将两个元素连接成数组中的字符串类型元素

Question

在对象数组中sortedsorted = [{…}, {…}, {…}](3) [{…}, {…}, {…}]0: {firstName: "john", lastName: "Doe", age: 36, gender: "male"}1: {lastName: "Latt", age: 40, gender: "male"}2: {firstName: "Tom", age: 22, gender: "male"}如何返回一个数组，"firstName+' '+lastName"如果两者都存在，则返回一个数组，如果firstName没有lastName给出，则返回现有值之一？我想用.map.filter(true).join(' ')例子：["john Doe", "Latt", "Tom"]

Answer 1

您不需要过滤和连接 - 这不会对性能产生不必要的影响。

join如果您需要使用和功能的答案，filter@Ever Dev 的答案非常好

var sorted = [ {firstName: "john", lastName: "Doe", age: 36, gender: "male"},

{lastName: "Latt", age: 40, gender: "male"},

{firstName: "Tom", age: 22, gender: "male"}

]

var result = sorted.map(val => (`${val.firstName ? val.firstName: ''} ${val.lastName? val.lastName: ''}`).trim() )

//.join(' ');

console.log(result);

Answer 2

const sort = [

    {firstName: "john", lastName: "Doe", age: 36, gender: "male"},

    {lastName: "Latt", age: 40, gender: "male"},

    {firstName: "Tom", age: 22, gender: "male"}

];

console.log( 

    sort.map(item => 

        [item.firstName, item.lastName].filter(v => !!v).join(' ')

))

Answer 3

你要找的是地图()

>     sorted.map(user => {

>         if (user.firstName && user.lastName) return user.firstName + " " + 

>           user.lastName

>         if (user.firstName) return user.firstName

>         if (user.lastName) return user.lastName

>         return ""

>     })

Answer 4

您将需要.map()每个项目并应用您的逻辑

sorted.map(person => {
  return `${person.firstName}${person.firstName && person.lastName ? ' ' : ''}${person.lastName}`;
});