Springfox

Question

现在 http://localhost:8080/myContextRoot/swagger-ui.html

提供 swagger

我可以配置 swagger 及其资源以在 http://localhost:8080/myContextRoot/swagger/swagger-ui.html

下提供服务吗

感谢您的帮助！

Answer 1

我采用了这种方法：

依赖性：

implementation 'org.tuckey:urlrewritefilter:4.0.4'

UrlRewrite Class:

@Component
public class SwaggerUrlRewriteFilter extends UrlRewriteFilter {
  private static final String CONFIG_LOCATION = "/urlrewrite.xml";

  @Value("classpath:/urlrewrite.xml")
  private Resource resource;

  @Override
  protected void loadUrlRewriter(FilterConfig filterConfig) throws ServletException {
    try {
      //Create a UrlRewrite Conf object with the injected resource
      Conf conf = new Conf(filterConfig.getServletContext(), resource.getInputStream(), resource.getFilename(), "@@yourOwnSystemId@@");
      checkConf(conf);
    } catch (IOException ex) {
      throw new ServletException("Unable to load URL rewrite configuration file from " + CONFIG_LOCATION, ex);
    }
  }
}

并在资源文件夹中 urlrewrite.xml

<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE urlrewrite PUBLIC "-//tuckey.org//DTD UrlRewrite 4.0//EN"
        "http://www.tuckey.org/res/dtds/urlrewrite4.0.dtd">
<urlrewrite>
    <rule>
        <from>/swagger/swagger-ui.html</from>
        <to type="passthrough">/swagger-ui.html</to>
    </rule>

    <rule>
        <from>/swagger/webjars/(.*)</from>
        <to type="passthrough">/webjars/</to>
    </rule>

    <rule>
        <from>/swagger/v2/api-docs</from>
        <to type="passthrough">/v2/api-docs</to>
    </rule>

    <rule>
        <from>/swagger/configuration/(.*)</from>
        <to type="passthrough">/configuration/</to>
    </rule>

    <rule>
        <from>/swagger/swagger-resources</from>
        <to type="passthrough">/swagger-resources</to>
    </rule>
</urlrewrite>

之后的招摇之路是：

http://localhost:8080/myContextRoot/swagger/swagger-ui.html

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