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。
如果你没有得到这个。道具…根据其他答案,您可能需要使用withthrouter (docs v4):
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 TwitterSsoButton = withRouter(ShowTheLocation)
// This gets around shouldComponentUpdate
withRouter(connect(...)(MyComponent))
// This does not
connect(...)(withRouter(MyComponent))
据我所知,有三种方法可以做到。
1.使用正则表达式获取查询字符串。
2.您可以使用浏览器api。
图片当前的url是这样的:
http://www.google.com.au?token=123
我们只想得到123;
第一个
const query = new URLSearchParams(this.props.location.search);
Then
const token = query.get('token')
console.log(token)//123
使用第三个名为“query-string”的库。
首先安装它
NPM I查询字符串
然后导入到当前的javascript文件中:
导入query-string
下一步是在当前url中获取'token',请执行以下操作:
const value=queryString.parse(this.props.location.search);
const token=value.token;
console.log('token',token)//123
2019年2月25日更新
4.
如果当前url如下所示:
http://www.google.com.au?app=home&act=article&aid=160990
我们定义一个函数来获取参数:
function getQueryVariable(variable)
{
var query = window.location.search.substring(1);
console.log(query)//"app=article&act=news_content&aid=160990"
var vars = query.split("&");
console.log(vars) //[ 'app=article', 'act=news_content', 'aid=160990' ]
for (var i=0;i<vars.length;i++) {
var pair = vars[i].split("=");
console.log(pair)//[ 'app', 'article' ][ 'act', 'news_content' ][ 'aid', '160990' ]
if(pair[0] == variable){return pair[1];}
}
return(false);
}
我们可以通过以下方式获得“援助”:
getQueryVariable('aid') //160990