Cannot create JDBC driver of class '' for connect URL 'null" - HikariCP, Tomcat8, PostgreSQL - postgresql

Firstly I have never setup a JDBC Pool before (So i am sorry if this seems mundane). I have been trying for some time to solve the problem but to no avail. I have explored the suggested options from various other stackoverflow posts but none have been successful.
I have tried:
Including the 'pgjdbc-ng' driver only in the /lib of tomcat and moving the context to the catalina home conf/
Swapping to using the example given by Hikari (but received same error) https://github.com/brettwooldridge/HikariCP/wiki/JNDI-DataSource-Factory-(Tomcat,-etc.)
I am using HikariCP with a PostgreSQL(pgjdbc-ng) driver. Tomcat8 deploys my war file which i build using maven. web.xml, context.xml, java code.
web.xml
<?xml version="1.0" encoding="UTF-8"?>
<web-app>
<display-name>Restful Web Application</display-name>
<resource-ref>
<res-ref-name>jdbc/postgresHikari</res-ref-name>
<res-type>javax.sql.DataSource</res-type>
<res-auth>Container</res-auth>
</resource-ref>
<servlet>
<servlet-name>Jersey Web Application</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.seng402.rest</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web Application</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
context.xml
<?xml version="1.0" encoding="UTF-8"?>
<Context>
<Resource name="jdbc/postgresHikari" auth="Container"
factory="com.zaxxer.hikari.HikariJNDIFactory"
type="javax.sql.DataSource"
minimumIdle="5"
maximumPoolSize="10"
connectionTimeout="300000"
dataSource.implicitCachingEnabled="true"
dataSource.user="docker"
dataSource.password="docker"
jdbcUrl="jdbc:postgresql://192.168.59.103:5432/docker"
driverClassName="com.impossibl.postgres.jdbc.PGDataSource"/>
</Context>
java code
Context initCtx = null;
try {
initCtx = new InitialContext();
Context envCtx;
try {
envCtx = (Context) initCtx.lookup("java:comp/env");
// Look up our data source
DataSource ds = (DataSource) envCtx.lookup("jdbc/postgresHikari");
try {
// Allocate and use a connection from the pool
Connection conn = ds.getConnection(); // throws the error
conn.close();
System.out.println("Connected!");
} catch (SQLException e) {
System.out.println("Failed to Connect!");
e.printStackTrace();
}
} catch (NamingException e) {
System.out.println("Failed to Lookup!");
e.printStackTrace();
}
} catch (NamingException e1) {
System.out.println("Fail to get Context!");
e1.printStackTrace();
}
Output:
Failed to Connect!
java.sql.SQLException: Cannot create JDBC driver of class '' for connect URL 'null'
ds.getConnection() throws the error.
Any suggestions as to how i could fix this or debug further would be greatly appreciated.
SOLVED !!! - context.xml was not being put into the META-INF folder

context.xml was not being put into the META-INF folder

With the JNDI provider, do not use the dataSourceClassName (or dataSource.URL) properties. Use jdbcUrl, and driverClassName if necessary (but try without it first).

Related

How to get Websphere to prompt for a database JNDI connection

I have an EAR application, containing an EJB and a WAR.
In the EJB I am using JPA to connect to the database, and then the WAR deals with the REST endpoints and calls methods defined in the EJB:
e.g - EJB Stateless Bean:
#PersistenceContext
private EntityManager em;
and EJB persistence.xml:
<?xml version="1.0" encoding="UTF-8"?>
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd"
version="2.0">
<persistence-unit name="WOTISEJB">
<jta-data-source>java:comp/env/jdbc/appds</jta-data-source>
<class>com...</class>
</persistence-unit>
</persistence>
Web class:
#Path("/getByType")
#ManagedBean
public class GetByTypeResource
{
#EJB
EjbDao ejbDao;
#GET
#Produces(MediaType.APPLICATION_JSON)
public String getData(#Context HttpServletRequest request, #PathParam("type") Type type)
{
JsonObject response = ejbDao.getByType(type);
StringWriter sw = new StringWriter();
try (JsonWriter jw = Json.createWriter(sw)) {
jw.write(response);
}
String retVal = sw.toString();
return retVal;
}
}
I presume I don't need anything in my web.xml as all of the EntityManager definitions are in the EJB code (and just called from the WEB application) - but just in case it's relevant:
<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_3_1.xsd"
version="3.1">
<servlet>
<servlet-name>appname</servlet-name>
<servlet-class>com.ibm.websphere.jaxrs.server.IBMRestServlet</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.app.ApplicationConfig</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>appname</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
<display-name>App Name</display-name>
</web-app>
When I install the application onto Websphere 9.0.5, I was expecting to be prompted as part of the installation to map the jndi name java:comp/env/jdbc/appds to the JDBC datasource I have already created in the console; however it is just using the default datasource in WAS instead.
What configuration elements am I missing to get the installation process to prompt me for the DSN mapping? In order words I don't want to put the jbdc name directly in the config, but have an indirect lookup which is configured on deploy.
Additionally, am I missing any configuration in my web.xml so that it also knows about the DSN when it calls the methods on the EJB classes, or will that happen automatically?

