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>
当前回答
在没有第三方库或复杂的解决方案的情况下，在一行中完成这一切。以下是如何
let myVariable = new URLSearchParams(history.location.search).get('business');
你唯一需要改变的是你自己的参数名称的单词“business”。
业务= url.com例子吗?你好
myVariable的结果将是hello
其他回答
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。
容易解构分配URLSearchParams
测试尝试如下:
1 扫描:https://www.google.com/?param1=apple¶m2=banana
2 右键单击>页，单击Inspect > goto Console选项卡 然后粘贴下面的代码:
const { param1, param2 } = Object.fromEntries(new URLSearchParams(location.search));
console.log("YES!!!", param1, param2 );
输出:
YES!!! apple banana
你可以扩展params，如param1, param2，想扩展多少就扩展多少。
你可以检查react-router，简单地说，你可以使用代码获取查询参数，只要你在路由器中定义:
this.props.params.userId
在没有第三方库或复杂的解决方案的情况下，在一行中完成这一切。以下是如何
let myVariable = new URLSearchParams(history.location.search).get('business');
你唯一需要改变的是你自己的参数名称的单词“business”。
业务= url.com例子吗?你好
myVariable的结果将是hello
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)
