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React Memo


使用 memo 将导致 React 跳过渲染组件,如果其 props 没有改变。

这可以提高性能。

本节使用 React Hooks。有关 Hooks 的更多信息,请参阅 React Hooks 部分。


问题

在此示例中,即使 todos 没有更改,Todos 组件也会重新渲染。

示例

index.js:

import { useState } from "react";
import ReactDOM from "react-dom/client";
import Todos from "./Todos";

const App = () => {
  const [count, setCount] = useState(0);
  const [todos, setTodos] = useState(["todo 1", "todo 2"]);

  const increment = () => {
    setCount((c) => c + 1);
  };

  return (
    <>
      <Todos todos={todos} />
      <hr />
      <div>
        Count: {count}
        <button onClick={increment}>+</button>
      </div>
    </>
  );
};

const root = ReactDOM.createRoot(document.getElementById('root'));
root.render(<App />);

Todos.js:

const Todos = ({ todos }) => {
  console.log("child render");
  return (
    <>
      <h2>My Todos</h2>
      {todos.map((todo, index) => {
        return <p key={index}>{todo}</p>;
      })}
    </>
  );
};

export default Todos;

运行示例 »

单击递增按钮时,Todos 组件会重新渲染。

如果此组件很复杂,则可能会导致性能问题。


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解决方案

要解决此问题,我们可以使用 memo

使用 memo 防止 Todos 组件不必要地重新渲染。

Todos 组件导出包装在 memo

示例

index.js:

import { useState } from "react";
import ReactDOM from "react-dom/client";
import Todos from "./Todos";

const App = () => {
  const [count, setCount] = useState(0);
  const [todos, setTodos] = useState(["todo 1", "todo 2"]);

  const increment = () => {
    setCount((c) => c + 1);
  };

  return (
    <>
      <Todos todos={todos} />
      <hr />
      <div>
        Count: {count}
        <button onClick={increment}>+</button>
      </div>
    </>
  );
};

const root = ReactDOM.createRoot(document.getElementById('root'));
root.render(<App />);

Todos.js:

import { memo } from "react";

const Todos = ({ todos }) => {
  console.log("child render");
  return (
    <>
      <h2>My Todos</h2>
      {todos.map((todo, index) => {
        return <p key={index}>{todo}</p>;
      })}
    </>
  );
};

export default memo(Todos);

运行示例 »

现在,只有当传递给它的 props 中的 todos 更新时,Todos 组件才会重新渲染。


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