Java jersey REST project not giving response

Java jersey REST project not giving response - java

I am developing REST API. I am using Jersey framework. I set up my project correctly. But when I hit URLs to the browser, no response is coming from server. Also nothing is logged in the server console. No warning no error.
My project hierarchy is
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_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>com.sun.rest</display-name>
<welcome-file-list>
<welcome-file>readme.html</welcome-file>
<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>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>com.sun.rest</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
</web-app>
Status.java
package com.sun.rest;
import javax.ws.rs.*;
import javax.ws.rs.core.MediaType;
#Path("/v1/status/*")
public class Status {
private static final String api_version = "00.01.00"; //version of the api
#GET
#Produces(MediaType.TEXT_HTML)
public String returnTitle() {
return "<p>Java Web Service</p>";
}
#Path("/version")
#GET
#Produces(MediaType.TEXT_HTML)
public String returnVersion() {
return "<p>Version:</p>" + api_version;
}
}
Response

You don't need /*/ in your path. If you remove that from your path, that would resolve your issue.

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Angular 4 and Java BackEnd Project in a single WAR

I created a angular 4 project and using JAVA REST API Project as a BackEnd. I need that I can move my angular 4 project as a staic content inside JAVA war file.
I created the production build for angular project and copy the files from dist folder and paste it inside webcontent of java project. I am able to read the index file of angular project but other files are giving 404.
Please help how can I move my angular project inside my java project to get a single war file for production.
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>userInformation</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>Jersey RESTful Application</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>com.newgen.ap2</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.spi.container.ContainerResponseFilters</param-name>
<param-value>com.newgen.ap2.CrossOrigin</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>Jersey RESTful Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>

Change your web.xml to the following:
<?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>userInformation</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>Jersey RESTful Application</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>com.newgen.ap2</param-value>
</init-param>
<init-param>
<param-name>com.sun.jersey.spi.container.ContainerResponseFilters</param-name>
<param-value>com.newgen.ap2.CrossOrigin</param-value>
</init-param>
</servlet>
<servlet>
<servlet-name>default</servlet-name>
<servlet-class>org.apache.catalina.servlets.DefaultServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Jersey RESTful Application</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>default</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
The idea is to use a different Servlet to serve your static resources.
Here are some more SO posts that might be helpful:
Servlet filters' order of execution
How are Servlet url mappings in web.xml used?
What happens if I have two servlet mappings in web.xml that match a request?

Java: REST service URL mapping

I am trying to create a very simple REST web service using Java, but I can't get the mapping right...
This is my service:
package rest;
#Path("/square/{num}")
public class SquareNumberRest {
#GET
#Produces(MediaType.APPLICATION_JSON)
public String squareNum(#PathParam("num") String num) {
int n = Integer.parseInt(num);
int squared = n * n;
JSONObject jo = new JSONObject(squared);
return jo.toString();
}
}
My web.xml descriptor:
<?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>TestProject</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>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>rest</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 try to consume the service on the following URL:
http://localhost:8080/MyProjectName/square/3
but I'm getting a 404 error.

Did you define the URL context for the project? It's possible that the URL root for the project is right on localhost:8080, which means you'll need to do
http://localhost:8080/square/3 instead.

HTTP Status 404 error in Rest WS concept

I am just Rest API concept with hello world program. I'm following some tutorial videos and trying the same program But I'm not getting the expected result.
Here is my Book.java
package com.book;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/book") //URI
public class Book {
#GET
#Produces(MediaType.TEXT_XML)
public String sayHelloXML() {
String response = "<?xml version='1.0'?>" + "<hello>Hello World</hello>";
return response;
}
}
and the web.xml
<?xml version="1.0" encoding="UTF-8"?>
<element>
<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">
<display-name>WSdemo</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>book</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>JAVA WS</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
</element>
And I'm using Tomcat v8.0 server.
But as I run the application, I am getting out put as
Can somebody give me key ideas for learning the concept of RESTfull web service?

There should be correct configuration in web.xml
I've edited my web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<!-- <element> -->
<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">
<display-name>WSdemo</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>com.book</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>JAVA WS</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
<!-- </element> -->
And I got the correct output as
Thank You.

HTTP method GET is not supported by this URL. (Java rest api with jersey)

im following one of the online tutorials for making rest api in java. I work in Eclipse 4.5.2. When i enter http://localhost:7001/ds2/api/v1/status i got mentioned in a title error. I included Jersey 1.19. Im using Oracle WebLogic Server 12c R2 (12.2.1). Starting page(index) works fine, but anything i would enter after /api/ gives me the same error.
My V1_status.java class:
package ds2.status;
import javax.ws.rs.*;
import javax.ws.rs.core.*;
#Path("/v1/status")
public class V1_status {
#GET
#Produces(MediaType.TEXT_HTML)
public String returnTitle() {
return "<p>Java Web Service</p>";
}
}
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="3.1">
<display-name>ds2</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>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>ds2</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/api/*</url-pattern>
</servlet-mapping>
</web-app>
weblogic.xml:
<?xml version="1.0" encoding="UTF-8"?>
<wls:weblogic-web-app xmlns:wls="http://xmlns.oracle.com/weblogic/weblogic-web-app" 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 http://xmlns.oracle.com/weblogic/weblogic-web-app http://xmlns.oracle.com/weblogic/weblogic-web-app/1.8/weblogic-web-app.xsd">
<wls:weblogic-version>12.2.1</wls:weblogic-version>
<wls:context-root>ds2</wls:context-root>
</wls:weblogic-web-app>
What can cause it or maybe how can i investigate by myslef whats wrong?

How to call rest with jax rs

I am new to jax rs web service. I was studying from this link- http://www.vogella.com/tutorials/REST/article.html
Now when I was trying to do my first rest service i faced some errors.
This is my service code
package de.vogella.jersey.first;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/hello")
public class Hello {
// This method is called if TEXT_PLAIN is request
#GET
#Produces(MediaType.TEXT_PLAIN)
public String sayPlainTextHello() {
return "Hello Jersey";
}
}
My web.xml is
<?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_3_0.xsd" id="WebApp_ID" version="3.0">
<display-name>de.vogella.jersey.first</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>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>de.vogella.jersey.first</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>
Now when I was running this code I faced HTTP Status 404 error. Please any one help me. I give all the jars from http://jersey.java.net/ Please help me.

If you are using jersey 2.x you Web.xml servlet should be as following
<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>de.vogella.jersey.first</param-value>
</init-param>
</servlet>
In jersey 2.x you need to refer "org.glassfish" and your resource URL would be
localhost:your_port/your_app_name/rest/hello

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Java jersey REST project not giving response - java

You don't need /*/ in your path. If you remove that from your path, that would resolve your issue.

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