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?
Related
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.
can someone tell me how to setup my project structure and my web.xml?
In my dynamic web project in the java resources folder i have a servlet (class name: RestFulService) which delivers a JSON file.
The web pages are in the WebContent-folder, for example index.jsp.
The project is published with a glassfish server.
This is my current 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">
<welcome-file-list>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<servlet-name>RestFulService</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.javasrc, com.jersey.jaxb, com.fasterxml.jackson.jaxrs.json</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>RestFulService</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
The link to access the JSON works well (http://localhost:8080/BCode/liveHelp/getText) but if i want to access the index.jsp via "http://localhost:8080/BCode/index.jsp" all i get is an error 404 page.
Can anyone please tell me where the problem is an what i have to change to get it working?
I guess the problem is in the web.xml but I don't have any idea what to change.
This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 5 years ago.
The xml file is located in WebContent/WEB-INF/web.xml of my project. I am using Eclipse and running Tomcat (which is not installed via Eclipse. I prefer it to be a separate installation).
<?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>EmployeeManagement</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>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
It doesnt work when the form page submits to the servlet. I am getting a 404 error everytime. I have been encountering this problem for a while. Somebody please help me.
You are missing <servlet>...</servlet> tag which is important to mapping. So use 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>EmployeeManagement</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>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
and you should give action value on your form like following:
<form action="/EmployeeManagement/WebContent/Registration" method="post">
//Some code here
</form>
and also note it down all values are case sensitive on the following code:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
Your servlet name Registration should be same on both tags <servlet>...</servlet> and <servlet-mapping>...</servlet-mapping> and also package name should be same where your servlet class is located.
you have not mapped servlet name to servlet class, It should be like given below,
In <servlet-class> give the path of your servlet
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.Registration<servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
Check your form action.
Is the path there
/EmployeeManagement/WebContent/Registration
or
YOURAPPCONTEXT/EmployeeManagement/WebContent/Registration
or
YOURAPPNAME/EmployeeManagement/WebContent/Registration
You have specifyed a servlet-mapping and used the name Registration in servlet-name without defining it before.
You need to define the servlet before using it in a servlet mapping
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>[fully qualifyied name of your servlet]</servlet-class>
</servlet>
You are missing another part to define the servlet in the web.xml
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>
package.path.to.RegistrationServlet
</servlet-class>
</servlet>
You forgot a vital part of the configuration. You should add this to your web.xml before servlet-mapping tag:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.name.of.your.servlet.class</servlet-class>
</servlet>
I have a running war servlet in tomcat6. It works fine in tomcat. Now I need to migrate it to glassfish and it doesn't work.
First it show a big amount of problems (context, path null, etc) So I decided to create a new project in eclipse and copy all my classes in java and libraries to the new one.
Now I have a problem because I can upload the new .war to glassfish but servlet doesn't answer to my queries.
I think I have a problem with my glassfish-web.xml and almost sure with other things which I don't know.
The autogenerated glassfish-web.xml is this one:
<!DOCTYPE glassfish-web-app PUBLIC "-//GlassFish.org//DTD GlassFish Application Server 3.1 Servlet 3.0//EN" "http://glassfish.org/dtds/glassfish-web-app_3_0-1.dtd">
<glassfish-web-app>
<context-root>/TEST</context-root>
</glassfish-web-app>
And the path to one of my resources is this one: /students (class path) /listNames (method resource inside class)
So I guess I should use this url to access to my resource: localhost:port/nameOfMyWar/TEST/students/listNames
But it shows a white screen in my browser and log doesn't show anything.
What do I have to modify from a tomcat servlet project to adapt to a glassfish project? Is there a guide or some webpage with clues?
This is my actual 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>SERVLET</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>/axis2-web/index.jsp</welcome-file>
</welcome-file-list>
<servlet>
<description>JAX-RS Tools Generated - Do not modify</description>
<servlet-name>JAX-RS Servlet</servlet-name>
<servlet-class>org.apache.wink.server.internal.servlet.RestServlet</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.my.example.servlet</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>JAX-RS Servlet</servlet-name>
<url-pattern>/servlet/*</url-pattern>
</servlet-mapping>
<servlet>
<display-name>Apache-Axis Servlet</display-name>
<servlet-name>AxisServlet</servlet-name>
<servlet-class>org.apache.axis2.transport.http.AxisServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>AxisServlet</servlet-name>
<url-pattern>/servlet/AxisServlet</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>AxisServlet</servlet-name>
<url-pattern>*.jws</url-pattern>
</servlet-mapping>
<servlet-mapping>
<servlet-name>AxisServlet</servlet-name>
<url-pattern>/services/*</url-pattern>
</servlet-mapping>
<servlet>
<display-name>Apache-Axis Admin Servlet Web Admin</display-name>
<servlet-name>AxisAdminServlet</servlet-name>
<servlet-class>org.apache.axis2.webapp.AxisAdminServlet</servlet-class>
<load-on-startup>100</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>AxisAdminServlet</servlet-name>
<url-pattern>/axis2-admin/*</url-pattern>
</servlet-mapping>
</web-app>
This question already has answers here:
Servlet returns "HTTP Status 404 The requested resource (/servlet) is not available"
(19 answers)
Closed 5 years ago.
The xml file is located in WebContent/WEB-INF/web.xml of my project. I am using Eclipse and running Tomcat (which is not installed via Eclipse. I prefer it to be a separate installation).
<?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>EmployeeManagement</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>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
It doesnt work when the form page submits to the servlet. I am getting a 404 error everytime. I have been encountering this problem for a while. Somebody please help me.
You are missing <servlet>...</servlet> tag which is important to mapping. So use 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>EmployeeManagement</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>
<context-param>
<param-name>name</param-name>
<param-value>Pramod</param-value>
</context-param>
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
</web-app>
and you should give action value on your form like following:
<form action="/EmployeeManagement/WebContent/Registration" method="post">
//Some code here
</form>
and also note it down all values are case sensitive on the following code:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.yourPackageName.yourServletName</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
Your servlet name Registration should be same on both tags <servlet>...</servlet> and <servlet-mapping>...</servlet-mapping> and also package name should be same where your servlet class is located.
you have not mapped servlet name to servlet class, It should be like given below,
In <servlet-class> give the path of your servlet
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.Registration<servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Registration</servlet-name>
<url-pattern>/EmployeeManagement/WebContent/Registration</url-pattern>
</servlet-mapping>
Check your form action.
Is the path there
/EmployeeManagement/WebContent/Registration
or
YOURAPPCONTEXT/EmployeeManagement/WebContent/Registration
or
YOURAPPNAME/EmployeeManagement/WebContent/Registration
You have specifyed a servlet-mapping and used the name Registration in servlet-name without defining it before.
You need to define the servlet before using it in a servlet mapping
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>[fully qualifyied name of your servlet]</servlet-class>
</servlet>
You are missing another part to define the servlet in the web.xml
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>
package.path.to.RegistrationServlet
</servlet-class>
</servlet>
You forgot a vital part of the configuration. You should add this to your web.xml before servlet-mapping tag:
<servlet>
<servlet-name>Registration</servlet-name>
<servlet-class>com.name.of.your.servlet.class</servlet-class>
</servlet>