Spring MVC REST nohandler error

Im new to spring MVC and REST.. I'm having an issue with a simple test controller I've put together from example I've found here and from the spring docs..
When I hit the url http://localhost:8080/test-api/user/14 I get the error below
Im getting the error:
Sep 23, 2015 11:26:55 AM org.springframework.web.servlet.PageNotFound noHandlerFound
WARNING: No mapping found for HTTP request with URI [/test-api/user/14] in DispatcherServlet with name 'testapi'
Im using xml to config.. Im not ready to move to java config.
web.xml
Spring Web MVC Application
<servlet>
<servlet-name>springtest</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>springtest</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/testapi-servlet.xml</param-value>
</context-param>
<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>
testapi-servlet.xml - only contains the component scan and annotation driven elements
<context:component-scan base-package="com.springtest.testapi" />
<mvc:annotation-driven />
SpringTest.java
package com.springtest.testapi.api;
#RestController
public class SpringTest {
#RequestMapping(value="/user/{id}", method = RequestMethod.GET)
public User getUser(#PathVariable int id) {
User u = new User(id,"Test","Me");
return u;
}
What handler should I be defining.. None of the examples or docs state that a handler needs to be defined..
Remove the contextConfigLocation.
Replace the following:
<servlet-mapping>
<servlet-name>springtest</servlet-name>
<url-pattern>/test-api</url-pattern>
</servlet-mapping>
Make sure your xml file is test-api-servlet.xml and not testapi-servlet.xml
I found my issue. I mispelled the package in the component scan. The sample code I have was edited so It didn't fully represent what I had and was actually correct.

