Question

我正在尝试根据该对象属性值检索对象数组中的对象。这是我的阵列：

const address_components = [
  {
    long_name: "11237",
    short_name: "11237",
    types: ["postal_code"]
  },
  {
    long_name: "Kings County",
    short_name: "Kings County",
    types: ["political, administrative_area_level_2"]
  },
  {
    long_name: "New York",
    short_name: "NY",
    types: ["political", "administrative_area_level_1"]
  },
  {
    long_name: "United States",
    short_name: "US",
    types: ["political", "country"]
  }
];

我想检索包含＆＃34; country＆＃34;的对象在types属性中。我试着做了

var result = address_components.filter( obj => {
      return obj.types === "country";
    });
console.log(result);
// logs undefined

控制台记录undefined。但是，当我使用另一个属性尝试此过滤器方法时，我得到了该对象。

var result = address_components.filter( obj => {
  return obj.short_name === "US";
});
console.log(result);
// logs object

有谁知道如何根据types属性中的项目过滤数组？

Answer 1

type是一个数组，您可以使用includes检查数组是否包含字符串。

const address_components = [{
    long_name: "11237",
    short_name: "11237",
    types: ["postal_code"]
  },
  {
    long_name: "Kings County",
    short_name: "Kings County",
    types: ["political", "administrative_area_level_2"]
  },
  {
    long_name: "New York",
    short_name: "NY",
    types: ["political", "administrative_area_level_1"]
  },
  {
    long_name: "United States",
    short_name: "US",
    types: ["political", "country"]
  }
];

var result = address_components.filter(obj => {
  return obj.types.includes("country");
});
console.log(result);

使用多个搜索参数，您可以使用some()

const address_components = [{long_name: "11237",short_name: "11237",types: ["postal_code"]}, {long_name: "Kings County",short_name: "Kings County",types: ["political", "administrative_area_level_2"]},{long_name: "New York",short_name: "NY",types: ["political", "administrative_area_level_1"]},{long_name: "United States",short_name: "US",types: ["political", "country"]}];

var result = address_components.filter(obj => {
  return obj.types.some(o => ['country', 'postal_code'].includes(o));
});
console.log(result);

Doc：includes()

Answer 2

属性.short_name是一个字符串，而types是一个数组。检查等效性（==）适用于字符串，但您需要检查数组的成员资格。

此外，如果你的箭头函数很简单，你可以省略大括号：

let result = address_components.filter(obj => obj.types.includes('country'))

Answer 3

由于types是一个数组，因此您应该使用indexOf方法。

indexOf（）方法返回调用String中的索引第一次出现指定值的对象，开始在fromIndex搜索。如果找不到值，则返回-1。

indexOf优于includes的优点是旧版浏览器支持indexOf而includes不支持。{/ p>

const address_components = [{
    long_name: "11237",
    short_name: "11237",
    types: ["postal_code"]
  },
  {
    long_name: "Kings County",
    short_name: "Kings County",
    types: ["political, administrative_area_level_2"]
  },
  {
    long_name: "New York",
    short_name: "NY",
    types: ["political", "administrative_area_level_1"]
  },
  {
    long_name: "United States",
    short_name: "US",
    types: ["political", "country"]
  }
];

var result = address_components.filter(obj => {
  return obj.types.indexOf("country") > -1;
});
console.log(result);

Answer 4

这里使用不同的方法http://jsben.ch/jmHQg

进行不同的表演

最快的是使用.filter()和.includes()有趣的是.map()和.includes()表现最差，至少在此实施中是这样的：

var result = [];
  address_components.map(obj => {

      if(obj.types.includes("country")){
          result.push(obj);
      }
  });
  console.log(result);

如何根据数组porp的值在数组中查找元素

4 个答案: