swagger 2.6.1 /swagger-resources/configuration/ui 404 - spring

Question

I am trying to add Swagger to the Spring Framework, but I get a 404 error.
My project setting
Spring 4.2.5 + Spring Security 4.2.3
pom.xml
<!-- Swagger2 -->
<dependency>
<groupId>io.springfox</groupId>
<artifactId>springfox-swagger2</artifactId>
<version>2.6.1</version>
</dependency>
<!-- Swagger-UI -->
<dependency>
<groupId>io.springfox</groupId>
<artifactId>springfox-swagger-ui</artifactId>
<version>2.6.1</version>
</dependency>
SwaggerConfig.java
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.ComponentScan;
import org.springframework.context.annotation.Configuration;
import org.springframework.web.servlet.config.annotation.ResourceHandlerRegistry;
import org.springframework.web.servlet.config.annotation.WebMvcConfigurationSupport;
import springfox.documentation.builders.PathSelectors;
import springfox.documentation.builders.RequestHandlerSelectors;
import springfox.documentation.service.ApiInfo;
import springfox.documentation.spi.DocumentationType;
import springfox.documentation.spring.web.plugins.Docket;
import springfox.documentation.swagger2.annotations.EnableSwagger2;
#Configuration
#EnableSwagger2
public class SwaggerConfig extends WebMvcConfigurationSupport {
#Bean public Docket api() {
return new Docket(DocumentationType.SWAGGER_2)
.select()
.apis(RequestHandlerSelectors.any())
.paths(PathSelectors.any())
.build()
.apiInfo(apiInfo())
.useDefaultResponseMessages(false);
}
/** API Info */
private ApiInfo apiInfo() {
ApiInfo apiInfo = new ApiInfo("Swagger Sample", "APIs Sample", "Sample Doc 0.1v", "", "Author Name", "This sentence will be display.", "/");
return apiInfo;
}
/** Swagger UI 를 Resource Handler 에 등록 */
#Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
registry.addResourceHandler("swagger-ui.html") .addResourceLocations("classpath:/META-INF/resources/");
registry.addResourceHandler("/webjars/**") .addResourceLocations("classpath:/META-INF/resources/webjars/");
}
}
context-security.xml
<intercept-url pattern="/swagger-ui.html" />
<intercept-url pattern="/swagger-resources" />
<intercept-url pattern="/webjars/** " />
<intercept-url pattern="/v2/api-docs" />
Please help me.

Answer 1

I hope this can help you. I use Spring Mvc, swagger2 and don´t use spring security.
Separate the configuration into 2 parts
#Configuration
#EnableSwagger2
#ComponentScan("com.demo.controller")
public class SwaggerConfig{
#Bean
public Docket apiSwagger(){
return new Docket(DocumentationType.SWAGGER_2)
.select()
.apis(RequestHandlerSelectors.any())
.paths(PathSelectors.any())
.build();
}
}
Next, I create another class of configuration
#Configuration
#EnableWebMvc
#Import(SwaggerConfig.class)
public class AppSwaggerConfig extends WebMvcConfigurerAdapter {
#Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
//enabling swagger-ui part for visual documentation
registry.addResourceHandler("swagger-ui.html").addResourceLocations("classpath:/META-INF/resources/");
registry.addResourceHandler("/webjars/**").addResourceLocations("classpath:/META-INF/resources/webjars/");
}
}
Finally, in my WebAppInitializer class, add the AppSwaggerConfig class in the Dispatcher
public class WebAppInitializer implements WebApplicationInitializer {
#Override
public void onStartup(ServletContext container) throws ServletException {
// Create the 'root' Spring application context
AnnotationConfigWebApplicationContext rootContext = new AnnotationConfigWebApplicationContext();
rootContext.register(SwaggerConfig.class);
// Manage the lifecycle of the root application context
container.addListener(new ContextLoaderListener(rootContext));
try {
loadProperties(rootContext);
} catch (IOException e) {
throw new ServletException("Properties could not be loaded", e);
}
// Create the dispatcher servlet's Spring application context
AnnotationConfigWebApplicationContext dispatcherServlet = new AnnotationConfigWebApplicationContext();
dispatcherServlet.register(AppSwaggerConfig.class);
// Register and map the dispatcher servlet
ServletRegistration.Dynamic dispatcher = container.addServlet("dispatcher",
new DispatcherServlet(dispatcherServlet));
dispatcher.setLoadOnStartup(1);
dispatcher.addMapping("/");
}

Answer 2

I seem, new Docket requires also defining few parameters. In my old project:
#Bean
public Docket configSwaggerApi() {
return new Docket(DocumentationType.SWAGGER_2)
.useDefaultResponseMessages(false)
.apiInfo(new ApiInfo(environment.getProperty("api.title"), environment.getProperty("api.description"), environment.getProperty("api.version"),
environment.getProperty("api.terms"), new Contact(environment.getProperty("api.contact.name"), environment.getProperty("api.contact.url"),
environment.getProperty("api.contact.mail")), environment.getProperty("api.license"), environment.getProperty("api.license.url")))
.host(environment.getProperty("platform.url"))
.pathMapping(environment.getProperty("server.context-path"))
.protocols(newHashSet("http", "https"))
.produces(Arrays.stream(new String[] {"application/json"}).collect(Collectors.toSet()))
.tags(new Tag("public", "public tools"), new Tag("user","user tools"));
}
If you use Spring Security, don't forget to provide non-secured access to swagger:
#Override
public void configure(WebSecurity web) throws Exception {
web.ignoring().antMatchers("/favicon.ico", "/swagger*", "/swagger.json", "/swagger-resources/**", "/swagger-*/**", "/webjars/**");
}
Default path /v2/api-docs does not require any interceptors and configurations. If you want to change it, add to application.properties new parameter springfox.documentation.swagger.v2.path=/new/path.
Resource handlers are used if you don't use Spring Boot.
If these advices won't help you, I can describe fully workable version with oauth security access.