I have Main class, there I doing api request and integrate with database.
I also create servlet where I want to get data from clients and put it in database.
it is the main method in Main class:
public static void main(String[] args) throws IOException {
serverSocket = new ServerSocket(8888); // Server port
System.out.println("Server started. Listening to the port 8888");
initProviderList();
initNewsAppDB();
Thread newFeedsUpdate = new Thread(new NewFeedsUpdate(providerList));
newFeedsUpdate.start();
}
it is the servlet:
#WebServlet(name = "GetClientTokenServlet")
public class GetClientTokenServlet extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String token = request.getParameter("token");
System.out.println(token);
}
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_4_0.xsd"
version="4.0">
<servlet>
<servlet-name>GetClientTokenServlet</servlet-name>
<servlet-class>GetClientTokenServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>GetClientTokenServlet</servlet-name>
<url-pattern>/GetClientTokenServlet</url-pattern>
</servlet-mapping>
</web-app>
how I can setup the GetClientTokenServlet (to be able listen to clients calls) into main method?
In most cases web applications in Java don't have main methods. The main method is implemented by a servlet container, such as Tomcat, and that's what you actually run. The servlet container discovers your application's classes and web.xml through some method, often by finding them in a WAR file that you have dropped in a directory defined by the servlet container, such as Tomcat's webapps directory. The servlet container then instantiates the servlets identified in your web.xml file.
That said, there are some web servers that you can instantiate as components within your own application. A server that is commonly used for this purpose is Jetty. Jetty is a web server that passes incoming requests to "handlers" that you define. You can have Jetty load your entire web application from your WAR file and instantiate the servlets defined in your web.xml, or you can use ServletHandler to register servlets manually; in this case you don't need web.xml.
Related
I created a new dynamic web project called TestWeb in Eclipse. I added a single index.html to the WebContent folder and created a single web servlet, making the servlet available at /Test.
I can access the index.html file at http://localhost:8080/TestWeb, however, I cannot access the servlet at http://localhost:8080/TestWeb/Test. Please assist.
Here is the code for the Servlet:
#WebServlet("/Test")
public class Test extends HttpServlet {
private static final long serialVersionUID = 1L;
public Test() {
super();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
response.getWriter().append("Served at: ").append(request.getContextPath());
}
}
Here is my web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/javaee http://xmlns.jcp.org/xml/ns/javaee/web-app_3_1.xsd" id="WebApp_ID" version="4.0">
<display-name>TestWeb</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
</web-app>
Firstly check your web.xml (deployment descriptor) file, make sure it has correct xml schema defined in it and if everything is good there then change you annotation like this
#WebServlet("/Test/*")
I soled the issue by renaming the project. My project name consisted of two words and one space. "Password Services". Once I renamed the project to "Server" I could reach the servlet.
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
I am new to java programming in the web environment and am having trouble understanding the flow.
For an assignment coming up I need to build a web application accessible with an API via get/post requests. For the tutorials that I have followed here is the flow I understand.
User visits top domain->
Per configuration user is directed to a jsp page->
Jsp contains javascrip and html. To access server code (for database, computations and other processes) the jsp page can use RCP to make async requests to a java servlet->
Java servlet does server handling and returns response to jsp page
Is this the required flow or can a user directly acess a servlet, and can that servlet handle get/post, or do I have to handle at the jsp and foward to the servlet?
Servlets can be accessed directly. You just need to extend HttpServlet and implement doGet and/or doPost. For example:
public class MyServlet extends HttpServlet {
#Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException {
Integer param = null;
try {
param = Integer.parseInt(req.getParameter("param"));
}
catch(NumberFormatException e) {
}
}
}
You also need to map your servlet to url in web.xml:
<servlet>
<servlet-name>MyServlet</servlet-name>
<servlet-class>com.adam.test.server.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>MyServlet</servlet-name>
<url-pattern>/my_servlet</url-pattern>
</servlet-mapping>
Now you can access your servlet using url like this:
http://domain.com/my_servlet?param=123
Not duplicate of this question
Here is my code
HelloWorld.java
// Import required java libraries
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
// Extend HttpServlet class
public class HelloWorld extends HttpServlet {
private String message;
public void init() throws ServletException
{
// Do required initialization
message = "Hello World";
}
public void doGet(HttpServletRequest request,
HttpServletResponse response)
throws ServletException, IOException
{
// Set response content type
response.setContentType("text/html");
// Actual logic goes here.
PrintWriter out = response.getWriter();
out.println("<h1>" + message + "</h1>");
}
public void destroy()
{
// do nothing.
}
}
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"
id="WebApp_ID" version="3.0">
<display-name>Servlet</display-name>
<servlet>
<servlet-name>HelloWorld</servlet-name>
<servlet-class>HelloWorld</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>HelloWorld</servlet-name>
<url-pattern>/HelloWorld</url-pattern>
</servlet-mapping>
</web-app>
there is no compilation error, however facing this issue.
Please help!!
