【发布时间】:2015-08-15 02:51:02
【问题描述】:
Swagger 配置:
@EnableSwagger
@Configuration
public class SwaggerConfig {
private SpringSwaggerConfig springSwaggerConfig;
@Autowired
public void setSpringSwaggerConfig(SpringSwaggerConfig springSwaggerConfig) {
this.springSwaggerConfig = springSwaggerConfig;
}
@Bean
public SwaggerSpringMvcPlugin swaggerSpringMvcPlugin() {
return new SwaggerSpringMvcPlugin(springSwaggerConfig)
.swaggerGroup("sample-app")
.includePatterns(
"/account/*"
)
.apiInfo(apiInfo())
.build();
}
private ApiInfo apiInfo() {
ApiInfo apiInfo = new ApiInfo(
"sample-app",
"sample-app doc",
"",
"support@sample-app",
"",
""
);
return apiInfo;
}
休息控制器
@RestController
@RequestMapping(value = "/account")
@Api(value = "Change Account details", description = "")
public class ChangeAccountController {
@ApiOperation(value = "Change address")
@RequestMapping(value = "/addresschange", method = RequestMethod.POST)
public String addressChange(HttpServletRequest httpRequest, HttpServletResponse httpResponse,
@Valid @RequestBody User user) throws ServletException, IOException {
// logic and return something!
}
}
参考:部分信息已从此处参考:http://java.dzone.com/articles/how-configure-swagger-generate
问题/疑问:
在SwaggerConfig.java,在includePatterns()方法中,当我给出模式时
作为/account/*,API 没有出现在 Swagger 输出页面中,而,
如果我将模式包含为/account/.*,它就会出现。
为什么?在这个用例中/account/* 和/account/.* 有什么区别?
更新:
另一个用例
@RestController
@RequestMapping(value = "/score")
@ApiOperation(value = "All score", notes = "")
@RequestMapping(value = "", method = RequestMethod.GET)
public @ResponseBody ActionResult allScores(HttpServletRequest httpRequest,
HttpServletResponse httpResponse) {
}
如果我将模式添加为/score/*,则 API 将出现在 Swagger 中。
我不需要把模式写成/score/.*
【问题讨论】:
标签: java swagger swagger-ui spring-restcontroller swagger-maven-plugin