我需要遍历组件的children
并仅在子组件为特定类型时才做一些事情:
React.Children.forEach(this.props.children, child => {
if (...) {
console.log('Suitable component!');
}
});
答案 0 :(得分:4)
正如康斯坦丁·斯莫拉宁(Konstantin Smolyanin)所指出的那样,公认的答案可能与react-hot-loader不太匹配。
希望更安全的替代方法,建议here:
import MyComponent from './MyComponent';
const myComponentType = (<MyComponent />).type;
this.props.children.forEach(child => {
if (child.type === myComponentType ) {
console.log('This child is <MyComponent />');
}
});
答案 1 :(得分:2)
其他解决方案:
import MyComponent from './MyComponent';
// const myComponentType = (<MyComponent />).type;
// It's the same than
const myComponentType = React.createElement(MyComponent).type;
this.props.children.forEach(child => {
if (child.type === myComponentType) {
console.log('This child is <MyComponent />');
}
});
答案 2 :(得分:1)
这是您应该做的:
import MyComponent from './MyComponent';
this.props.children.forEach(child => {
if (child.type === MyComponent) {
console.log('This child is <MyComponent />');
}
});
答案 3 :(得分:0)
我当前的解决方案是
import MyComponent from 'MyComponent';
...
React.Children.forEach(this.props.children, child => {
if (child.type.name === MyComponent.name) {
console.log('Suitable component!');
}
});
但是我看到React元素的type
属性没有记录(或者我在阅读文档时不好?)。因此,我正在寻找更可靠的解决方案。
更新:
实际上,代码if (child.type.name === MyComponent.name)
应该简化为if (child.type === MyComponent)
。但这对我没有用。我发现这是由react-hot-loader
引起的。禁用它可以修复我的应用。
react-hot-loader
GitHub上的相关问题:https://github.com/gaearon/react-hot-loader/issues/304