj = obj2.toLowerCase();
   return firstObj.DepartmentName > secondObj.DepartmentName;
});

Javascript - 在数组的一个属性上按字母顺序对数组中的对象进行排序

14 个答案:

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