Question

这是一个非常简单的JSON结构，类似于我的数据类型。

我需要返回整个人员数组。我觉得有一种适当的方法可以执行此操作，而无需创建新列表和新信息数组，以及用新数组拼接出旧数组。我只是想将新对象添加或添加到旧数组。

**JSON People Data** = [

{name: "n1", 
info: [ {address: "a1", phone: "p1"}, {address: "a2", phone: "p2"} ]
},

{name: "n2", 
info: [ {address: "a1", phone: "p1"}, {address: "a2", phone: "p2"} ]

]

问题：

如何将（地址和电话）的Info对象添加到第二个info数组的第二人称数组并返回整个people数组？

理想结果：

**JSON People Data** = [

{name: "n1", 
info: [ {address: "a1", phone: "p1"}, {address: "a2", phone: "p2"} ]
},

{name: "n2", 
info: [ {address: "a1", phone: "p1"}, {address: "a2", phone: "p2"}, {address: "a3", phone: "p3"} ]

]

我第一次尝试：

**const newListConcat** = people[1].info.concat({address: "a3", phone: "p3"})

尝试后：

**const newListPush** = people[1].info.push({address: "a3", phone: "p3"})

但要返回：

// console.log  - [ {address: "a1", phone: "p1"}, {address: "a1", phone: "p1"}, {address: "a1", phone: "p1"} ]

Answer 1

我纠正了一些错别字，但这可以按需工作：

const people = [
{
name: "n1", 
info: [ {address: "a1", phone: "p1"}, {address: "a2", phone: "p2"} ]
},
{
name: "n2", 
info: [ {address: "a1", phone: "p1"}, {address: "a2", phone: "p2"} ]
}
];

people[1].info.push({address: "a3", phone: "p3"});

console.log(people);

**const newListPush** = people[1].info.push({address: "a3", phone: "p3"})

Answer 2

我不确定您的期望。致电info.push()

时，您做对了

const people = [
  {
    name: 'n1',
    info: [
      { address: 'a1', phone: 'p1' },
      { address: 'a2', phone: 'p2' },
    ],
  },
  {
    name: 'n2',
    info: [
      { address: 'a1', phone: 'p1' },
      { address: 'a2', phone: 'p2' },
    ],
  },
];

people[1].info.push({address: "a3", phone: "p3"});

console.log(JSON.stringify(people, null, 2));

如何使用索引将对象添加到嵌套数组？

2 个答案: