This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 6 years ago.
I am new to servlets .I am using eclipse juno for this.I am having a trouble in running my program..My code is
package sTraining;
import java.io.*;
import javax.servlet.*;
public class Servlet1 implements Servlet{
ServletConfig config=null;
public void init(ServletConfig config){
this.config=config;
System.out.println("servlet is initialized");
}
public void service(ServletRequest req,ServletResponse res)
throws IOException,ServletException{
res.setContentType("text/html");
PrintWriter out=res.getWriter();
out.print("<html><body>");
out.print("<b>hello simple servlet</b>");
out.print("</body></html>");
}
public void destroy(){System.out.println("servlet is destroyed");}
public ServletConfig getServletConfig(){return config;}
public String getServletInfo(){return "copyright 2007-1010";}
}
I am getting this error[http://localhost:8080/Test/WEB-INF/classes/sTraining/Servlet1.java][1]
although i have this thing in my web .xml file
<servlet>
<description></description>
<display-name>Servlet1</display-name>
<servlet-name>Servlet1</servlet-name>
<servlet-class>servlet.Servlet1</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Servlet1</servlet-name>
<url-pattern>/Servlet1</url-pattern>
</servlet-mapping>
why this is not running? My code is fine. First time when I run this page it run, but running this program after my second program it did not run and that second program also not run.
Why are you accessing
http://localhost:8080/Test/WEB-INF/classes/sTraining/Servlet1.java
? You should be accessing
http://localhost:8080/Test/Servlet1
Read the above as
[protocol or scheme] :// [host] : [port] / [context] / [servlet mapping]
Also, according to the source code you've posted. The Servlet1 class is in package sTraining. Your web.xml should therefore have
<servlet>
<description></description>
<display-name>Servlet1</display-name>
<servlet-name>Servlet1</servlet-name>
<servlet-class>sTraining.Servlet1</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Servlet1</servlet-name>
<url-pattern>/Servlet1</url-pattern>
</servlet-mapping>
A Servlet Container will not make anything in the WEB-INF folder available to client requests.
What you are doing is not great practice. Your class should probably extend HttpServlet to get some standard HTTP behavior. You also shouldn't be writing HTML in Java code. Try reading the tutorial and references we have on Stackoverflow, here.
Put ./Servlet1 in your form action attribute
<form action="./Servlet1">
....
</form>
and check your web.xml
your package name is different
<servlet>
<description></description>
<display-name>Servlet1</display-name>
<servlet-name>Servlet1</servlet-name>
<servlet-class>sTraining.Servlet1</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Servlet1</servlet-name>
<url-pattern>/Servlet1</url-pattern>
</servlet-mapping>
Related
I have this class in my Spring Web model-view-controller (MVC) framework. The version of the Spring Web model-view-controller (MVC) framework is 3.2.8.
I have this web.xml file.
...
<servlet-mapping>
<servlet-name>ecolabelWeb</servlet-name>
<url-pattern>*.do</url-pattern>
<url-pattern>/newdesign/manage/manageapplications</url-pattern>
<url-pattern>/newdesign/manage/manageapplications/</url-pattern>
<url-pattern>/newdesign/manage/manageapplications/*</url-pattern>
<url-pattern>/newdesign/manage/home</url-pattern>
<url-pattern>/newdesign/manage/home/</url-pattern>
<url-pattern>/newdesign/manage/home/*</url-pattern>
<!-- Explicitly mention /welcome.do for usage as welcome page -->
<url-pattern>/welcome/welcome.do</url-pattern>
</servlet-mapping>
...
and this controller:
/**
*
*/
#RequestMapping(value = { "/newdesign/manage/home",
"/newdesign/manage/home/",
"/newdesign/manage/manageapplications",
"/newdesign/manage/manageapplications/"}, method = {RequestMethod.GET})
public String manageApplications (#ModelAttribute("aplicationListForm") final AplicationListForm aplicationListForm,
HttpServletRequest request,
Model model ) throws ExecutionException {
User sessionUser = (User)request.getSession().getAttribute(Const.SESSION_USER);
..
}
this URL works properly
/newdesign/manage/manageapplications
but with this one newdesign/manage/home
I got this error
WARNING: No mapping found for HTTP request with URI [/devices/newdesign/manage/home]
I am really getting crazy !
I also tried to put it in another method with the same result
#RequestMapping(value = { "/newdesign/manage/home",
"/newdesign/manage/home/"}, method = {RequestMethod.GET})
public String cbHome (Model model ) throws ExecutionException {
..
