RESTful web service is not starting in Eclipse IDE

RESTful web service is not starting in Eclipse IDE - java

I have written a simple RESTful web service in JAVA but when I run it using Run As --> Run on server option from Eclipse IDE, I am getting HTTP Status 404 – Not Found error. Please tell what can be the issue in my code?
Response:
web.xml
<?xml version="1.0" encoding="UTF-8"?> <!--?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>RESTfulWebServiceExample</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>org.arpit.javapostsforlearning.webservice</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
FeetToInchAndInchToFeetConversionService.java
package org.arpit.javapostsforlearning.webservice;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("ConversionService")
public class FeetToInchAndInchToFeetConversionService {
#GET
#Path("/InchToFeet/{i}")
#Produces(MediaType.TEXT_XML)
public String convertInchToFeet(#PathParam("i") int i) {
int inch=i;
double feet = 0;
feet =(double) inch/12;
return ""
+ "" + inch + ""
+ "" + feet + ""
+ "";
}
#Path("/FeetToInch/{f}")
#GET
#Produces(MediaType.TEXT_XML)
public String convertFeetToInch(#PathParam("f") int f) {
int inch=0;
int feet = f;
inch = 12*feet;
return ""
+ "" + feet + ""
+ "" + inch + ""
+ "";
}
}

You're trying to go to localhost:8080/RESTfulWebServiceExample when, according to your web.xml, RESTfulWebServiceExample is your display name. You need to go to localhost:8080/rest/ to hit your servlet context root (as defined in your servlet-mapping section of the web.xml).

Related

What would be my Configuring File (Web.xml)?

I need to configure This file i tried to run this file but error was file not found : 404. so what would be the configure file I need to configure This file i tried to run this file but error was file not found : 404. so what would be the configure file
LoginService.java
package com.webService;
import java.util.ArrayList;
import javax.ws.rs.Consumes;
import javax.ws.rs.FormParam;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.core.MediaType;
import com.pojo.UserVO;
import com.model.SecurityManager;;
#Path("/WebService")
public class LoginService
{
#POST
#Path("/login")
#Consumes(MediaType.APPLICATION_FORM_URLENCODED)
public String login(#FormParam("username") String username, #FormParam("password") String password)
{
return getAllUsersList(username, password);
}
public String getAllUsersList(String username, String password)
{
String userListData=null;
ArrayList<UserVO> userList = null;
SecurityManager sm = new SecurityManager();
userList =sm.getAllUsersList();
for(UserVO userVO: userList)
{
if(userVO.getUsername().equals(username))
{
if(userVO.getPassword().equals(password))
{
return "Logged In User :" + username;
}
}
}
return "You Are not a Valid User";
}
}
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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>FirstProject</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>ServletAdaptor</servlet-name>
<servlet-class>com.sun.jersey.server.impl.container.servlet.ServletAdaptor</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>REST</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>

How to add parameters in rest api for specific response in code?

I have created my very first basic rest api. Its working find and give me output in xml using this url
http://localhost:8080/Webservice/Test/RestClient
Now, I'm thinking If I want my api to ask server only specific data then I how can I use parameters for the same in api url.
Here is my little updated code
package RestClient;
//import javax.servlet.annotation.WebServlet;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.xml.bind.annotation.XmlRootElement;
#Path("/RestClient")
#XmlRootElement
public class Restwebclient {
#GET
#Produces(MediaType.TEXT_XML)
public String sayXMLHello(#PathParam("x") int x, #PathParam("y") int y) {
String output = "";
if (x == 1 && y == 1) {
output = "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
}
else if (x == 2 && y == 2) {
output = "<?xml version=\"1.0\"?>" + "<hello> Hello Jersey" + "</hello>";
}
return output;
}
}
And web.xml is here
<?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" version="3.1">
<display-name>Webservice</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>JAVA WS</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>RestClient</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>JAVA WS</servlet-name>
<url-pattern>/Test/*</url-pattern>
</servlet-mapping>
</web-app>
But when run the url passing parameters like this nothing comes up in output
http://localhost:8080/Webservice/Test/RestClient?x=1&y=1
Note: Expected output from above parameters in url should be first if condition
May I know What I'm missing to achieve this ?

