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 java, I just tried to read initialization parameters from Deployment Descriptor file (web.xml), But got above error?
My web.xml and java file coding coding in snap attached.
My directrory structure is
c:\....tomcat\webapps\dd\web-inf\classes
No error in java class file.
Java file code which is compiled successfully
import java.io.*;
import java.net.*;
import javax.servlet.*;
import javax.servlet.http.*;
public class MyServlet2 extends HttpServlet {
String fileName;
public void init(ServletConfig config) throws ServletException {
super.init(config);
fileName = config.getInitParameter("logfilename");
}
protected void doGet(HttpServletRequest request,HttpServletResponse response)throws ServletException, IOException {
processRequest(request, response);
}
protected void doPost(HttpServletRequest request,HttpServletResponse response)throws ServletException, IOException {
processRequest(request, response);
}
protected void processRequest(HttpServletRequest request,HttpServletResponse
response)throws ServletException, IOException {
PrintWriter out = response.getWriter();
out.println(fileName);
out.close();
}
}
web.xml
<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app>
<servlet>
<servlet-name>MyServlet2</servlet-name>
<servlet-class>MyServlet2</servlet-class>
<init-param>
<param-name>logfilename</param-name>
<param-value>value1</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>MyServlet2</servlet-name>
<url-pattern>/mc11</url-pattern>
</servlet-mapping>
</web-app>
Other detail of my directory and error page i think that my web.xml not working
There are two problems I can see at the moment...
Servlet init parameters
You currently have:
//defining param1
param1
value1
That's not how you define the parameter. You should specify a param-name element containing the name of the parameter, rather than using the name of the parameter as an XML element name.
<init-param>
<param-name>logfilename</param-name>
<param-value>...</param-value>
</init-param>
Also note that // isn't how you write comments in XML - if you wanted a comment, you should have:
<!-- Define the first parameter -->
<init-param>
<param-name>logfilename</param-name>
<param-value>...</param-value>
</init-param>
(The param-value element should have been a hint - if you could really just specify your own element, I'd have expected <logfilename>value in here</logfilename> - having the name specified as an element name, but the value specified with a fixed element name of param-value would have been an odd scheme.)
Servlet mapping
Currently your mapping is:
<servlet-name> FormServlet</servlet-name>
<url-pattern>/ss/</url-pattern>
</servlet-mapping>
I suspect that mapping won't match http://localhost:8080/dd/ss/s.html because you don't have any wildcards in there - it you may well find that it matches exactly http://localhost:8080/dd/ss/. It's not clear where the dd part comes in, but I assume that's a separate part of your configuration. You should try:
<!-- I would recommend removing the space from the servlet
- name *everywhere*. -->
<servlet-name>FormServlet</servlet-name>
<url-pattern>/ss/*</url-pattern>
</servlet-mapping>
If that doesn't work for http://localhost:8080/dd/ss/s.html, see whether it maps http://localhost:8080/ss/s.html - it may be that your engine isn't configured the way you expect elsewhere.
There is no problem with code,I try net beans tool and done assignment perfectly with the help of above code.Before that may be problem in tomcat.
Related
I have a simple application to test the communication between html and jsp. My jsp is located in
WEB-INF/test.jsp
Here is the structure of my files:
ProjectA
src
irstServlet.java
Web-Content
test1.html
WEB-INF
test.jsp
Here is the code from servlet
protected void doPost(HttpServletRequest request, response) throws ServletException, IOException {
RequestDispatcher dispatcher = getServletContext().getRequestDispatcher("/WEB-INF/test.jsp");
request.setAttribute("userName", request.getParameter("userName"););
dispatcher.forward(request, response);
}
First I have deploy in tomcat start my test1.html: It take me to the servlet: FirstServlet.java and I can enter userName there.
But after i enter the values in and press enter I expect it to forward me to test.jsp which is not working. I get the error:
The origin server did not find a current representation for the target resource or is not willing to disclose that one exists.
Edited:
In my html I am trying to use it like:
<form method="POST" name="XX" action="/HelloWorldServlet">
Still not working.
Please can someone help me?
Your code does not look like it would compile at all.
Parameter response has no type - should be HttpServletResponse
There is a semicolon (;) after request.getParameter("userName")
Also I'm not sure why you're getting RequestDispatcher from servlet context rather than from the request - then again I've never checked if it makes any difference.
Anyway, I would rewrite doPost method like this:
#Override
protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
req.setAttribute("userName", request.getParameter("userName"));
req.getRequestDispatcher("/WEB-INF/test.jsp").forward(req, resp);
}
EDIT:
I'm assuimng you have either a correct servlet mapping in your web.xml:
<servlet>
<servlet-name>HelloWorldServlet</servlet-name>
<servlet-class>FirstServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorldServlet</servlet-name>
<url-pattern>/HelloWorldServlet</url-pattern>
</servlet-mapping>
or your servlet is annotated with #WebServlet annotation:
#WebServlet("/HelloWorldServlet")
public class FirstServlet extends HttpServlet {
//your code
}
. If neither of those is true, that's your problem right there.
