I have a simple Servlet that looks like this:
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class Bla extends HttpServlet {
private static final long serialVersionUID = 16252534;
#Override
public void init() throws ServletException {
System.out.println("init");
}
#Override
public void doGet(HttpServletRequest req, HttpServletResponse resp)
throws ServletException, IOException {
System.out.println("doGet");
resp.setContentType("text/html");
PrintWriter out = resp.getWriter();
out.println("<html><h1>It works!!</h1></html>");
}
#Override
public void service(ServletRequest req, ServletResponse resp) throws ServletException, IOException {
System.out.println("service");
}
#Override
public void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
System.out.println("doPost");
}
#Override
public void destroy() {
System.out.println("Destroy servlet");
}
}
and a web.xml that looks like this:
<?xml version="1.0" encoding="UTF-8" ?>
<web-app>
<display-name>Archetype Created Web Application</display-name>
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>Bla</servlet-name>
<servlet-class>instrurental_proj.servlets.Bla</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Bla</servlet-name>
<url-pattern>/bla</url-pattern>
</servlet-mapping>
</web-app>
When I visit the url http://localhost:8080/instrurental/bla, the following is printed out in the console:
init
service
but not doGet as I expect. Also, nothing is printed out in the browser! (I'm expecting it to say "It Works").
I've been struggling with this issue since yesterday. Does anyone have any suggestions, what could the problem be?
Why are u overriding the service method. There is no need of that. Remove it or else call
super.service(request,response);
REASON
Try to see the source of HttpServlet class. There you will see that depending on the method that is used to called the servlet i.e. GET/POST the necessary method doGet() or doPost() is called. And when the container actually receives a request it starts a new thread and and serves the client by calling service() method. So if you override it and don't call the super class' service method or define your own strategyhow GET will be called, doGet() method will never be called. Your request never calls doGet() method, its the service() method which calls it.
Related
We have an application developed in Spring MVC and deployed on a Jetty 9 Standalone (JDK 1.8) and served in principle without Apache / Nginx.
The application makes, internally, requests to obtain external xml and xsd files to make a series of validations.
We would need to be able to intercept the calls, especially for the response as we need to be able to modify the xmls and xsds (to remove a number of characters).
The challenge is that we can NOT modify the application. That is we cannot add something to the source code and recompile.
The first strategy has been to include a jar with a Filter / ServletFilter but only to log the requests, so that we could check if the GET calls that we suppose the controller makes to the xml / xsd appear.
Filter code
package com.htmlfilter;
import javax.servlet.Filter;
import javax.servlet.FilterChain;
import javax.servlet.FilterConfig;
import javax.servlet.ServletException;
import javax.servlet.ServletRequest;
import javax.servlet.ServletResponse;
import javax.servlet.http.HttpServletRequest;
public class HtmlFilter implements Filter {
#Override
public void init(FilterConfig filterConfig) throws ServletException {
System.out.println("HtmlFilter: init method");
}
#Override
public void doFilter(ServletRequest request, ServletResponse response, FilterChain chain)
throws IOException, ServletException {
System.out.println("HtmlFilter: doFilter method");
HttpServletRequest httpServletRequest = (HttpServletRequest) request;
String path = httpServletRequest.getRequestURI().substring(httpServletRequest.getContextPath().length());
System.out.println(path);
chain.doFilter(request, response);
}
#Override
public void destroy() {
System.out.println("HtmlFilter: destroy method");
}
}
ServletFilter code
package com.htmlfilter;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class HtmlServletFilter extends HttpServlet {
public void init() throws ServletException {
System.out.println("HtmlServletFilter: init");
}
public void doGet(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException {
System.out.println("HtmlServletFilter: doGet");
String path = request.getRequestURI();
System.out.println(path);
response.getWriter().println("Hello This is GET
Response.");
}
}
web.xml
<filter>
<filter-name>HtmlFilter</filter-name>
<filter-class>com.htmlfilter.HtmlFilter</filter-class>
</filter>
<filter-mapping>
<filter-name>HtmlFilter</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>HtmlServletFilter</servlet-name>
<servlet-class>com.htmlfilter.HtmlServletFilter</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>HtmlServletFilter</servlet-name>
<url-pattern>*.xsd</url-pattern>
</servlet-mapping>
PrintNamesServlet.java:
This servlet prints the entered name of the user
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
#WebServlet(name = "PrintNamesServlet")
public class PrintNamesServlet extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("Ram Dhakal");
}
}
CounterServlet.java:
Counts the number of hits or visit in the page
import javax.servlet.ServletException; import
javax.servlet.annotation.WebServlet; import
javax.servlet.http.HttpServlet; import
javax.servlet.http.HttpServletRequest; import
javax.servlet.http.HttpServletResponse; import java.io.IOException;
import java.io.PrintWriter;
#WebServlet(name = "CounterServlet") public class CounterServlet
extends HttpServlet {
int totalHits;
public void init() throws ServletException{
totalHits = 0;
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
PrintWriter out = response.getWriter();
out.print("Total visit count: " + totalHits++);
} public void destroy(){
}
}
web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns="http://xmlns.jcp.org/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd"
version="3.1">
<servlet>
<servlet-name>PrintNamesServlet</servlet-name>
<servlet-class>PrintNamesServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>PrintNamesServlet</servlet-name>
<url-pattern>/PrintNamesServlet</url-pattern>
</servlet-mapping>
<servlet>
<servlet-name>CounterServlet</servlet-name>
<servlet-class>CounterServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>CounterServlet</servlet-name>
<url-pattern>/CounterServlet</url-pattern>
</servlet-mapping>
</web-app>
I am getting error:
No webpage was found for the web address: http://localhost:8080/
As, I am trying to run the servlet for the first time, I am not getting what is wrong with my code. I typed http://localhost:8080/PrintNamesServlet in the url.
