Question

如何基于对象内函数的用户输入来访问对象属性？我想称重为“马丁”作为参数，然后返回“马丁”的体重。以这个为例：

const data = {
  info: [
    {name: "martin", weight: "2kg"},
  ],
  weight(name){
    return this.name[name].weight;
  },
};

console.log(data.weight("martin"));

Answer 1

基于OP的评论：

我希望能够访问数据的name属性。

为此，您将需要重组数据并使其成为对象。在这种情况下，您也不能使用相同的名称。下面是一个示例。

const data = {
  info: {
    martin:{weight: "2kg"},
    jack:{weight: "5kg"},
  },
  weight(name){
    return this.info[name].weight
  },
};

console.log(data.weight("martin"));
console.log(data.weight("jack"));

Answer 2

如果要按名称访问它，则可以重组数据并按以下方式使用名称：

 const data = {
  info: {
    martin: { weight: "2kg"},
  },
  weight: function (name) {
    return this.info[name].weight || null;
  },
};

console.log(data.weight("martin"));

Answer 3

如问题所述

我想用“马丁”作为参数调用体重，并返回“马丁”的体重

为此，您将需要使用http -v https://docker.elastic.co/v2/elasticsearch/elasticsearch/tags/list 'Authorization: Bearer some-long-token' HTTP/1.1 200 OK Connection: keep-alive Content-Length: 1765 Content-Type: application/json; charset=utf-8 Date: Thu, 09 May 2019 15:26:18 GMT Docker-Distribution-Api-Version: registry/2.0 X-Content-Type-Options: nosniff { "name": "elasticsearch/elasticsearch", "tags": [ "5.0.0-731e78df", "5.0.0-86a0b164", "5.0.0-alpha5", "5.0.0-beta1", "5.0.0-ccd69424", "5.0.0-rc1", "5.0.0", ... ... ...，因为.find是一个数组。

如果找到它，则可以访问info。

此行.weight很重要，因此不会出现错误return info ? info.weight : null;

Unable to access property weight of null or undefined

Answer 4

如果要返回以“马丁”作为名称的每个对象的重量：

const data = {
  info: [
    {name: "martin", weight: "2kg"},
    {name: "alex", weight: "3kg"},
    {name: "martin", weight: "12kg"},
  ],
  getAll(name) {
    return this.info.filter(obj => obj.name == name);
  },
  weight(name){
    return this.getAll(name).map(obj => obj.weight);
  },
};

console.log(data.weight("martin"));

如何基于对象内函数的用户输入访问对象属性？

4 个答案: