配置响应以在 Spring Boot 应用程序上提供服务

Question

到目前为止，我一直在为我的 Spring boot 应用程序编写一堆端点，没有 html UI 端。现在，我想提供包含反应代码的 HTML 文件和 js 文件。

当我访问 http://localhost:8443 时，我得到：

Whitelabel Error Page

This application has no explicit mapping for /error, so you are seeing this as a fallback.

Sun Nov 19 11:54:01 PST 2017
There was an unexpected error (type=Not Found, status=404).
Not Found

到目前为止我做了什么：

1.添加了一个扩展WebMvcConfigurerAdapter的Web配置类：

@EnableWebMvc
@Configuration
public class WebConfig extends WebMvcConfigurerAdapter {
    @Bean
    public ViewResolver internalResourceViewResolver() {
        InternalResourceViewResolver bean = new InternalResourceViewResolver();
        bean.setPrefix("/resources/static/");
        bean.setSuffix(".html");
        return bean;
    }
}

2.增加了一个休息控制器端点：

@RestController
public class HomeController {

    @RequestMapping(value = "/", method = RequestMethod.GET)
    public ModelAndView getdata() {
        ModelAndView model = new ModelAndView("index");
        return model;
    }
}

我的项目中有myproject/src/main/resources/static/index.html 文件。

Answer 1

您知道，在 Spring Boot 2.1.4 中，新配置是：

package com.simplicity.resourceserver.configs;

import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.web.servlet.ViewResolver;
import org.springframework.web.servlet.config.annotation.*;
import org.springframework.web.servlet.view.InternalResourceViewResolver;

@Configuration
@EnableWebMvc
public class WebMvcConfiguration implements WebMvcConfigurer {

    private static final String[] CLASSPATH_RESOURCE_LOCATIONS = {
            "classpath:/META-INF/resources/", "classpath:/resources/",
            "classpath:/static/", "classpath:/public/" };



    @Override
    public void addResourceHandlers(ResourceHandlerRegistry registry) {
        registry.addResourceHandler("/**")
                .addResourceLocations(CLASSPATH_RESOURCE_LOCATIONS);

        registry.addResourceHandler("/vanilla/**")
                .addResourceLocations("classpath:/vanilla/");

            registry.addResourceHandler("/react/**")
                    .addResourceLocations("classpath:/react/")
                .resourceChain(true);
    }

    @Bean
    public ViewResolver getViewResolver() {
        InternalResourceViewResolver resolver = new InternalResourceViewResolver();
        resolver.setSuffix(".html");
        return resolver;
    }

    @Override
    public void configureDefaultServletHandling(DefaultServletHandlerConfigurer configurer) {
        configurer.enable();
    }

    @Override
    public void configurePathMatch(PathMatchConfigurer configurer) {
        configurer.setUseTrailingSlashMatch(true);
    }

    @Override
    public void addViewControllers(ViewControllerRegistry registry) {
        registry.addViewController("/react")
                .setViewName("forward:/react/index.html");
        registry.addViewController("/react/{path:[^\\.]*}")
                .setViewName("forward:/react/index.html");
        registry.addViewController("/react/**/{path:[^\\.]*}")
                .setViewName("forward:/react/index.html");

    }
}

Spring Boot 应该被定义为基于“视频扑克”的编程：尝试数千次（例如当您玩视频扑克时），但细节很少不同，直到它起作用。

就我而言，我在 src/main/resources 下的两个单独文件夹中有两个不同的应用程序

• src/main/resources/vanilla
• src/main/resources/react

在我的例子中，它们是两个证明概念，但它可能是一个微前端架构。

Answer 2

这真的取决于您的设置。假设您想要类似的东西：

让我们看一个简单的案例：没有 thymeleaf 模板或 spring 静态文件。 Spring 用于提供 rest api，其余的则由响应来决定。但是您可以在任何请求映射 url 上使用控制器。

一种选择是使用ResourceResolver 并像这样配置它：

@Configuration
public class Config implements WebMvcConfigurerAdapter {

    @Override
    public void addResourceHandlers(ResourceHandlerRegistry registry) {
        ResourceResolver resolver = new ReactResourceResolver();
        registry.addResourceHandler("/**")
                .resourceChain(true)
                .addResolver(resolver);

        // Can try to play with
        // registry.addResourceHandler("/**")
        //        .addResourceLocations("classpath:/static/");
        // But this option can't map every path to index.html
        // Can try https://stackoverflow.com/a/42998817/1032167
        // to resolve this, but then you loose /api/** => rest
        // and to be honest it is some regex madness, so
        // it was easier for me to setup custom resource resolver


    }

    public class ReactResourceResolver implements ResourceResolver {
        // root dir of react files
        // example REACT_DIR/index.html
        private static final String REACT_DIR = "/static/";

        // this is directory inside REACT_DIR for react static files
        // example REACT_DIR/REACT_STATIC_DIR/js/
        // example REACT_DIR/REACT_STATIC_DIR/css/
        private static final String REACT_STATIC_DIR = "static";

        private Resource index = new ClassPathResource(REACT_DIR + "index.html");
        private List<String> rootStaticFiles = Arrays.asList("favicon.io",
                "asset-manifest.json", "manifest.json", "service-worker.js");

        @Override
        public Resource resolveResource(
            HttpServletRequest request, String requestPath,
            List<? extends Resource> locations, ResourceResolverChain chain) {

            return resolve(requestPath, locations);
        }

        @Override
        public String resolveUrlPath(
            String resourcePath, List<? extends Resource> locations,
            ResourceResolverChain chain) {

            Resource resolvedResource = resolve(resourcePath, locations);
            if (resolvedResource == null) {
                return null;
            }
            try {
                return resolvedResource.getURL().toString();
            } catch (IOException e) {
                return resolvedResource.getFilename();
            }
        }

        private Resource resolve(
            String requestPath, List<? extends Resource> locations) {

            if (requestPath == null) return null;

            if (rootStaticFiles.contains(requestPath)
                    || requestPath.startsWith(REACT_STATIC_DIR)) {
                return new ClassPathResource(REACT_DIR + requestPath);
            } else
                return index;
        }

    }
}

这是 Spring 2.0.0.M4 的完整工作演示： https://github.com/varren/SpringBootReactExample

---

我遇到了类似的问题，但设置略有不同：Spring single page for every url route and subroute "/a/** => /a/index.html except /a/static/**"。

还有一个选项可以使用正则表达式 Spring catch all route for index.html 来部分解决问题，但我对这种方法没有运气

Answer 3

Spring Boot 可以自动处理静态文件（按约定），只需将所有 html、js、css 等文件放到src/main/resources/static，删除您的 ViewResolver 和 Controller for '/' 就可以了，@987654322 @ 也将被 Spring Boot 映射到 /。

除此之外，您当然可以通过在您的@RestControllers 上使用正确的@RequestMapping 来制作带有api 前缀的REST 端点

Answer 4

你需要一个返回 ModelAndView 的@Controller

    @Controller
    public class HomeController {

        @RequestMapping(value = {"/", "/index.html"})
        public ModelAndView sellerHome(HttpServletRequest request, 
                                       HttpServletResponse response) {
             return new ModelAndView("../static/index");
        }

    }

然后您可以访问http://localhost:8443，并且如果您正确配置了端口，应该会看到您的索引页面。