我试图获取弹出模块的单独网址,我使用按钮重定向:
trigger={ <Link href="/exercisesadd">
<Button style={style} waves='light' > Add new</Button></Link>
}>
但我收到404错误,即使我在路由器中有这条路线:
export default () => {
return <Route path="/"component={Container} auth={auth}>
<IndexRoute component={Home}/>
<Route path="/portfolio" onEnter={requireAuth} component={Portfolio} />
<Route path="/Login" component={Home} />
<Route path="/home" component={Home} />
<Route path="/edit" component={Edits}/>
<Route path="/exercises" component={Exercises}/>
<Route path="/exercisesadd" component={Exercises}/>
</Route>
};
答案 0 :(得分:5)
Link组件采用'to'prop而不是href。所以你想要
<Link to="/exerciseadd">