I changed url to
http://localhost:8080/Servlet/HelloWorld
and now facing exception
Folder Structure
You should provide the class name with the package name.
However, I would recommend to try using #WebServlet(name = "HelloWorld", urlPatterns = "/helloWorld") above the servlet class.
so if owuld be :
#WebServlet(name = "Welcome", urlPatterns = "/testPage")
public class HelloWorld extends HttpServlet {
//blah blah
}
There are a couple of things to check :
- first that the application is deployed corectly on the server. You should have a .war project that is deployed. You should see a successful deployment on you server logs
- Build your url. This is composed of server IP, in your case local host, than port 8080 is usually the default, than context root (if not defined than it should be the name of the application) and last the servlet name: http://localhost:8080/appname/helloworld
There is also an possible error on your Web.xml. The servlet class tag you defined your class but has no package. Probably you have the class also in a package.
package. HelloWorld
Verify if there is another application working on the same port as Apache Tomcat.
Try to clean your Apache Tomcat server.
this is my first time that I work on a Java project that use HttpServlet.
So I know that an HttpServlet is a program that run on a Web Application server and act as a middle layer between a request coming from a Web browser or other HTTP client and databases or applications on the HTTP server. So the servlet extend the competence of my application server.
I have some doubt to understand how exactly work this servlet founded into my project, into web.xml file I found this configuration:
<?xml version="1.0" encoding="ISO-8859-1"?>
<!DOCTYPE web-app PUBLIC
"-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN"
"http://java.sun.com/dtd/web-app_2_3.dtd">
<web-app>
<display-name>My Project</display-name>
<listener>
<listener-class>it.sistinf.ediweb.quartz.QuartzListener</listener-class>
</listener>
<servlet>
<servlet-name>edimon</servlet-name>
<servlet-class>it.sistinf.ediweb.monitor.servlets.Monitoraggio</servlet-class>
<load-on-startup>0</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>edimon</servlet-name>
<url-pattern>/edimon.do/*</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>/logon.jsp</welcome-file>
</welcome-file-list>
<taglib>
<taglib-uri>displaytag</taglib-uri>
<taglib-location>/WEB-INF/displaytag-11.tld</taglib-location>
</taglib>
</web-app>
So reading some documentation it seem to understand that I have to tell the servlet container (or application server) what servlets to deploy, and what URL's to map the servlets to.
In the previous case I am configuring a servlet named edimon implemented by the Monitoraggio class.
Then it is mapped the servlet to a URL or URL pattern. In this case the edimon servlet is mapping with the /edimon.do/* URL pattern. So when it is called something that match with the previous pattern the edimon servlet is performed.
Then into my Monitoraggio class that implement the HttpServlet I found the service() method:
public void service(HttpServletRequest req, HttpServletResponse res) throws ServletException, IOException {
LoggerMDC.setup(req, res);
Logger logger = (Logger) Logger.getStdLogger(Monitoraggio.class); // do not declare 'logger' as static field in order to work with MDC
String service = req.getParameter("serv");
char serviceId = Utility.getServizio(req.getParameter("serv"));
if (checkSession(req, serviceId) == false) {
gotoPage(ConfigurationFactory.getPropertiesPages().getProperty("pagina_errore_session"), req, res);
return;
}
LoggerWatch loggerWatch = new LoggerWatch(Monitoraggio.class, Long.valueOf(System.getProperty(Constants.Keys.CONFIG_STATS_WARNING_THRESHOLD, String.valueOf(LoggerWatch.DEFAULT_WARNING_THRESHOLD))).longValue());
if (logger.isTraceEnabled())
logger.trace("lanciaServizio() | logger threshold: " + loggerWatch.getWarningThreshold());
loggerWatch.start();
loggerWatch.info(new StringBuffer("service() | servizio: [").append(service).append("] | service start").toString());
String paginaDaLanciare = lanciaServizio(serviceId, req, res);
String executionTime = loggerWatch.getInfoTime();
//Modifica per export
if (req.getSession().getAttribute("export") == null) {
gotoPage(paginaDaLanciare, req, res);
}
loggerWatch.info(new StringBuffer("service() | servizio: [").append(service).append("] | [").append(executionTime).append("] after forward to ").append(paginaDaLanciare).toString(), true);
loggerWatch.stop();
req.getSession().removeAttribute("export");
req.getSession().removeAttribute("stringaXML");
req.getSession().removeAttribute("downbyte");
return;
}
Reading on the documentation it receives standard HTTP requests from the public service method and dispatches them to the doXXX methods defined in this class
So what exatly do this method? I can't understand how the servlet load the JSP
The documentation you read describes what the service() method of HttpServlet does by default. Since your servlet overrides the service() method and provides a different implementation, it doesn't do that anymore. Instead, it does... what the code in the method does.
A servlet doesn't "load a JSP". I don't see how JSPs have any relation to the servlet code you posted. Maybe gotoPage() does tell the container to forward the request to a JSP. You should look at the documentation and/or code of that method to know what it does.