Spring Integration with RESTEasy

In our existing integration, we are planning to replace Queue ( the entry point to our integration processing) with RESTEasy services.
We are processing the HTTP requests as below:
1) Asynchronous HTTP Request Processing for GET
2) Asynchronous Job Service for POST
I understand that spring integration provides and for HTTP requests. But this is not something we want, as the request processing is handled by RESTEasy.
Software stack:
RESTEasy 3.0.9 Framework
Spring Integration 4.1.2.RELEASE
JBOSS EAP 6.4.
Is there a component that we could use to integrate RESTEasy services with spring integration ?
There is no explicit component exist, it is all API work to be done. You need to use dependent jar files and integration code
Below are the minimal Jar files to be path in workspace environment using ant or maven:
org.jboss.resteasy:resteasy-jaxrs:3.0.10.Final
org.jboss.resteasy:resteasy-spring:3.0.10.Final
org.springframework.boot:spring-boot-starter-web:1.2.2.RELEASE
org.jboss.resteasy:resteasy-jackson2-provider:3.0.10.Final
Following listener entries in web.xml to be done:
<context-param>
<param-name>resteasy.servlet.mapping.prefix</param-name>
<param-value>/project</param-value>
</context-param>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/dispatcher-servlet.xml</param-value>
</context-param>
<listener>
<listener- class>org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap</listener-class>
</listener>
<listener>
<listener-class>org.jboss.resteasy.plugins.spring.SpringContextLoaderListener</listener-class>
</listener>
<servlet>
<servlet-name>RESTEasyService</servlet-name>
<servlet-class>org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>com.concretepage.Application</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>RESTEasyService</servlet-name>
<url-pattern>/project/*</url-pattern>
</servlet-mapping>
Dispatch Servlet can be put into WEB-INF
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="
http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<context:component-scan base-package="com.concretepage" />
</beans>
Sample Java Service Code:
import java.util.Map;
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.Response;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Component;
#Path("/manage" )
#Component
public class ExService {
#Autowired
private ExRepository repository;
#GET
#Path("/{id}")
#Produces("application/json")
public Response getEmp(#PathParam("id") String id) {
Map<String,String> map = repository.getEmpDetail(id);
return Response.ok(map).build();
}
}
For more detail You may refer to http://docs.jboss.org/resteasy/docs/3.0.9.Final/userguide/html_single/index.html#RESTEasy_Spring_Integration
Use <int:gateway> to do the integration with spring.
<int:gateway id="providerGateway" service-interface="com.stack.overflow.TestInterface"
default-request-channel="requestChannel">
<int:method name="getDataByID" request-channel="requestChannel"/>
<int:method name="postDataByID" request-channel="requestChannel"/>
</int:gateway>
Where com.stack.overflow.TestInterface is the Resource Interface see below:
#Path(RestConstants.SERVICES)
public interface TestInterface {
#Path(RestConstants.TEST1)
#GET
#Produces({ MediaType.APPLICATION_XML, MediaType.APPLICATION_JSON })
public Response getDataByID();
#Path(RestConstants.TEST2)
#POST
#Produces({ MediaType.TEXT_PLAIN })
public Response postDataByID(String edi);
}
You can have different message channel if desired (see the gateway above) for a request. e.g. a request to getDataByID, will be put on the requestChannel. You can read from this channel and do the required processing as you require and then send a response back.

REST Web-services eclipse

I am new to REST Web services and was following a tutorial in Youtube. Although i followed the exact steps, I am unable to successfully run it.
Here is what i did
web.xml file
<context-param>
<param-name>resteasy.scan</param-name>
<param-value>true</param-value>
</context-param>
<!-- this need same with resteasy servlet url-pattern -->
<context-param>
<param-name>resteasy.servlet.mapping.prefix</param-name>
<param-value>/resteasy</param-value>
</context-param>
<listener>
<listener-class>org.jboss.resteasy.plugins.server.servlet.ResteasyBootstrap</listener-class>
</listener>
<servlet>
<servlet-name>resteasy-servlet</servlet-name>
<servlet-class>org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>resteasy-servlet</servlet-name>
<url-pattern>/resteasy/*</url-pattern>
</servlet-mapping>
Here is the service class that i created.
#Path("service")
public class Service {
private static Map<Integer , DoctorInformation> doctorInformations = new HashMap<Integer , DoctorInformation>();
static{
Location location = new Location();
location.setCity("Roorkee");
location.setState("Uttarakhand");
DoctorInformation doctorInformation = new DoctorInformation();
doctorInformation.setID(1);
doctorInformation.setFirstName("Sanchit");
doctorInformation.setLastName("Jain");
doctorInformation.setAddressLine1("ABSGDHD");
doctorInformation.setAddressLIne2("NBAJS");
doctorInformation.setPincode(247667);
doctorInformation.setEmail("#gmail.com");
doctorInformation.setSpecialization("ENT");
doctorInformation.setLocation(location);
doctorInformations.put(1 , doctorInformation);
}
//Method that will return single person object in XML
#GET
#Path ("/getPersonByIdXML/{id}")
#Produces(MediaType.APPLICATION_JSON)
public DoctorInformation getPersonByIdXML(#PathParam("id") int id) {
return doctorInformations.get(id);
}
I am using Maven to build the project , and finally when the Web project is deployed in localhost i am using the following url
http://localhost:8080/Project_Name/service/getPersonByIdXML/1
Please let me know what i am doing wrong. I think the <url-patter> in my web.xml is not proper, but when i changed it even the index.html page that loads on http://localhost:8080/Project_Name/ stopped working. In many places i have seen that <init-param> is defined along with ...somthing like this...
init-param>
<param-name>com.sun.jersey.config.property.packages</param-name>
<param-value>com.package.whatever</param-value>
</init-param>
but in my example i am using org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher as my servlet class. How will it make the difference?
The exception displayed on console is :
SEVERE: failed to execute
javax.ws.rs.NotFoundException: Could not find resource for full path: http://localhost:8080/TestWS/resteasy/service/getPersonByIdXML/1
at org.jboss.resteasy.core.registry.ClassNode.match(ClassNode.java:73)
at org.jboss.resteasy.core.registry.RootClassNode.match(RootClassNode.java:48)
at org.jboss.resteasy.core.ResourceMethodRegistry.getResourceInvoker(ResourceMethodRegistry.java:444)
at org.jboss.resteasy.core.SynchronousDispatcher.getInvoker(SynchronousDispatcher.java:234)
at org.jboss.resteasy.core.SynchronousDispatcher.invoke(SynchronousDispatcher.java:171)
at org.jboss.resteasy.plugins.server.servlet.ServletContainerDispatcher.service(ServletContainerDispatcher.java:220)
at org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher.service(HttpServletDispatcher.java:56)
at org.jboss.resteasy.plugins.server.servlet.HttpServletDispatcher.service(HttpServletDispatcher.java:51)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:233)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:857)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
at java.lang.Thread.run(Thread.java:745)