}
This URL is working
http://127.0.0.1:7001/devices/newdesign/manage/manageapplications
not this one (?!)
http://127.0.0.1:7001/devices/newdesign/manage/home
Your configuration doesn't work because in your web.xml you have limited your application to work only with these url:
<url-pattern>/newdesign/manage/manageapplications</url-pattern>
<url-pattern>/newdesign/manage/manageapplications/</url-pattern>
<url-pattern>/newdesign/manage/manageapplications/*</url-pattern>
This is a tipical web.xml:
<servlet>
<servlet-name>dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
With this configuration you are telling Spring to use the dispatcher servlet mechanism to handle the incoming requests and the view rendering. This way he can handle all the url you are defining in controllers, like /newdesign/manage/home.
You need to create the dispatcher-servlet.xml file where you will configure the dispatcher.
See this link for more detailed informations: http://www.mkyong.com/spring-mvc/spring-mvc-hello-world-example/
I have PersonController as below :
#Controller
#RequestMapping("person")
public class PersonController {
#RequestMapping(value= "/{personId}", method = RequestMethod.GET, produces={"application/json"})
public #ResponseBody Map<String, Object> getPerson(#PathVariable("personId") Integer personId) {
// code to get person
}
Tomcat starts up fine, I see this in the console :
Mapped "{[/person/{personId}],methods=[GET],params=[],headers=[] ,consumes=[],produces=[application/json],custom=[]}" onto public java.util.Map<java.lang.String, java.lang.Object> com.test.web.controller.PersonController.getPerson(java.lang.Integer)
But if I hit the url http://localhost:8080/sample/person/1 I get
HTTP Status 404 - /sample/person/1
In the web.xml
<servlet>
<servlet-name>app</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<!--init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/servlet-context.xml</param-value>
</init-param-->
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/sample/*</url-pattern>
</servlet-mapping>
I copy/pasted your PersonController class and it worked fine here.
So I did check your web.xml and your app servlet is mapping the pattern "/sample/*".
If I am corret, I suspect your project is called "sample" in Eclipse. In that case, you have to access your site as follows:
http://localhost:8080/sample/sample/person/1
The mapping in your web.xml will always start from your root context, and that is why you are getting 404 error.
If you want to access your controller from the root domain (in this case it is your actual Eclipse project name by default, but it can be configured too) you can use your servlet mapping as follows:
<servlet-mapping>
<servlet-name>app</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
I recommend that you use /rest/* or other mark since it will scale better for other types of content.
Let me know if it worked.
I have no compilation errors and my app launches fine on my testing server. However, I get an error when trying a GET request:
[1/2/14 10:23:13:248 EST] 00000022 RequestProces I org.apache.wink.server.internal.RequestProcessor logException The following error occurred during the invocation of the handlers chain: WebApplicationException (404 - Not Found) with message 'null' while processing GET request sent to http://localhost:9081/IDMWorkflowServices/resources/workflow
Here is my web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<display-name>IDMWorkflowServices</display-name>
<servlet>
<description>
JAX-RS Tools Generated - Do not modify</description>
<servlet-name>JAX-RS Servlet</servlet-name>
<servlet-class>com.ibm.websphere.jaxrs.server.IBMRestServlet</servlet-class>
<init-param>
<param-name>javax.ws.rs.core.Application</param-name>
<param-value>com.psg.itim.workflow.WorkflowResourceApplication</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>JAX-RS Servlet</servlet-name>
<url-pattern>
/resources/*</url-pattern>
</servlet-mapping>
</web-app>
Here is WorkflowResource:
import javax.ws.rs.GET;
import javax.ws.rs.Produces;
import javax.ws.rs.Path;
// The Java class will be hosted at the URI path "/workflow"
#Path("/workflow")
public class WorkflowResource {
#GET
#Produces("text/plain")
public String getClichedMessage() {
// Return some cliched textual content
return "Hello World";
}
}
Here is WorflowResourceApplication:
import javax.ws.rs.core.Application;
import java.util.HashSet;
import java.util.Set;
public class WorkflowResourceApplication extends Application{
#Override
public Set<Class<?>> getClasses() {
Set<Class<?>> classes = new HashSet<Class<?>>();
classes.add(WorkflowResource.class);
return classes;
}
}
If it's not painfully obvious, this is my first attempt using JAX-RS. I'm not exactly sure what I do or do not need from the above code to get this to work. It seems simple, but when I go to this url
http://localhost:9081/IDMWorkflowServices/resources/workflow
the 404 happens. Any ideas of what I am doing wrong?
Resolved! The only thing that was wrong was this line:
<param-name>javax.ws.rs.core.Application</param-name>
I changed it to:
<param-name>javax.ws.rs.Application</param-name>
I assumed it should have been the same Class I was calling from WorflowResourceApplication.java, but this was not the case. Everything works fine now. Apparently the application recognized the class error as a client side issue and registered a 404.
The first step to debug this would be to see if you have the correct port. So do this
- Try accessing just - http: //localhost:9081 . see if you get to default page or blank page or hello page if there is default index.jsp. If you get 404 then that means that you serever is running but your port number is incorrect.
