Gle*_*mad 5 reactjs react-hooks
Hooks 的第一条规则是仅在顶层调用 Hooks,即“Don\xe2\x80\x99t 在循环、条件或嵌套函数内调用 Hooks”。该文档通过一个在条件内调用 Hooks 的示例非常清楚地解释了这一点,但不适用于其他两种情况:循环和嵌套函数。
\n\n是否有在循环和嵌套函数内调用 Hooks 时可能出错的示例?此外,自定义钩子不就是一个嵌套函数吗?
\n\n\n可变长度循环的示例:
const {useState} = React
const WrongLoop = () => {
const [count, setCount] = useState(1)
for (let i = 0; i < count; i++) {
const [x, setX] = useState(i)
}
const [unknownOrder, setUnknownOrder] = useState('some state')
return <button onClick={() => setCount(c => c+ 1)}>{count} {unknownOrder}</button>
}
ReactDOM.render(<WrongLoop />, document.getElementById('root'))Run Code Online (Sandbox Code Playgroud)
<script src="https://unpkg.com/react@16/umd/react.development.js"></script>
<script src="https://unpkg.com/react-dom@16/umd/react-dom.development.js"></script>
<div id="root"></div>Run Code Online (Sandbox Code Playgroud)
eslint 很难检测到的嵌套函数示例(自定义钩子是一种特殊情况 - 当use...在函数名称中使用前缀时,可以通过 eslint 静态分析它在组件中的用法):
const {useState} = React
const fn = () => {
const [x, setX] = useState() // is it OK to use hooks unconditionally here?
}
const WrongFn = () => {
const [count, setCount] = useState(1)
if (count === 1) {
fn() // OK to use normal functions conditionally.. but what if there's a hook inside?
}
const [unknownOrder, setUnknownOrder] = useState('some state')
return <button onClick={() => setCount(c => c+ 1)}>{count} {unknownOrder}</button>
}
ReactDOM.render(<WrongFn />, document.getElementById('root'))Run Code Online (Sandbox Code Playgroud)
<script src="https://unpkg.com/react@16/umd/react.development.js"></script>
<script src="https://unpkg.com/react-dom@16/umd/react-dom.development.js"></script>
<div id="root"></div>Run Code Online (Sandbox Code Playgroud)
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