Mixing REST and JSP in Spring MVC, Cannot find JSP

I'm sure this is a noob question, and I've spent the better part of an hour trawling stackoverflow for an answer but nobody seems to have my case so here we go...
I have a new webapp that uses Spring MVC. Most of the app (99%) is pure REST, so it doesn't have a "view" as such but rather simply sends JSON back down the wire, or sends an alternate HTTP Status for errors etc.
The exception is the login page which needs to be an actual JSP, but somehow the configuration I am using to map my REST controllers is leaving me in a state where normal JSP mappings fail.
Here's what I've got:
In my dispatcher servlet config, the relevant portions are:
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/pages/"/>
<property name="suffix" value=".jsp"/>
</bean>
In my attempts to get it working, I have also added a mapping to the "HomeController" which currently just redirects to my login JSP:
<bean name="/" class="com.somepackage.HomeController"/>
Now, in the web.xml I have:
<servlet>
<servlet-name>spring-dispatcher</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<listener>
<listener-class>
org.springframework.web.context.ContextLoaderListener
</listener-class>
</listener>
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
/WEB-INF/spring-dispatcher-servlet.xml
</param-value>
</context-param>
This works fine for my RESTful controllers, which look like this:
#Controller
#RequestMapping(value = "/api/user")
public class BlahBlahController {...
My "HomeController", which just looks like this:
#Controller
#RequestMapping(value = "/")
public class HomeController extends AbstractController {
#Override
protected ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
return new ModelAndView("login");
}
}
IS triggered when I hit the "/" url, but I get this error in the logs:
WARNING: No mapping found for HTTP request with URI [/WEB-INF/pages/login.jsp] in DispatcherServlet with name 'spring-dispatcher'
Now I get what it's saying, it doesn't know how to resolve /WEB-INF/pages/login.jsp (this page does exist btw), but I'm stuck as to how I need to alter things to get this to work.
I'm a little confused on how it's supposed to work. Anyone got any clues?
Thanks.
OK.. I found the answer, it's the url-pattern in the dispatcher config.
Instead of:
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
It should be
<servlet-mapping>
<servlet-name>spring-dispatcher</servlet-name>
<url-pattern>/</url-pattern>
</servlet-mapping>
I had actually found this answer elsewhere and tried it but "thought" it wasn't working, then realized the reason I thought this was unrelated to the root cause.
No idea why this would work and the other wouldn't.. but one problem at a time...
Put RequestMapping at the method level, I tried at my code and it worked. You don't need to define HomeController bean if you are using #Controller and having a proper "context:component-scan"
#Override
#RequestMapping(value = "/")
protected ModelAndView handleRequestInternal(HttpServletRequest request, HttpServletResponse response) throws Exception {
return new ModelAndView("login");
}
you can also use below code if you just want to redirect a login view for all "/" access.
<mvc:view-controller path="/" view-name="login"/>
Checkout the mvc show case project from github for a helpful reference.