比较两个不同 JSON 数组中的值：分步指南

Question

   

 const fruits = [{id: '1', name: 'Apple'},
    {id: '2', name: 'Orange'},
    {id: '3', name: 'Cherry'}];

    const food=[{id: '1', food_name: 'Orange', deleted:"0"},
    {id: '2', food_name: 'Bread' ,deleted:"0"},
    {id: '3', food_name: 'Cheese', deleted:"0"},
    {id: '4', food_name: 'Apple', deleted:"1"},
    {id: '5', food_name: 'Salt',deleted:"0"}
    ]
    //Code that I tried:
    var dep_data = [];
var foodSet = new Set(food.map(item => item.food_name));

for (var j = 0; j < fruits.length; j++) {
  if (!foodSet.has(fruits[j].name) && fruits[j].deleted !== "1") {
    dep_data.push({ id: fruits[j].id, name: fruits[j].name });
  }
}
    console.log(dep_data)

我想比较两个数组，获取食物中不存在且删除不等于1的水果的id和名称，然后将结果保存到新数组中。

例如，食物数组中存在橙子，结果应存储食物中不存在的水果的 id 和名称，并删除！=1。 （苹果、樱桃）。

Answer 1

您的代码有语法错误，这是更新的错误：

const fruits = [
  { id: '1', name: 'Apple' },
  { id: '2', name: 'Orange' },
  { id: '3', name: 'Cherry' }
];

const food = [
  { id: '1', food_name: 'Orange', deleted: "0" },
  { id: '2', food_name: 'Bread', deleted: "0" },
  { id: '3', food_name: 'Cheese', deleted: "0" },
  { id: '4', food_name: 'Apple', deleted: "1" },
  { id: '5', food_name: 'Salt', deleted: "0" }
];

var dep_data = [];
var foodSet = new Set(food.map(item => item.food_name));

for (var j = 0; j 
结果将返回一个包含 cherry 的数组对象