我有 yaml 格式的 openapi 3.0 規范和從中生成代碼的應用程序。除了生成 swagger ui 外，一切正常。我使用 spring-fox 來生成它，但它似乎從控制器生成 swagger ui 2.0 版本，這些控制器是從 openapi 規范生成的。如何直接從我的 3.0 規范而不是從 3.0 openapi 規范生成的控制器生成 swagger ui？
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TA貢獻1780條經驗 獲得超5個贊
好吧，我已經解決了這個問題（雖然解決方案很麻煩）。
首先，我添加了 swagger ui webjar -
<plugin>
<!-- Download Swagger UI webjar. -->
<artifactId>maven-dependency-plugin</artifactId>
<version>${maven-dependency-plugin.version}</version>
<executions>
<execution>
<phase>prepare-package</phase>
<goals>
<goal>unpack</goal>
</goals>
<configuration>
<artifactItems>
<artifactItem>
<groupId>org.webjars</groupId>
<artifactId>swagger-ui</artifactId>
<version>${swagger-ui.version}</version>
</artifactItem>
</artifactItems>
<outputDirectory>${project.build.directory}/classes</outputDirectory>
</configuration>
</execution>
</executions>
</plugin>
然后我將我的 yaml 規范轉換為 json 格式并將其復制到 swagger-ui webjar 目錄：
<plugin>
<groupId>org.openapitools</groupId>
<artifactId>openapi-generator-maven-plugin</artifactId>
<version>4.0.0-beta3</version>
<executions>
<execution>
<id>generate-spec</id>
<goals>
<goal>generate</goal>
</goals>
<configuration>
<inputSpec>${openapi-spec-file-location}</inputSpec>
<validateSpec>true</validateSpec>
<generatorName>openapi</generatorName>
<output>${project.build.directory}/classes/META-INF/resources/webjars/swagger-ui/${swagger-ui.version}</output>
</configuration>
</execution>
</executions>
</plugin>
接下來我們需要在 swagger-ui 中設置規范路徑。根據swagger-ui API，我們可以傳遞specJSON 變量而不是 url。因此，要初始化此spec變量并編輯 swagger ui 渲染，我在 maven 中使用了替換插件：
<plugin>
<!-- Replace the OpenAPI specification example URL with the local one. -->
<groupId>com.google.code.maven-replacer-plugin</groupId>
<artifactId>replacer</artifactId>
<version>1.5.3</version>
<executions>
<execution>
<phase>prepare-package</phase>
<goals>
<goal>replace</goal>
</goals>
</execution>
</executions>
<configuration>
<includes>
<!-- Static index html with swagger UI rendering and OAS in JSON format. -->
<include>${project.build.directory}/classes/META-INF/resources/webjars/swagger-ui/${swagger-ui.version}/index.html</include>
<include>${project.build.directory}/classes/META-INF/resources/webjars/swagger-ui/${swagger-ui.version}/openapi.json</include>
</includes>
<regexFlags>
<regexFlag>CASE_INSENSITIVE</regexFlag>
<regexFlag>MULTILINE</regexFlag>
</regexFlags>
<replacements>
<!-- This replacement imports spec json variable into static html page. -->
<replacement>
<token><script></token>
<value><script src="./openapi.json"> </script><script></value>
</replacement>
<!-- This part replaces url input variable with spec variable. -->
<replacement>
<token>url:\s"https:\/\/petstore\.swagger\.io\/v2\/swagger\.json"</token>
<value>spec: spec</value>
</replacement>
<replacement>
<!-- This replacement initializes spec variable, that will be passed to swagger ui index.html. -->
<token>^\{</token>
<value>spec = {</value>
</replacement>
</replacements>
</configuration>
</plugin>
所以在構建后的這一步，我們得到了帶有 swagger ui 的靜態資源。最后要做的就是用 Spring 來服務它。
@Configuration
@EnableWebMvc
public class SwaggerConfiguration implements WebMvcConfigurer {
@Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
registry.addResourceHandler("/**")
.addResourceLocations("classpath:/META-INF/resources/webjars/swagger-ui/3.22.0/");
}
//this method was introduced just for convenient swagger ui access. Without it swagger ui can be accessed with /index.html GET call
@Override
public void addViewControllers(ViewControllerRegistry registry) {
registry.addViewController("/swagger-ui.html").setViewName("forward:/index.html");
}
}
所以就是這樣。如果您評論此答案并指出如何簡化此過程，那就太好了。
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