这里有一个简洁明了的答案。我不是百分百确定这能行，但看起来还行。请让我知道，如果一个库是必需的，但我不认为是。另外，如果这在x浏览器中不起作用，请告诉我。我在Chrome IE11和Edge上尝试了这个功能，它们似乎都能正常工作。

    var Students = [
        { ID: 1, FName: "Ajay", LName: "Test1", Age: 20},
        { ID: 2, FName: "Jack", LName: "Test2", Age: 21},
        { ID: 3, FName: "John", LName: "Test3", age: 22},
        { ID: 4, FName: "Steve", LName: "Test4", Age: 22}
    ]

    Students.forEach(function (Student) {
        if (Student.LName == 'Test1') {
            Student.LName = 'Smith'
        }
        if (Student.LName == 'Test2') {
            Student.LName = 'Black'
        }
    });

    Students.forEach(function (Student) {
        document.write(Student.FName + " " + Student.LName + "<BR>");
    });

输出应该如下所示

Ajay史密斯

杰克·布莱克

约翰Test3

史蒂夫Test4

// using higher-order functions to avoiding mutation var projects = [ { value: "jquery", label: "jQuery", desc: "the write less, do more, JavaScript library", icon: "jquery_32x32.png" }, { value: "jquery-ui", label: "jQuery UI", desc: "the official user interface library for jQuery", icon: "jqueryui_32x32.png" }, { value: "sizzlejs", label: "Sizzle JS", desc: "a pure-JavaScript CSS selector engine", icon: "sizzlejs_32x32.png" } ]; // using higher-order functions to avoiding mutation index = projects.findIndex(x => x.value === 'jquery-ui'); [... projects.slice(0,index), {'x': 'xxxx'}, ...projects.slice(index + 1, projects.length)];

您需要知道要更改的对象的索引。这很简单

projects[1].desc= "new string";

let thismoth = moment(new Date()).format('MMMM');
months.sort(function (x, y) { return x == thismoth ? -1 : y == thismoth ? 1 : 0; });

根据以下数据，我们想用西瓜替换summerFruits列表中的浆果。

const summerFruits = [
{id:1,name:'apple'}, 
{id:2, name:'orange'}, 
{id:3, name: 'berries'}];

const fruit = {id:3, name: 'watermelon'};

有两种方法。

第一种方法:

//create a copy of summer fruits.
const summerFruitsCopy = [...summerFruits];

//find index of item to be replaced
const targetIndex = summerFruits.findIndex(f=>f.id===3); 

//replace the object with a new one.
summerFruitsCopy[targetIndex] = fruit;

第二种方法:使用map和spread:

const summerFruitsCopy = summerFruits.map(fruitItem => 
fruitItem .id === fruit.id ? 
    {...summerFruits, ...fruit} : fruitItem );

summerFruitsCopy列表现在将返回一个更新对象的数组。

你必须像这样在数组中搜索:

function changeDesc( value, desc ) {
   for (var i in projects) {
     if (projects[i].value == value) {
        projects[i].desc = desc;
        break; //Stop this loop, we found it!
     }
   }
}

然后像这样使用它

var projects = [ ... ];
changeDesc ( 'jquery-ui', 'new description' );

更新:

为了更快:

var projects = {
   jqueryUi : {
      value:  'lol1',
      desc:   'lol2'
   }
};

projects.jqueryUi.desc = 'new string';

(根据Frédéric的评论，你不应该在对象键中使用连字符，或者你应该使用“jquery-ui”和项目["jquery-ui"]符号。)