contextpath

问题 I'm using Spring Web Services to expose a service as a web service. In my spring configuration xml file, I have a bean which is an instance of DefaultWsdl11Definition. One of the properties that needs to be set is the locationUri . This needs to be a fully qualified Uri, but I don't want to have to change this value when the application gets promoted from dev to uat and to production. Spring knows what the web application context root is, so is there a variable that I can specify in my

问题 I'm using tomcat to deploy my applications, I would like to deploy as one single war to should server multiple context paths. Currently I have app1.war and app2.war, but both has same contents only name is different to access via http:///app1 and http:///app1.war. But I would like to do this, Deploy app.war and access it by both app1 and app2 context path. Is it possible to do in tomcat? Please help me out of this 回答1: This can be done in several ways. I think this way is the most

问题 I have Spring Boot application which I deploy to Weblogic container. After it is deployed, I can access it using http://localhost:7001/my-application/ When I use http://localhost:7001/my-application - without trailing slash at the end, I'd like to redirect it to http://localhost:7001/my-application/ I know that Tomcat already has such feature as there is a flag you can change: /** * Whether requests to the context root should be redirected by appending a / to * the path. */ private Boolean

问题 I've got a struts2 application running under a contextpath "/path" on my local tomcat without problems. When I deploy it on a webserver (using a proxy to redirect from "http://www.domain.com" to "myserver:8080/path/") Struts does all kinds of strange things. First, it includes the context in -tags. That can be turned off by an attribute. But sadly, it also includes the path in the action attributes of my forms, so a login form points to "http://www.domain.com/path/login.action" instead of

问题 I have a webapp that is working swimmingly in Tomcat 6. Let's say it's running on server:8282/MyApplication . I have a context xml that looks like this: <Context crossContext="true" debug="5" docBase="MyApplication" path="MyApplication" reloadable="true"> In my application, my links look like server:8282/MyApplication/myAction.do?params=blah When I switch it to Tomcat 7, it seems to change the link to: server8282/myAction.do?params=blah I haven't changed -anything- in the application, it's

问题 I created a web application using netbeans and glassfish server. I created a new java file inside that application. I want to find the current application path in that java file. 回答1: You can get path information from a servlet with methods of HttpServletRequest class: public class RequestInfoExample extends HttpServlet { public void doGet(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException { String requestURI = request.getRequestURI(); String

问题 I am trying to implement a "context" system similar to the one used by GitHub. For example, a Post may be created belonging either to the User or one of the Companies the User belongs to depending on whether to User is in the "User" context or a context that refers to one of the Companies. As a part of this, I'd like to be able to do routing based on the user's current context. For example, if the User is in their own context, /dashboard should route to users/show , but if they are in the

问题 I have been trying to parse the London Underground Linestatus XML "feed" - with little success. I would have expected this to ne "easy" using XPath, but I am getting empty nodes. I am fairly sure sure that I am not dealing with the uk namespace correctly. Here is my (rather simple code): import libxml2 from urllib2 import urlopen data = urlopen('http://cloud.tfl.gov.uk/TrackerNet/LineStatus').read() try: doc = libxml2.parseDoc(data) except (libxml2.parserError, TypeError): print "Problems

问题 I have a web application where the context name is different from my root name. For example my root Dir name is myapp and the context path is app . Then how to get the root name (myapp) from the Java web application? 回答1: If the application is not deployed with a war file you can use this code that will return the whole address where the application is running. //request variable is the Java HTTP servlet request object received from an action request.getSession().getServletContext()

问题 I've got a webapp in my Tomcat/webapps directory. My app's directory is "site" and is showing as http:localhost:8080/site I'd like to have the site show as http://localhost:8080/ I've read the docs and tried creating a ROOT.xml file in my Tomcat/conf/Catalina/localhost directory to create a ROOT context path but the connection isn't being made. Can someone nudge me in the right direction? Here's the content of that ROOT.xml file. <Context docBase="site" path="/"> </Context> 回答1: The easiest