Java 使用Spring在Swagger UI上接收404错误
我正在将swagger UI与Spring boot应用程序集成。当我点击swagger-ui.html时。我得到了404错误。我的配置类如下:Java 使用Spring在Swagger UI上接收404错误,java,spring-boot,swagger,swagger-ui,Java,Spring Boot,Swagger,Swagger Ui,我正在将swagger UI与Spring boot应用程序集成。当我点击swagger-ui.html时。我得到了404错误。我的配置类如下: @Configuration @EnableSwagger2 //@Import(SwaggerConfiguration.class) public class SwaggerConfig { @Bean public Docket api() { return new Docket(Document
@Configuration
@EnableSwagger2
//@Import(SwaggerConfiguration.class)
public class SwaggerConfig {
@Bean
public Docket api() {
return new Docket(DocumentationType.SWAGGER_2)
.select()
.apis(RequestHandlerSelectors.any())
.paths(PathSelectors.any())
.build();
}
...
错误消息:
{"status":404,"message":"HTTP 404 Not Found","link":"https://jersey.java.net/apidocs/2.8/jersey/javax/ws/rs/NotFoundException.html"}
希望这有帮助,请在下面找到我的工作招摇配置:
@Configuration
@EnableSwagger2
@Profile({"!production"})
public class SwaggerConfiguration extends WebMvcConfigurerAdapter {
@Autowired
private ServletContext servletContext;
@Bean
public Docket api() {
return new Docket(DocumentationType.SWAGGER_2)
.host("localhost")
.directModelSubstitute(LocalDate.class, Date.class)
.pathProvider(new RelativePathProvider(servletContext) {
@Override
public String getApplicationBasePath() {
return "/";
}
})
.select()
.apis(RequestHandlerSelectors.any())
.paths(PathSelectors.any())
.build();
}
}
您的配置类中是否有@EnableWebMvc?如果是这样,您必须手动进行资源配置,如下所示:
@Override
public void addResourceHandlers (ResourceHandlerRegistry registry) {
registry.addResourceHandler("/swagger-ui.html**")
.addResourceLocations("classpath:/META-INF/resources/swagger-ui.html");
registry.
addResourceHandler("/webjars/**")
.addResourceLocations("classpath:/META-INF/resources/webjars/");
}
我遇到了一个类似的问题,/swagger-ui.html(或新的端点版本/swagger-ui/)返回404,但/v2/api文档返回了有效的json。解决方案是从3.0.0版下调swagger版本。到2.9.2.大家好,请帮助我。我想默认url是。你能进入吗?或