下面的代码来自jQuery UI自动完成:

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"
    }
];

例如,我想更改jquery-ui的desc值。我该怎么做呢?

此外,是否有更快的方法来获取数据?我的意思是给对象一个名字来获取它的数据,就像数组中的对象一样?比如jquery-ui。jquery-ui。desc = ....


当前回答

假设您希望在修改期间运行更复杂的代码,您可能会使用if-else语句而不是三元操作符方法

// original 'projects' array;
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"
    }
];
// modify original 'projects' array, and save modified array into 'projects' variable
projects = projects.map(project => {
// When there's an object where key 'value' has value 'jquery-ui'
    if (project.value == 'jquery-ui') {

// do stuff and set a new value for where object's key is 'value'
        project.value = 'updated value';

// do more stuff and also set a new value for where the object's key is 'label', etc.
        project.label = 'updated label';

// now return modified object
        return project;
    } else {
// just return object as is
        return project;
    }
});

// log modified 'projects' array
console.log(projects);

其他回答

let users = [
    {id: 1, name: 'Benedict'},
    {id: 2, name: 'Myles'},
    {id: 3, name: 'Happy'},
]

 users.map((user, index) => {
 if(user.id === 1){
  users[index] = {id: 1, name: 'Baba Benny'};    
 }
 
 return user
})


console.log(users)

这段代码所做的是映射对象,然后匹配所需的 使用if语句,

if(user.id === 1) 

一旦有匹配的地方使用它的索引交换

 users[index] = {id: 1, name: 'Baba Benny'};

对象,然后返回修改后的数组

upsert(array, item) { 
        const i = array.findIndex(_item => _item.id === item.id);
        if (i > -1) {
            let result = array.filter(obj => obj.id !== item.id);
            return [...result, item]
        }
        else {
            return [...array, item]
        };
    }

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。

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

我们也可以使用Array的map函数来使用Javascript修改数组的对象。

function changeDesc(value, desc){
   projects.map((project) => project.value == value ? project.desc = desc : null)
}

changeDesc('jquery', 'new description')