This question already has an answer here:
Change root context for a servlet in an IntelliJ IDEA project
(1 answer)
Closed 4 years ago.
I'm trying to run my simple servlet "Hello".
I have installed tomcat 9.0.6
Than i create a new JavaEE web project in Idea (called test)... after i create new package in src (called servlet), after new servlet file "myServlet".
Project Structure
Here is code:
public class myServlet extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("<html><body>");
out.println("<h2>Hello my first Servlet</h2>");
out.println("<br/>");
out.println("Time on the server is: " + new java.util.Date());
out.println("</body></html>");
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
doGet(request, response);
}
Than i open web.xml and add next lines for servlet and servlet-mapping:
<?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_4_0.xsd"
version="4.0">
<servlet>
<servlet-name>myServlet</servlet-name>
<servlet-class>servlet.myServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>myServlet</servlet-name>
<url-pattern>/myServlet</url-pattern>
</servlet-mapping>
</web-app>
After successfully running tomcat ("Artifact is deployed successfully") i try to visit url: http://localhost:8080/test/myServlet Result was page 404... But if i change url to: http://localhost:8080/myServlet result was correct.
Result
What's wrong with url: http://localhost:8080/test/myServlet ???
Need some libraries or what ???
Where is my mistake ???
Also i try with #WebServlet("/myServlet") annotation without servlet and servlet-mapping lines in web.xml - result the same.
You need to define the context path in context.xml - please see this link
http://tomcat.apache.org/tomcat-9.0-doc/config/context.html#Defining_a_context
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Hi all I am new in servlet and jsp I am not clear with servletconfig interface and servletcontext interface I started jsp again I encounter the term PageContext.
So any body explain me these term with the nice example.
servletconfig interface and servletcontext interface and PageContext in jsp
ServletConfig
ServletConfig object is created by web container for each servlet to pass information to a servlet during initialization.This object can be used to get configuration information from web.xml file.
when to use :
if any specific content is modified from time to time.
you can manage the Web application easily without modifing servlet through editing the value in web.xml
Your web.xml look like :
<web-app>
<servlet>
......
<init-param>
<!--here we specify the parameter name and value -->
<param-name>paramName</param-name>
<param-value>paramValue</param-value>
</init-param>
......
</servlet>
</web-app>
This way you can get value in servlet :
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
//getting paramValue
ServletConfig config=getServletConfig();
String driver=config.getInitParameter("paramName");
}
ServletContext
web container create one ServletContext object per web Application. This object is used to get information from web.xml
when to use :
If you want to share information to all sevlet, it a better way to make it available for all servlet.
web.xml look like :
<web-app>
......
<context-param>
<param-name>paramName</param-name>
<param-value>paramValue</param-value>
</context-param>
......
</web-app>
This way you can get value in servlet :
public void doGet(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException
{
//creating ServletContext object
ServletContext context=getServletContext();
//Getting the value of the initialization parameter and printing it
String paramName=context.getInitParameter("paramName");
}
PageContext
Is class in jsp, its implicit object pageContext is used to set , get or remove attribute from following scope:
1.page
2.request
3.session
4.application
ServletConfig is implemented by GenericServlet (which is a superclass of HttpServlet). It allows the application deployer to pass parameters to the servlet (in the web.xml global config file), and servlet to retrieve those parameters during its initialization.
For example, your web.xml could look like :
<servlet>
<servlet-name>MyServlet</servlet-name>
<servlet-class>com.company.(...).MyServlet</servlet-class>
<init-param>
<param-name>someParam</param-name>
<param-value>paramValue</param-value>
</init-param>
</servlet>
In you servlet, the "someParam" param can then be retrieved like this :
public class MyServlet extends GenericServlet {
protected String myParam = null;
public void init(ServletConfig config) throws ServletException {
String someParamValue = config.getInitParameter("someParam");
}
}
ServletContext is a bit different. It is quite badly named, and you'd better think of it as "Application scope".
It is an application-wide scope (think "map") that you can use to store data that is not specific to any user, but rather belongs to the application itself. It is commonly used to store reference data, like the application's configuration.
You can define servlet-context parameters in web.xml :
<context-param>
<param-name>admin-email</param-name>
<param-value>admin-email#company.com</param-value>
</context-param>
And retrieve them in your code like this in your servlet :
public void doGet(HttpServletRequest req, HttpServletResponse res) throws IOException {
String adminEmail = getServletContext().getInitParameter("admin-email"));
}
This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 6 years ago.
To be honest i am a learner and this is my first ever servlet program.
