javascript - Typescript 过滤两个对象数组

Question

我必须过滤掉一个对象中存在但另一个中不存在的数据

例如对象1

0:Object {Name: "Java", ResourceCount: 3}
1:Object {Name: "DotNet", ResourceCount: 4}
2:Object {Name: "Hadoop", ResourceCount: 1}
3:Object {Name: "Pega", ResourceCount: 2}
4:Object {Name: "Oracle", ResourceCount: 1}
5:Object {Name: "ETL", ResourceCount: 1}

对象 2

0:"DotNet"
1:"ETL"
2:"Hadoop"
3:"Java"
4:"Oracle"
5:"Pega"
6:"Mainframe"

我需要从对象 2 返回“MainFrame”，因为它不存在于对象 1 中。

这是我迄今为止尝试过的方法，但没有成功。

  const filteredList = object2.filter(item1 => 
          object1.find(item2 => item1.Name != item2.Name));

这只是返回对象 2 中的所有行

Answer 1

基本上，创建一个名称数组来帮助过滤对象2； (数组2) 如果存在多个名称，我已使用 加入结果

let object1 = [{Name: "Java", ResourceCount: 3}, {Name: "DotNet", ResourceCount: 4}, {Name: "Hadoop", ResourceCount: 1}, {Name: "Pega", ResourceCount: 2}, {Name: "Oracle", ResourceCount: 1}, {Name: "ETL", ResourceCount: 1}]
let object2 = ["DotNet", "ETL", "Hadoop", "Java", "Oracle", "Pega", "Mainframe"]

let object1Names = object1.map(obj => obj.Name); // for caching the result
results = object2.filter(name => !object1Names.includes(name)).join(',');
console.log(results);