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React-Router-Dom V5中
function useQeury() {
const [query, setQeury] = useState({});
const search = useLocation().search.slice(1);
useEffect(() => {
setQeury(() => {
const query = new URLSearchParams(search);
const result = {};
for (let [key, value] of query.entries()) {
result[key] = value;
}
setQeury(result);
}, [search]);
}, [search, setQeury]);
return { ...query };
}
// you can destruct query search like:
const {page , search} = useQuery()
// result
// {page : 1 , Search: "ABC"}
其他回答
this.props.params。Your_param_name将工作。
这是从查询字符串中获取参数的方法。 请执行console.log(this.props);探索所有的可能性。
如果你的路由器是这样的
<Route exact path="/category/:id" component={ProductList}/>
你会得到这样的id
this.props.match.params.id
实际上，没有必要使用第三方库。我们可以用纯JavaScript。
考虑以下URL:
https://example.com?yourParamName=yourParamValue
现在我们得到:
const url = new URL(window.location.href);
const yourParamName = url.searchParams.get('yourParamName');
简而言之
const yourParamName = new URL(window.location.href).searchParams.get('yourParamName')
另一个智能解决方案(推荐)
const params = new URLSearchParams(window.location.search);
const yourParamName = params.get('yourParamName');
简而言之
const yourParamName = new URLSearchParams(window.location.search).get('yourParamName')
注意:
对于有多个值的参数，使用“getAll”而不是“get”
https：//example.com？yourParamName[]=yourParamValue1&yourParamName[]=yourParamValue2
const yourParamName = new URLSearchParams(window.location.search).getAll('yourParamName[]')
结果如下:
["yourParamValue1", "yourParamValue2"]
在React-Router-Dom V5中
function useQeury() {
const [query, setQeury] = useState({});
const search = useLocation().search.slice(1);
useEffect(() => {
setQeury(() => {
const query = new URLSearchParams(search);
const result = {};
for (let [key, value] of query.entries()) {
result[key] = value;
}
setQeury(result);
}, [search]);
}, [search, setQeury]);
return { ...query };
}
// you can destruct query search like:
const {page , search} = useQuery()
// result
// {page : 1 , Search: "ABC"}
从v4开始，React路由器不再直接在其location对象中提供查询参数。原因是
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".
包含了这个之后，只解析location会更有意义。在需要查询对象的视图组件中搜索。
你可以通过覆盖react-router中的withRouter来实现这一点
customWithRouter.js
import { compose, withPropsOnChange } from 'recompose';
import { withRouter } from 'react-router';
import queryString from 'query-string';
const propsWithQuery = withPropsOnChange(
['location', 'match'],
({ location, match }) => {
return {
location: {
...location,
query: queryString.parse(location.search)
},
match
};
}
);
export default compose(withRouter, propsWithQuery)