You are sending parameters in query so you should use #QueryParam and not #PathParam. change method to:
public String sayXMLHello(#QueryParam("x") int x, #QueryParam("y") int y) {
Query Params:
http://localhost:8080/Webservice/Test/RestClient?x=1&y=1
Path Params need to change Path:
#Path("{x}/{y}/RestClient")
http://localhost:8080/Webservice/Test/1/1/RestClient

JAVA JSON Restfull WebService No 'Access-Control-Allow-Origin' header is present on the requested resource [duplicate]

This question already has answers here:
How to handle CORS using JAX-RS with Jersey
(5 answers)
Closed 6 years ago.
I have a JAVA RESTful webservice which will return JSON string and it was written in Java. My problem is when I send request to that webservice with below URL
http://localhost:8080/WebServiceXYZ/Users/insert
it's giving me the below error message
XMLHttpRequest cannot load http://localhost:8080/WebServiceXYZ/Users/insert/. No 'Access-Control-Allow-Origin' header is present on the requested resource. Origin 'http://localhost:8100' is therefore not allowed access. The response had HTTP status code 500.
Here is my Code
package com.lb.jersey;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import org.json.JSONException;
import org.json.JSONObject;
#Path("/Users")
public class RegistrationService
{
#POST
#Produces(MediaType.APPLICATION_JSON)
#Path("/insert")
public String InsertCredentials (String json) throws JSONException
{
java.util.Date dt = new java.util.Date();
java.text.SimpleDateFormat sdf = new java.text.SimpleDateFormat("yyyy-MM-dd HH:mm:ss");
String currentTime = sdf.format(dt);
String phone = null;
JSONObject returnJson = new JSONObject();
try
{
JSONObject obj = new JSONObject(json);
JSONObject result1 = obj.getJSONObject("Credentials");
phone = result1.getString("phone");
DBConnection conn = new DBConnection();
int checkUserID = conn.GetUserIDByPhone(phone);
if(checkUserID <= 0)
{
DBConnection.InsertorUpdateUsers(phone, currentTime);
}
int userID = conn.GetUserIDByPhone(phone);
int otp = (int) Math.round(Math.random()*1000);
DBConnection.InsertorUpdateCredentials(userID, otp, currentTime);
JSONObject createObj = new JSONObject();
createObj.put("phone", phone);
createObj.put("otp", otp);
createObj.put("reqDateTime", currentTime);
returnJson.put("Credentials", createObj);
System.out.println(returnJson);
}
catch (Exception e)
{
}
return returnJson.toString();
}
}
My web.xml code
<?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" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>WebServiceXYZ</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>com.sun.jersey.server.impl.container.servlet.ServletAdaptor</servlet-class>
<init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.lb.jersey</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
I already read many articles but no progress, so please let me know, How can I handle this issue?

this is assuming you are using jetty as your server. hope it helps...
add the following code to your web.xml
<filter>
<filter-name>cross-origin</filter-name>
<filter-class>org.eclipse.jetty.servlets.CrossOriginFilter</filter-class>
<init-param>
<param-name>allowedOrigins</param-name>
<param-value>*</param-value>
</init-param>
<init-param>
<param-name>allowedMethods</param-name>
<param-value>GET,POST,DELETE,PUT,HEAD</param-value>
</init-param>
<init-param>
<param-name>allowedHeaders</param-name>
<param-value>origin, content-type, accept</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>cross-origin</filter-name>
<url-pattern>/*</url-pattern>
</filter-mapping>
and the following dependency in your pom.xml
<dependency>
<groupId>org.eclipse.jetty</groupId>
<artifactId>jetty-servlets</artifactId>
<version>8.0.0.M0</version>
</dependency>
the link to configure tomcat is link... here is another link2 modify accordingly :)

You can try setting the header for the HttpServletResponse
import javax.servlet.http.HttpServletResponse;
#Path("/Users")
public class RegistrationService
{
#Context
private HttpServletResponse servletResponse;
private void allowCrossDomainAccess() {
if (servletResponse != null){
servletResponse.setHeader("Access-Control-Allow-Origin", "*");
}
}
#POST
#Produces(MediaType.APPLICATION_JSON)
#Path("/insert")
public String InsertCredentials (String json) throws JSONException
{
allowCrossDomainAccess();
// your code here
}
}

