默认情况下,Togglz 管理控制台在应用程序端口上运行(由server.port
属性配置)。我想把它暴露在management.port
. 我的问题:有可能吗?
问问题
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1 回答
4
如果您使用 Togglz >= 2.4.0,则此功能开箱即用。
对于旧版本的解决方案如下:
我设法management.port
通过用MvcEndpoint
. 使用 Spring Cloud 模块为您完成所有工作的最简单方法(例如在 HystrixStreamEndpoint 中):
public class HystrixStreamEndpoint extends ServletWrappingEndpoint {
public HystrixStreamEndpoint() {
super(HystrixMetricsStreamServlet.class, "hystrixStream", "/hystrix.stream",
true, true);
}
}
不幸的是,由于它从请求 URI 中提取前缀的方式,TogglzConsoleServlet
还有一个与路径有关的 hack,所以整个解决方案看起来有点难看:
@Component
class TogglzConsoleEndpoint implements MvcEndpoint {
private static final String ADMIN_CONSOLE_URL = "/togglz-console";
private final TogglzConsoleServlet togglzConsoleServlet;
@Autowired
TogglzConsoleEndpoint(final ServletContext servletContext) throws ServletException {
this.togglzConsoleServlet = new TogglzConsoleServlet();
togglzConsoleServlet.init(new DelegatingServletConfig(servletContext));
}
@Override
public String getPath() {
return ADMIN_CONSOLE_URL;
}
@Override
public boolean isSensitive() {
return true;
}
@Override
public Class<? extends Endpoint> getEndpointType() {
return null;
}
@RequestMapping("**")
public ModelAndView handle(HttpServletRequest request, HttpServletResponse response) throws Exception {
HttpServletRequestWrapper requestWrapper = new HttpServletRequestWrapper(request) {
@Override
public String getServletPath() {
return ADMIN_CONSOLE_URL;
}
};
togglzConsoleServlet.service(requestWrapper, response);
return null;
}
private class DelegatingServletConfig implements ServletConfig {
private final ServletContext servletContext;
DelegatingServletConfig(final ServletContext servletContext) {
this.servletContext = servletContext;
}
@Override
public String getServletName() {
return TogglzConsoleEndpoint.this.togglzConsoleServlet.getServletName();
}
@Override
public ServletContext getServletContext() {
return servletContext;
}
@Override
public String getInitParameter(final String name) {
return servletContext.getInitParameter(name);
}
@Override
public Enumeration<String> getInitParameterNames() {
return servletContext.getInitParameterNames();
}
}
}
于 2016-11-29T14:56:40.303 回答