[英]React propTypes component class?
How can I validate that the supplied prop is a component class (not instance)?如何验证提供的 prop 是组件类(不是实例)?
eg例如
export default class TimelineWithPicker extends React.PureComponent {
static propTypes = {
component: PropTypes.any, // <-- how can I validate that this is a component class (or stateless functional component)?
};
render() {
return (
<this.props.component {...this.props} start={this.state.start}/>
);
}
}
For anyone using PropTypes >= 15.7.0
a new PropTypes.elementType
was added in this pull request and was released on february 10, 2019 .对于使用
PropTypes >= 15.7.0
的任何人,此拉取请求中添加了新的PropTypes.elementType
并于2019年2 月 10 日发布。
This prop type supports all components (native components, stateless components, stateful components, forward refs React.forwardRef
, context providers/consumers).这个 prop 类型支持所有组件(原生组件、无状态组件、有状态组件、前向引用
React.forwardRef
、上下文提供者/消费者)。
And it throws a warning when is not any of those elements, it also throws a warning when the prop passed is an element ( PropTypes.element
) and not a type.当不是这些元素中的任何一个时,它会发出警告,当传递的道具是元素(
PropTypes.element
)而不是类型时,它也会发出警告。
Finally you can use it like any other prop type:最后,您可以像使用任何其他道具类型一样使用它:
const propTypes = {
component: PropTypes.elementType,
requiredComponent: PropTypes.elementType.isRequired,
};
EDITED: Added React's FancyButton example to codesandbox as well as a custom prop checking function that works with the new React.forwardRef
api in React 16.3.编辑:向代码和框添加了 React 的FancyButton示例,以及与 React 16.3 中的新
React.forwardRef
api 一起使用的自定义道具检查功能。 The React.forwardRef
api returns an object with a render
function. React.forwardRef
api 返回一个带有render
函数的对象。 I'm using the following custom prop checker to verify this prop type.我正在使用以下自定义道具检查器来验证此道具类型。 - Thanks for Ivan Samovar for noticing this need.
- 感谢 Ivan Samovar 注意到这一需求。
FancyButton: function (props, propName, componentName) {
if(!props[propName] || typeof(props[propName].render) != 'function') {
return new Error(`${propName}.render must be a function!`);
}
}
You'll want to use你会想要使用
. PropTypes.element
. Actually...
PropType.func
works for both stateless functional components and class components.实际上...
PropType.func
适用于无状态功能组件和类组件。
I've made a sandbox to prove that this works... Figured this was needed considering I gave you erroneous information at first.我做了一个沙箱来证明这是有效的......考虑到我一开始给你错误的信息,我认为这是必要的。 Very sorry about that!
对此非常抱歉!
Working sandbox example !工作沙箱示例!
Here is the code for the test in case link goes dead:这是测试链接失效时的代码:
import React from 'react';
import { render } from 'react-dom';
import PropTypes from "prop-types";
class ClassComponent extends React.Component {
render() {
return <p>I'm a class component</p>
}
}
const FancyButton = React.forwardRef((props, ref) => (
<button ref={ref} className="FancyButton">
{props.children}
</button>
));
// You can now get a ref directly to the DOM button:
const ref = React.createRef();
<FancyButton ref={ref}>Click me!</FancyButton>;
const FSComponent = () => (
<p>I'm a functional stateless component</p>
);
const Test = ({ ClassComponent, FSComponent, FancyButton }) => (
<div>
<ClassComponent />
<FSComponent />
<FancyButton />
</div>
);
Test.propTypes = {
ClassComponent: PropTypes.func.isRequired,
FSComponent: PropTypes.func.isRequired,
FancyButton: function (props, propName, componentName) {
if(!props[propName] || typeof(props[propName].render) != 'function') {
return new Error(`${propName}.render must be a function!`);
}
},
}
render(<Test
ClassComponent={ ClassComponent }
FSComponent={ FSComponent }
FancyButton={ FancyButton } />, document.getElementById('root'));
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