Question

pluck

函数签名: pluck(properties: ...args): Observable

选择属性来发出。

示例

示例 1: 提取对象属性

( StackBlitz | jsBin | jsFiddle )

// RxJS v6+
import { from } from 'rxjs';
import { pluck } from 'rxjs/operators';

const source = from([{ name: 'Joe', age: 30 }, { name: 'Sarah', age: 35 }]);
// 提取 name 属性
const example = source.pipe(pluck('name'));
// 输出: "Joe", "Sarah"
const subscribe = example.subscribe(val => console.log(val));

示例 2: 提取嵌套的属性

( StackBlitz | jsBin | jsFiddle )

// RxJS v6+
import { from } from 'rxjs';
import { pluck } from 'rxjs/operators';

const source = from([
  { name: 'Joe', age: 30, job: { title: 'Developer', language: 'JavaScript' } },
  // 当找不到 job 属性的时候会返回 undefined
  { name: 'Sarah', age: 35 }
]);
// 提取 job 中的 title 属性
const example = source.pipe(pluck('job', 'title'));
// 输出: "Developer" , undefined
const subscribe = example.subscribe(val => console.log(val));

其他资源

• pluck :newspaper: - 官方文档

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源码: https://github.com/ReactiveX/rxjs/blob/master/src/internal/operators/pluck.ts