Error on a JERSEY REST project (and tomcat)

I met a surious problem when I want to develop an API to call an "exterior" method (this method is called "createClientNode" which is a simple request for neo4j).
Anyways, this is my 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_2_5.xsd" id="WebApp_ID" version="2.5">
<display-name>API</display-name>
<servlet>
<servlet-name>jersey-serlvet</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>API</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>jersey-serlvet</servlet-name>
<url-pattern>/API/*</url-pattern>
</servlet-mapping>
</web-app>
And this my class :
package API;
import javax.ws.rs.DefaultValue;
import javax.ws.rs.GET;
import javax.ws.rs.POST;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.QueryParam;
import javax.ws.rs.core.MediaType;
import org.neo4j.graphdb.Node;
import function.query.weceipt.GraphManagement;
#Path("/API")
public class MyService {
#POST
#Path("/node")
#Produces({ MediaType.APPLICATION_JSON })
public Node createNode(#DefaultValue("default_id") #QueryParam("ID") String id) {
GraphManagement gm = new GraphManagement();
Node node = gm.createClientNode(id);
return node;
}
#GET
#Path("/node/{ID}")
#Produces({ MediaType.APPLICATION_JSON })
public Node getNode(#PathParam("ID") String id) {
GraphManagement gm = new GraphManagement();
Node node = gm.getClientNode(id);
return node;
}
#Path("/test")
#GET
#Produces(MediaType.TEXT_PLAIN)
public void afficher() {
System.out.println("test");
}
}
the error that I get is "the Servlet jersey-serlvet is not available".
I have the impression that I have-perhaps- a problem with a specific JAR or Tomcat, but I coudn't find the cause of this problem. So, can anyone help me please because I tried a loooong time with this!

Not able Publish RestFul webservices Using Jersey

I am trying to build simple webservices using Java, Rest Webservices(Jersey), Glass fish. project is successfully deploying in server without any errors, but when i access resources in browser it is giving 404-Not Found Error.
My resources class:
package org.student.services;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.PathParam;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/ConversionService")
public class FeetToInchAndInchToFeetConversionService {
#GET
#Path("/InchToFeet/{i}")
#Produces(MediaType.TEXT_XML)
public String convertInchToFeet(#PathParam("i") int i) {
int inch=i;
double feet = 0;
feet =(double) inch/12;
return "<InchToFeetService>"
+ "<Inch>" + inch + "</Inch>"
+ "<Feet>" + feet + "</Feet>"
+ "</InchToFeetService>";
}
#Path("/FeetToInch/{f}")
#GET
#Produces(MediaType.TEXT_XML)
public String convertFeetToInch(#PathParam("f") int f) {
int inch=0;
int feet = f;
inch = 12*feet;
return "<FeetToInchService>"
+ "<Feet>" + feet + "</Feet>"
+ "<Inch>" + inch + "</Inch>"
+ "</FeetToInchService>";
}
}
And Web.xml is:
<?xml version="1.0" encoding="UTF-8"?>
<web-app 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_3_0.xsd"
version="3.0">
<display-name>StudentPortal</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>org.student.services</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
using following Url with https//
mylocalhost:port/StudentPortal/rest/ConversionService/InchToFeet/2
Please Someone point what i am doing wrong

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RESTful web service is not starting in Eclipse IDE - java

You're trying to go to localhost:8080/RESTfulWebServiceExample when, according to your web.xml, RESTfulWebServiceExample is your display name. You need to go to localhost:8080/rest/ to hit your servlet context root (as defined in your servlet-mapping section of the web.xml).

Related

What would be my Configuring File (Web.xml)?

How to add parameters in rest api for specific response in code?

JAVA JSON Restfull WebService No 'Access-Control-Allow-Origin' header is present on the requested resource [duplicate]

Error on a JERSEY REST project (and tomcat)

Not able Publish RestFul webservices Using Jersey

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