If you are unsure what your port settings are(If I am correct then default port should be 9080) then follow this documentation - http://pic.dhe.ibm.com/infocenter/wasinfo/v7r0/index.jsp?topic=%2Fcom.ibm.websphere.migration.nd.doc%2Finfo%2Fae%2Fae%2Frmig_portnumber.html
This question already has answers here:
Getting server name during servlet initialization
(4 answers)
Closed 7 years ago.
How can I find the host and port that a Servlet is running on without using the HttpServletRequest.
I need to know this at time my servlet is being initialized i.e. in the Servlet#init method.
http://docstore.mik.ua/orelly/java-ent/servlet/ch04_01.htm#ch04-33108
A servlet uses the getInitParameter() method to get access to its init parameters:
public String ServletConfig.getInitParameter(String name)
you define the host init-param in your web.xml file:
<web-app>
<servlet>
<servlet-name>MyServletName</servlet-name>
<servlet-class>com.mycompany.MyServlet</servlet-class>
<init-param>
<param-name>host</param-name>
<param-value>myhost.mycompany.com</param-value>
</init-param>
</servlet>
</web-app>
and get it from within Servlet.init() like so:
public void init() throws ServletException {
getServletContext().log("init");
// Get the value of an initialization parameter
String value = getServletConfig().getInitParameter("host");
I'm looking at the jython servlet tutorial and have got everything working. How do I make the url be
localhost:8080/jythondemo/JythonServlet1
instead of
localhost:8080/jythondemo/JythonServlet1.py
http://seanmcgrath.blogspot.com/JythonWebAppTutorialPart1.html
Here is the relevant part from web.xml
<web-app>
<servlet>
<servlet-name>ServletTest</servlet-name>
<servlet-class>ServletTest</servlet-class>
</servlet>
<servlet>
<servlet-name>PyServlet</servlet-name>
<servlet-class>org.python.util.PyServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>ServletTest</servlet-name>
<url-pattern>/ServletTest</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>PyServlet</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
I've also tried with
<servlet-mapping>
<servlet-name>PyServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
It seems with the above changes pyservlet is getting the url JythonServlet1 but it does not know what to do with it. Here is the error message:
javax.servlet.ServletException: I can't guess the name of the class from /.metadata/.plugins/org.eclipse.wst.server.core/tmp0/wtpwebapps/testjython3/JythonServlet1
org.python.util.PyServlet.createInstance(PyServlet.java:202)
org.python.util.PyServlet.loadServlet(PyServlet.java:188)
org.python.util.PyServlet.getServlet(PyServlet.java:178)
org.python.util.PyServlet.service(PyServlet.java:155)
As noted by #Jigar there's a restriction in the actual Jython Servlet code. However, you may get around that issue by creating a simple URL translator. It consists in a Servlet that internally forwards the request to the Py Servlet.
Use the following code for web.xml:
<web-app>
<servlet>
<servlet-name>AliasServlet</servlet-name>
<servlet-class>juanal.AliasServlet</servlet-class>
</servlet>
<servlet>
<servlet-name>PyServlet</servlet-name>
<servlet-class>org.python.util.PyServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>AliasServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>PyServlet</servlet-name>
<url-pattern>*.py</url-pattern>
</servlet-mapping>
</web-app>
Create a new class, say, juanal.AliasServlet, with the following content:
package juanal;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class AliasServlet extends HttpServlet
{
#Override
protected void service(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
String URI = request.getRequestURI();
String newURI = URI + ".py";
getServletConfig().getServletContext().getRequestDispatcher(newURI).forward(request, response);
}
}
So, for a URL like this: localhost:8080/jythondemo/JythonServlet1, it will internally forward the request to JythonServlet1.py
Try this
<url-pattern>/*</url-pattern>
After that every request will be served by PyServlet
It looks like the problem is inherent to PyServlet.java. PyServlet uses regular expressions to find the name of the Python class to load based on the path of the request. The regular expression used is defined as follows (line 245 in my copy of the source):
private static final Pattern FIND_NAME = Pattern.compile("([^/]+)\\.py$");
Unfortunately, this regular expression will break if the input URL path doesn't contain a ".py" extension - this raises the error that you saw. From line 200:
Matcher m = FIND_NAME.matcher(file.getName());
if (!m.find()) {
throw new ServletException("I can't guess the name of the class from "
+ file.getAbsolutePath());
}
Since FIND_NAME is defined as private and final, I don't see a good way to override this behavior by just subclassing PyServlet - I think you'll have to make a copy of PyServlet and redefine this behavior in a fresh class.
I'm new to python. But if you have Apache, you could rewrite all .py URLs!
Just for the record: #Juanal AliasServlet trick works nicely.
I just had to use request.getServletPath() instead of getRequestURI().
(working with Eclipse)