我认为这样比较好

const index = projects.findIndex(project => project.value==='jquery-ui');
projects[index].desc = "updated desc";

让你想更新数组[2]= "data"的值

    for(i=0;i<array.length;i++){
      if(i == 2){
         array[i] = "data";
        }
    }

ES6方式，不改变原始数据。

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"
}];

//find the index of object from array that you want to update
const objIndex = projects.findIndex(obj => obj.value === 'jquery-ui');

// Make sure to avoid incorrect replacement
// When specific item is not found
if (objIndex === -1) {
  return;
}

// make new object of updated object.   
const updatedObj = { ...projects[objIndex], desc: 'updated desc value'};

// make final new array of objects by combining updated object.
const updatedProjects = [
  ...projects.slice(0, objIndex),
  updatedObj,
  ...projects.slice(objIndex + 1),
];

console.log("original data=", projects);
console.log("updated data=", updatedProjects);

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

这是另一个涉及到发现的答案。 这取决于以下事实:

遍历数组中的每个对象，直到找到匹配的对象 每个对象都提供给你并且是可修改的

下面是关键的Javascript代码片段:

projects.find( function (p) {
    if (p.value !== 'jquery-ui') return false;
    p.desc = 'your value';
    return true;
} );

下面是相同Javascript的另一个版本:

projects.find( function (p) {
    if (p.value === 'jquery-ui') {
        p.desc = 'your value';
        return true;
    }
    return false;
} );

这里有一个更短的(有点邪恶的版本):

projects.find( p => p.value === 'jquery-ui' && ( p.desc = 'your value', true ) );

以下是完整的工作版本:

let 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" } ]; projects.find( p => p.value === 'jquery-ui' && ( p.desc = 'your value', true ) ); console.log( JSON.stringify( projects, undefined, 2 ) );

这很简单

使用findIndex方法查找对象的索引。 将索引存储在变量中。 做一个简单的更新，像这样:

//Initailize array of objects. let myArray = [ {id: 0, name: "Jhon"}, {id: 1, name: "Sara"}, {id: 2, name: "Domnic"}, {id: 3, name: "Bravo"} ], //Find index of specific object using findIndex method. objIndex = myArray.findIndex((obj => obj.id == 1)); //Log object to Console. console.log("Before update: ", myArray[objIndex]) //Update object's name property. myArray[objIndex].name = "Laila" //Log object to console again. console.log("After update: ", myArray[objIndex])