Javascript:如何在两个数组对象之间使用映射和筛选

Question

在使用.map和.filter时，我遇到了一些问题，无法获得两个对象中不相似的对象。要从arrayObjTwo中获取不寻常的对象，我需要做哪些更改。

码

const arrayObjOne = [{
          countryCode: "US",
          description: " Backyard of home",
          id: "1234",
          location: "US",
          name: "Backyard",
          }]
// Array Object 2
const arrayObjTwo =[
    { description: "Backyard of home", spaceName: "Backyard" },
    { description: "Frontyard of home", spaceName: "Frontyard"},
]
const object1Names = arrayObjOne.map(obj => obj.Name); // for caching the result
const results = arrayObjTwo.filter(name => !object1Names.includes(name));
console.log(results);

一个编译器代码：https://onecompiler.com/javascript/3xy92hpmp

预期结果：

const arrayObjTwo =[
        { description: "Frontyard of home", spaceName: "Frontyard" }
    ]

谢谢..

Answer 1

您对结果的筛选是错误的。您应该使用"object.spaceName“，而不是仅仅使用”名称“，因为您正在遍历对象数组而不仅仅是字符串

// So this line:
const results = arrayObjTwo.filter(name => !object1Names.includes(name));

// Should be:
const results = arrayObjTwo.filter(object => !object1Names.includes(object.spaceName));

// And you have a typo:
const object1Names = arrayObjOne.map(obj => obj.Name);

// Should be like so as keys and properties are case-sensitive:
const object1Names = arrayObjOne.map(obj => obj.name);

因此，这一切都应该是：

const arrayObjOne = [{
          countryCode: "US",
          description: " Backyard of home",
          id: "1234",
          location: "US",
          name: "Backyard",
          }]
// Array Object 2
const arrayObjTwo =[
    { description: "Backyard of home", spaceName: "Backyard" },
    { description: "Frontyard of home", spaceName: "Frontyard"},
]
const object1Names = arrayObjOne.map(obj => obj.name); // for caching the result
const results = arrayObjTwo.filter(object => !object1Names.includes(object.spaceName));
console.log(results);

Answer 2

你快到了！

您的map-语句中有大写名称，但是javascript区分大小写，所以您查找的“名称”字段不存在，因此您的"object1Names“返回未定义的字段。

你应该查一下obj.name

您还需要将其与arrayObjTwo的"spaceName"-field，而不是整个对象进行比较。

下面是您进行这些小更改的代码：

const arrayObjOne = [{
          countryCode: "US",
          description: " Backyard of home",
          id: "1234",
          location: "US",
          name: "Backyard",
          }]
// Array Object 2
const arrayObjTwo =[
    { description: "Backyard of home", spaceName: "Backyard" },
    { description: "Frontyard of home", spaceName: "Frontyard"},
]
const object1Names = arrayObjOne.map(obj => obj.name); // for caching the result
const results = arrayObjTwo.filter(obj => !object1Names.includes(obj.spaceName));
console.log(results);

Answer 3

你需要这个吗？

​

const arr1 = [{
          countryCode: "US",
          description: " Backyard of home",
          id: "1234",
          location: "US",
          name: "Backyard",
          }]
// Array Object 2
const arr2 =[
    { description: "Backyard of home", spaceName: "Backyard" },
    { description: "Frontyard of home", spaceName: "Frontyard"},
]
const names = arr1.map(item => item.name); // for caching the result
console.log(names);
const res = arr2.filter(item => !names.includes(item.name));
console.log(res);

​