SPA和Spring boot-serve index.html用于非API请求

Question

所有api调用都应使用前缀为/api的url，并应由REST控制器处理。

所有其他URL应该只提供index.html文件。我如何用Spring实现这一点？

Answer 1

实现您想要的最简单的方法是实现自定义404处理程序。

将这些参数添加到应用程序中。properties：

spring.resources.add-mappings=false
spring.mvc.throw-exception-if-no-handler-found=true

第一个属性删除所有默认的静态资源处理，第二个属性禁用Spring默认的whitelabel页面（默认情况下Spring捕获NoHandlerFoundException并提供标准的whitelabel页面）

import org.springframework.web.bind.annotation.ControllerAdvice;
import org.springframework.web.bind.annotation.ExceptionHandler;
import org.springframework.web.servlet.NoHandlerFoundException;
import javax.servlet.http.HttpServletRequest;

@ControllerAdvice
public class PageNotFoundController {
    @ExceptionHandler(NoHandlerFoundException.class)
    public String handleError404() {
            return "redirect:/index.html";
    }
}

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.EnableWebMvc;
import org.springframework.web.servlet.config.annotation.ResourceHandlerRegistry;
import org.springframework.web.servlet.config.annotation.WebMvcConfigurerAdapter;
import org.springframework.web.servlet.view.InternalResourceView;
import org.springframework.web.servlet.view.UrlBasedViewResolver;

@Configuration
@EnableWebMvc
public class WebConfig extends WebMvcConfigurerAdapter {

    @Override
    public void addResourceHandlers(ResourceHandlerRegistry registry) {
        registry.addResourceHandler("/index.html").addResourceLocations("classpath:/static/index.html");
        super.addResourceHandlers(registry);
    }

    @Bean
    public ViewResolver viewResolver() {
        UrlBasedViewResolver viewResolver = new UrlBasedViewResolver();
        viewResolver.setViewClass(InternalResourceView.class);
        return viewResolver;
    }

}