本文整理了Java中org.restlet.routing.Router.getNext
方法的一些代码示例,展示了Router.getNext
的具体用法。这些代码示例主要来源于Github
/Stackoverflow
/Maven
等平台,是从一些精选项目中提取出来的代码,具有较强的参考意义,能在一定程度帮忙到你。Router.getNext
方法的具体详情如下:
包路径:org.restlet.routing.Router
类名称:Router
方法名:getNext
[英]Returns the next Restlet if available.
[中]返回下一个Restlet(如果可用)。
代码示例来源:origin: org.restlet.osgi/org.restlet
@Override
public Restlet getNext(Request request, Response response) {
Restlet result = super.getNext(request, response);
if (result == null) {
getLogger()
.warning(
"The protocol used by this request is not declared in the list of client connectors. ("
+ request.getResourceRef()
.getSchemeProtocol()
+ "). In case you are using an instance of the Component class, check its \"clients\" property.");
}
return result;
}
代码示例来源:origin: org.restlet.osgi/org.restlet
/**
* Handles a call by invoking the next Restlet if it is available.
*
* @param request
* The request to handle.
* @param response
* The response to update.
*/
@Override
public void handle(Request request, Response response) {
super.handle(request, response);
Restlet next = getNext(request, response);
if (next != null) {
doHandle(next, request, response);
} else {
response.setStatus(Status.CLIENT_ERROR_NOT_FOUND);
}
}
代码示例来源:origin: org.restlet.jee/org.restlet.ext.apispark
TemplateRoute templateRoute = (TemplateRoute) router.getNext(request,
response);
代码示例来源:origin: org.restlet.jee/org.restlet.ext.platform
TemplateRoute templateRoute = (TemplateRoute) router.getNext(request,
response);
代码示例来源:origin: org.restlet.jse/org.restlet.ext.platform
TemplateRoute templateRoute = (TemplateRoute) router.getNext(request,
response);
代码示例来源:origin: org.restlet.gae/org.restlet.ext.platform
TemplateRoute templateRoute = (TemplateRoute) router.getNext(request,
response);
代码示例来源:origin: org.restlet.osgi/org.restlet.ext.platform
TemplateRoute templateRoute = (TemplateRoute) router.getNext(request,
response);
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