You have used Servlet 3.0 specification
Lets takes look at it
In Servlet 3.0, servlet metadata can be specified using #WebServlet
#WebServlet(name="mytest",
urlPatterns={"/myurl"})
public class TestServlet extends javax.servlet.http.HttpServlet {
....
}
In this way the servlet is accessed by using the url pattern specified in the annotation.
##WebServlet(name="mytest",urlPatterns={"/myurl"})
according to that servlet is acceseed using
http://localhost:8080/myurl
In you case you have only specified name you have to specify urlPatterns also, so you can able to call you servlet properly.
#WebServlet(name = "CounterServlet",urlPatterns={"/CounterServlet"}) public class CounterServlet extends HttpServlet {}
And you do not need to use web.xml file.
When you use annotations like this #WebServlet(name = "PrintNamesServlet") the web.xml mapping is not used.
You have to either remove these annotations or add urlMapping attribute to them.
I have the following Java servlet:
package com.controller;
import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.RequestDispatcher;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class Teste extends HttpServlet {
protected void processRequest(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html;charset=UTF-8");
request.setAttribute("teste", "Test");
request.getSession().setAttribute("teste", "Test Session");
RequestDispatcher rd = request.getRequestDispatcher("teste.jsp");
rd.forward(request, response);
}
#Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
}
#Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
processRequest(request, response);
}
#Override
public String getServletInfo() {
return "Short description";
}
}
And the following web.xml file:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.1" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd">
<servlet>
<servlet-name>Teste</servlet-name>
<servlet-class>com.controller.Teste</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Teste</servlet-name>
<url-pattern>/teste</url-pattern>
</servlet-mapping>
<session-config>
<session-timeout>
30
</session-timeout>
</session-config>
</web-app>
I can access the servlet through the url http://localhost:8084/myapp/teste
What I want is change the url-pattern to /teste/edit, but when I do that, and try to access the servlet through the url http://localhost:8084/myapp/teste/edit I get the following 404 error:
HTTP Status 404 - /TrabalhoPSW/teste/teste.jsp
type Status report
message /TrabalhoPSW/teste/teste.jsp
description The requested resource is not available. Apache
Tomcat/8.0.27
Why is this happening? How can I fix this?
The problem was the JSP file that the servlet was trying to find. when I changed from
request.getRequestDispatcher("teste.jsp");
to
request.getRequestDispatcher("/teste.jsp");
It worked normally.
if you need than your servlet works with /path and /path/to, Change the mapping of servlet in web.xml to:
<url-pattern>/teste/*</url-pattern>
This question already has answers here:
HTTP Status 405 - HTTP method is not supported by this URL
(2 answers)
Closed 6 years ago.
I began studying servlets.
Code Servlet:
package arver;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
/**
* Created by 35717 on 30.03.2016.
*/
public class MainServlet extends HttpServlet {
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
super.doGet(req, resp);
PrintWriter out = resp.getWriter();
out.print("servlet");
}
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
super.doPost(req, resp);
}
}
File web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<servlet>
<servlet-name>MainServlet</servlet-name>
<servlet-class>arver.MainServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>MainServlet</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
</web-app>
Server response:
HTTP Status 405 - HTTP method GET is not supported by this URL
type Status report
message HTTP method GET is not supported by this URL
description The specified HTTP method is not allowed for the requested resource.
Apache Tomcat/9.0.0.M4
why am I getting HTTP Status 405 - HTTP method GET is not supported by this URL error in this Program.
We extend the HttpServlet and #Override doPost but in our implementation we don't call its super since call to the super will give this message.
When you do super.doGet(request, response); in your Servlet's doGet() method, you actually call the doGet() of the HttpServlet class. So drop the super call. It's not needed.
Just remove these lines :
super.doGet(req, resp);
super.doPost(req, resp);
Use Either get or Post Method
MainServlet.java
package arver;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
/**
* Created by 35717 on 30.03.2016.
*/
public class MainServlet extends HttpServlet {
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException
{
PrintWriter out = resp.getWriter();
out.print("servlet");
}
}
web.xml file
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
version="3.0">
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>MainServlet</servlet-name>
<servlet-class>arver.MainServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>MainServlet</servlet-name>
<url-pattern>/MainServlet</url-pattern>
</servlet-mapping>
</web-app>
Url must be like this: http://localhost:8080/Project name/MainServlet
I am new in j2EE and now learning servlet.I display a simple "Hello World" string into the web page using servlet.
My code is :
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class HelloWorld extends HttpServlet {
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html");
PrintWriter pw = response.getWriter();
pw.println("<html>");
pw.println("<head><title>Hello World</title></title>");
pw.println("<body>");
pw.println("<h1>Hello World</h1>");
pw.println("</body></html>");
}
}
How can I give the name of this webpage like HelloWorld.html
It may helps you :
<servlet>
<servlet-name>HelloWorld</servlet-name>
<servlet-class>(with package name).HelloWorld</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorld</servlet-name>
<url-pattern>HelloWorld.html</url-pattern>
</servlet-mapping>