如何在我的路由中定义路由。jsx文件捕获__firebase_request_key参数值从一个URL生成的Twitter的单点登录过程后,从他们的服务器重定向?
http://localhost:8000/#/signin?_k=v9ifuf&__firebase_request_key=blablabla
我尝试了以下路由配置,但:redirectParam没有捕获提到的参数:
<Router>
<Route path="/" component={Main}>
<Route path="signin" component={SignIn}>
<Route path=":redirectParam" component={TwitterSsoButton} />
</Route>
</Route>
</Router>
React路由器v6
来源:在React路由器中获取查询字符串(搜索参数)
使用新的useSearchParams钩子和.get()方法:
const Users = () => {
const [searchParams] = useSearchParams();
console.log(searchParams.get('sort')); // 'name'
return <div>Users</div>;
};
使用这种方法,您可以读取一个或几个参数。
将参数作为一个对象:
如果你需要一次性获得所有的查询字符串参数,那么我们可以像这样使用Object.fromEntries:
const Users = () => {
const [searchParams] = useSearchParams();
console.log(Object.fromEntries([...searchParams])); // ▶ { sort: 'name', order: 'asecnding' }
return <div>Users</div>;
};
阅读更多和现场演示:在React路由器中获取查询字符串(搜索参数)
你可以使用下面的react钩子:
如果url改变,钩子状态会更新
SSR: typeof window === "undefined",只是检查窗口导致错误(尝试一下)
代理对象隐藏实现,因此返回undefined而不是null
这是获取搜索参数为对象的函数:
const getSearchParams = <T extends object>(): Partial<T> => {
// server side rendering
if (typeof window === "undefined") {
return {}
}
const params = new URLSearchParams(window.location.search)
return new Proxy(params, {
get(target, prop, receiver) {
return target.get(prop as string) || undefined
},
}) as T
}
然后像这样把它用作钩子:
const useSearchParams = <T extends object = any>(): Partial<T> => {
const [searchParams, setSearchParams] = useState(getSearchParams())
useEffect(() => {
setSearchParams(getSearchParams())
}, [typeof window === "undefined" ? "once" : window.location.search])
return searchParams
}
如果你的url是这样的:
/app?page=2&count=10
你可以这样读:
const { page, count } = useQueryParams();
console.log(page, count)
React路由器v3
使用React Router v3,你可以从this.props.location.search (?qs1=naisarg&qs2=parmar)获取查询字符串。例如,使用let params = queryString.parse(this.props.location.search),将给出{qs1: 'naisarg', qs2: 'parmar'}
React路由器v4
在React Router v4中,this.props.location.query不再存在。您需要使用this.props.location.search,并自己或使用现有的包(如query-string)解析查询参数。
例子
下面是一个使用React Router v4和query-string库的最小示例。
import { withRouter } from 'react-router-dom';
import queryString from 'query-string';
class ActivateAccount extends Component{
someFunction(){
let params = queryString.parse(this.props.location.search)
...
}
...
}
export default withRouter(ActivateAccount);
理性的
React Router团队移除query属性的理由是:
There are a number of popular packages that do query string parsing/stringifying slightly differently, and each of these differences might be the "correct" way for some users and "incorrect" for others. If React Router picked the "right" one, it would only be right for some people. Then, it would need to add a way for other users to substitute in their preferred query parsing package. There is no internal use of the search string by React Router that requires it to parse the key-value pairs, so it doesn't have a need to pick which one of these should be "right".
[...]
The approach being taken for 4.0 is to strip out all the "batteries included" kind of features and get back to just basic routing. If you need query string parsing or async loading or Redux integration or something else very specific, then you can add that in with a library specifically for your use case. Less cruft is packed in that you don't need and you can customize things to your specific preferences and needs.
你可以在GitHub上找到完整的讨论。
在React Router v4中,只有withRoute才是正确的方式
您可以通过withRouter高阶组件访问历史对象的属性和最近的匹配。withRouter将在包装组件呈现时将更新的匹配、位置和历史道具传递给它。
import React from 'react'
import PropTypes from 'prop-types'
import { withRouter } from 'react-router'
// A simple component that shows the pathname of the current location
class ShowTheLocation extends React.Component {
static propTypes = {
match: PropTypes.object.isRequired,
location: PropTypes.object.isRequired,
history: PropTypes.object.isRequired
}
render() {
const { match, location, history } = this.props
return (
<div>You are now at {location.pathname}</div>
)
}
}
// Create a new component that is "connected" (to borrow redux
// terminology) to the router.
const ShowTheLocationWithRouter = withRouter(ShowTheLocation)
https://reacttraining.com/react-router/web/api/withRouter