我有以下结构为我的React.js应用程序使用React路由器:

var Dashboard = require('./Dashboard');
var Comments = require('./Comments');

var Index = React.createClass({
  render: function () {
    return (
        <div>
            <header>Some header</header>
            <RouteHandler />
        </div>
    );
  }
});

var routes = (
  <Route path="/" handler={Index}>
    <Route path="comments" handler={Comments}/>
    <DefaultRoute handler={Dashboard}/>
  </Route>
);

ReactRouter.run(routes, function (Handler) {
  React.render(<Handler/>, document.body);
});

我想把一些属性传递给Comments组件。

(通常我会这样做,如<Comments myprop="value" />)

用React路由器最简单和正确的方法是什么?


当前回答

你也可以结合es6和无状态函数来得到一个更清晰的结果:

import Dashboard from './Dashboard';
import Comments from './Comments';

let dashboardWrapper = () => <Dashboard {...props} />,
    commentsWrapper = () => <Comments {...props} />,
    index = () => <div>
        <header>Some header</header>
        <RouteHandler />
        {this.props.children}
    </div>;

routes = {
    component: index,
    path: '/',
    childRoutes: [
      {
        path: 'comments',
        component: dashboardWrapper
      }, {
        path: 'dashboard',
        component: commentsWrapper
      }
    ]
}

其他回答

使用如下所示的解决方案,它可以在v3.2.5中工作。

<Route
  path="/foo"
  component={() => (
    <Content
      lang="foo"
      meta={{
        description: lang_foo.description
      }}
    />
  )}
/>

or

<Route path="/foo">
  <Content
    lang="foo"
    meta={{
      description: lang_foo.description
    }}
  />
</Route>

你可以像这样通过<RouterHandler/>传入props:

var Dashboard = require('./Dashboard');
var Comments = require('./Comments');

var Index = React.createClass({
  render: function () {
    var props = this.props; // or possibly this.state
    return (
        <div>
            <header>Some header</header>
            <RouteHandler {...props} />
        </div>
    );
  }
});

这样做的缺点是你不加区别地传递道具。因此,Comments可能最终接收到的道具实际上是针对不同组件的,这取决于您的路由配置。这不是什么大问题,因为props是不可变的,但如果两个不同的组件都期望一个名为foo的道具,但值不同,这就会出现问题。

React-router v4 alpha

现在有了一种新的方法,尽管和之前的方法很相似。

import { Match, Link, Miss } from 'react-router';
import Homepage from './containers/Homepage';

const route = {
    exactly: true,
    pattern: '/',
    title: `${siteTitle} - homepage`,
    component: Homepage
  }

<Match { ...route } render={(props) => <route.component {...props} />} />

附注:这只适用于alpha版本,并且在v4 alpha发布后被删除。在v4 latest中,同样是带有路径和精确props的。

React-lego是一个示例应用程序,它在react-router-4分支的routes.js中包含了这样做的代码

对于react-router 2.5.2,解决方案非常简单:

    //someConponent
...
render:function(){
  return (
    <h1>This is the parent component who pass the prop to this.props.children</h1>
    {this.props.children && React.cloneElement(this.props.children,{myProp:'value'})}
  )
}
...

使用自定义路由组件,这在React Router v3中是可能的。

var Dashboard = require('./Dashboard');
var Comments = require('./Comments');
var routes = (
  <Route path="/" handler={Index}>
    <MyRoute myprop="value" path="comments" handler={Comments}/>
    <DefaultRoute handler={Dashboard}/>
  </Route>
);

对于<MyRoute>组件代码,它应该类似于:

import React from 'react';
import { Route } from 'react-router';
import { createRoutesFromReactChildren } from 'react-router/lib//RouteUtils';

const MyRoute = () => <div>&lt;MyRoute&gt; elements are for configuration only and should not be rendered</div>;

MyRoute.createRouteFromReactElement = (element, parentRoute) => {
    const { path, myprop } = element.props;
    // dynamically add crud route
    const myRoute = createRoutesFromReactChildren(
        <Route path={path} />,
        parentRoute
    )[0];
    // higher-order component to pass myprop as resource to components
    myRoute.component = ({ children }) => (
        <div>
            {React.Children.map(children, child => React.cloneElement(child, { myprop }))}
        </div>
    );
    return myRoute;
};

export default MyRoute;

有关自定义路由组件方法的更多详细信息,请查看我关于该主题的博客文章:http://marmelab.com/blog/2016/09/20/custom-react-router-component.html