I made the basic servlet and intalled tomcat version 6 and even tomcat version 8.
the server starts up correctly and i am able to see the tomcat start up page on going to
http://localhost:8080
but after logging to tomcat manager when i click on my folder name it gives me an error saying
http status 404-/online/ (online is my folder created in webapps)
type Status report
message /online/
description The requested resource is not available.
here's my codes
web.xml-> (in folder online->WEB-INF)
- <web-app>
- <servlet>
<servlet-name>FirstServlet</servlet-name>
<servlet-class>FirstServlet</servlet-class>
</servlet>
- <servlet-mapping>
<servlet-name>FirstServlet</servlet-name>
<url-pattern>/FirstServlet</url-pattern>
</servlet-mapping>
</web-app>
FirstServlet.java->
import javax.servlet.*;
import java.io.*;
class FirstServelet implements Servlet
{
public void init(ServletConfig config)
{
}
public void service(ServletRequest request, ServletResponse response) throws ServletException, IOException
{
PrintWriter out;
out=response.getWriter();
out.println("hello");
out.println("<html>");
out.println("<head>");
out.println("<title>MY First Servlet</title>");
out.println("</head>");
out.println("<body>");
out.println("<marquee>ban ja tar pls</marquee>");
out.println("</body>");
out.println("</html>");
}
public String getServletInfo()
{
return null;
}
public ServletConfig getServletConfig()
{
return null;
}
public void destroy ()
{
}
}
please resolve the 404 error
The problem is you don't welcome-file-list, I think the default page is index.html which I suppose is not there in you folder. You can provide any html or jsp file as default file but NOT a servlet as below.
<welcome-file-list>
<welcome-file>myfile.html</welcome-file>
</welcome-file-list>
You can access your servlet by hitting http://localhost:8080/online/FirstServlet URL.
You can create a default page which will redirect to FirstServlet i.e.
myfile.html
<meta http-equiv="refresh" content="0; url=http://localhost:8080/online/FirstServlet" />
And also what #Braj said in the comment extend HttpServlet instead of implement Servlet.
Edit
You have a typo in servlet name. change the servlet name to FirstServlet from FirstServelet.
This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 1 year ago.
I am writing a Java Servlet, and I am struggling to get a simple HelloWorld example to work properly.
The HelloWorld.java class is:
package crunch;
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 {
PrintWriter out = response.getWriter();
out.println("Hello World");
}
}
I am running Tomcat v7.0, and have already read similar questions, with responses referring to changing the invoker servlet-mapping section in web.xml. This section actually doesn't exist in mine, and when I added it the same problem still occurred.
Try this (if the Java EE V6)
package crunch;
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
#WebServlet(name="hello",urlPatterns={"/hello"}) // added this line
public class HelloWorld extends HttpServlet {
public void doGet(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException {
PrintWriter out = response.getWriter();
out.println("Hello World");
}
}
now reach the servlet by http://127.0.0.1:8080/yourapp/hello
where 8080 is default Tomcat port, and yourapp is the context name of your applciation
You definitely need to map your servlet onto some URL. If you use Java EE 6 (that means at least Servlet API 3.0) then you can annotate your servlet like
#WebServlet(name="helloServlet", urlPatterns={"/hello"})
public class HelloWorld extends HttpServlet {
//rest of the class
Then you can just go to the localhost:8080/yourApp/hello and the value should be displayed. In case you can't use Servlet 3.0 API than you need to register this servlet into web.xml file like
<servlet>
<servlet-name>helloServlet</servlet-name>
<servlet-class>crunch.HelloWorld</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>helloServlet</servlet-name>
<url-pattern>/hello</url-pattern>
</servlet-mapping>
Writing Java servlets is easy if you use Java EE 7
#WebServlet("/hello-world")
public class HelloWorld extends HttpServlet {
#Override
public void doGet(HttpServletRequest request,
HttpServletResponse response) {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
out.println("Hello World");
out.flush();
}
}
Since servlet 3.0
The good news is the deployment descriptor is no longer required!
Read the tutorial for Java Servlets.
this is may be due to the thing that you have created your .jsp or the .html file in the WEB-INF instead of the WebContent folder.
Solution: Just replace the files that are there in the WEB-INF folder to the Webcontent folder and try executing the same - You will get the appropriate output
For those stuck with "The requested resource is not available" in Java EE 7 and dynamic web module 3.x, maybe this could help: the "Create Servlet" wizard in Eclipse (tested in Mars) doesn't create the #Path annotation for the servlet class, but I had to include it to access successfuly to the public methods exposed.
You have to user ../../projectName/Filename.jsp in your action attr. or href
../ = contains current folder simple(demo.project.filename.jsp)
Servlet can only be called with 1 slash forward to your project name..
My problem was in web.xml file. In one <servlet-mapping> there was an error inside <url-pattern>: I forgot to add / before url.