我目前需要使用Spring Boot来提供我的React应用程序。它适用于根URL - localhost:8080/
,但当然没有一个子路由被Spring Controller识别。
我不确定如何在没有硬编码的情况下将React Routing和Spring请求映射排成一行,以映射到index.html
- 然后一些路由具有可变的子路由。
以下是我HomeController
index.html
@Controller
public class HomeController {
@RequestMapping(value = "/")
public String index() {
return "index.html";
}
}
这是我的React App中的路由渲染
const Root = ({ store }) => (
<Provider store={store}>
<Router history={browserHistory}>
<Route path="/" component={App}>
<IndexRedirect to="users"/>
<Route path="/org" component={OrgSearch}>
{<Route path="/org/:orgId" component={Org}/>}
</Route>
<Route path="/users" component={UserSearch}>
{<Route path="/users/:userId" component={User} />}
</Route>
</Route>
</Router>
</Provider>
)
非常感谢任何帮助!
编辑:我尝试添加一些通配符功能,它呈现出一些奇怪的行为。以下是HomeController
的更新代码。
@Controller
public class HomeController {
@RequestMapping(value = {"/", "/users/**", "/org/**"})
public String index() {
return "index.html";
}
}
我可以访问/
和/users
但不能访问/users/
或/users/2545
。
当我尝试访问后者时出现错误。
2017-04-14 09:21:59.896 ERROR 58840 --- [nio-8080-exec-3] o.a.c.c.C.[.[.[/].[dispatcherServlet] : Servlet.service() for servlet [dispatcherServlet] in context with path [] threw exception [Circular view path [index.html]: would dispatch back to the current handler URL [/users/index.html] again. Check your ViewResolver setup! (Hint: This may be the result of an unspecified view, due to default view name generation.)] with root cause
javax.servlet.ServletException: Circular view path [index.html]: would dispatch back to the current handler URL [/users/index.html] again. Check your ViewResolver setup! (Hint: This may be the result of an unspecified view, due to default view name generation.)
at org.springframework.web.servlet.view.InternalResourceView.prepareForRendering(InternalResourceView.java:205) ~[spring-webmvc-4.3.7.RELEASE.jar:4.3.7.RELEASE]
at org.springframework.web.servlet.view.InternalResourceView.renderMergedOutputModel(InternalResourceView.java:145) ~[spring-webmvc-4.3.7.RELEASE.jar:4.3.7.RELEASE]
at org.springframework.web.servlet.view.AbstractView.render(AbstractView.java:303) ~[spring-webmvc-4.3.7.RELEASE.jar:4.3.7.RELEASE]
at org.springframework.web.servlet.DispatcherServlet.render(DispatcherServlet.java:1282) ~[spring-webmvc-4.3.7.RELEASE.jar:4.3.7.RELEASE]
at org.springframework.web.servlet.DispatcherServlet.processDispatchResult(DispatcherServlet.java:1037) ~[spring-webmvc-4.3.7.RELEASE.jar:4.3.7.RELEASE]
at org.springframework.web.servlet.DispatcherServlet.doDispatch(DispatcherServlet.java:980) ~[spring-webmvc-4.3.7.RELEASE.jar:4.3.7.RELEASE]
答案 0 :(得分:6)
对我来说,我在link中尝试了答案。 我没有控制器。
基本上是关于定义addViewController ...
@Configuration
public class WebConfiguration extends WebMvcConfigurerAdapter {
@Override
public void addViewControllers(ViewControllerRegistry registry) {
registry.addViewController("/{spring:\\w+}")
.setViewName("forward:/");
registry.addViewController("/**/{spring:\\w+}")
.setViewName("forward:/");
registry.addViewController("/{spring:\\w+}/**{spring:?!(\\.js|\\.css)$}")
.setViewName("forward:/");
}
}
答案 1 :(得分:1)
我遇到了类似的问题,并通过将带有“**”的 URL 分离为单独的带注释的方法来解决它,例如:
@Controller
public class HomeController {
@RequestMapping(value = {"/"})
public String index() {
return "index.html";
}
@RequestMapping(value = {"/users/{**}"})
public String users() {
return "/";
}
@RequestMapping(value = {"/org/{**}"})
public String org() {
return "/";
}
}
答案 2 :(得分:0)
我有同样的问题,这个映射对我有用:
@RequestMapping(value = {"/","/users/{id}"})