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Class not found even though it is clearly visible in Web App Libraries

Similar question exists but none of the answers helped.
Starting Tomcat Server 9 on my Eclipse Project gives the error
java.lang.ClassNotFoundException: org.glassfish.jersey.servlet.ServletContainer.class
at org.apache.catalina.loader.WebappClassLoaderBase.loadClass(WebappClassLoaderBase.java:1364)
at org.apache.catalina.loader.WebappClassLoaderBase.loadClass(WebappClassLoaderBase.java:1187)
at org.apache.catalina.core.DefaultInstanceManager.loadClass(DefaultInstanceManager.java:539)
Even though I can clearly see ServletContainer.class under org.glassfish.jersey.servlet under jersey-container-servlet-core.jar under Web App Libraries.
All the required jars were added to /src/main/webapp/WEB-INF/lib, they seem to be properly imported.
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_4_0.xsd" id="WebApp_ID" version="4.0">
<display-name>JavaAPI</display-name>
<servlet>
<servlet-name>JAVA API</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer.class</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>test</param-value>
</init-param>
<load-on-startup>1</load-on-startup> <!-- Tried Removing this, no difference -->
</servlet>
<servlet-mapping>
<servlet-name>JAVA API</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
</web-app>
Hello.java:
package test;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.ApplicationPath;
import javax.ws.rs.core.Application;
#Path("/hello")
public class Hello {
#GET
#Produces(MediaType.TEXT_XML)
public String sayHello() {
String resource = "<? xml version='1.0' ?>" +
"<hello>Hi Varun! This is the sayHello call.</hello>";
return resource;
}
#GET
#Produces(MediaType.APPLICATION_JSON)
public String sayHelloJSON() {
String resource = null;
return resource;
}
#GET
#Produces(MediaType.TEXT_HTML)
public String sayHelloHTML() {
String resource = "<h1>Hi Varun! This is the sayHelloHTML call.</h1>";
return resource;
}
}
Project Structure:
Using:
Eclipse: 2021-03
Tomcat: 9
Java: 1.8 (set under BuildPath, ProjectFacets)
JAX-RS 2.0 / Jersey 2.25.x
No Maven, Gradle involved, simply web project, with every jar hand-picked and imported.
Was following this tutorial step-by-step.

Remove the .class suffix
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
instead of
<servlet-class>org.glassfish.jersey.servlet.ServletContainer.class</servlet-class>
You specifiy the class name, not the class object.

Jersey web.xml application issues

Whenever I run my test app on Tomcat I can make it to my main page but whenever I try to go to my api/rest path I received a 500 error because my AppConfig could not be found. I believe this has to do with my directory set up.
My xml looks like this
<?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>FACHybrid</display-name>
<servlet>
<servlet-name>Jersey REST Service</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer</servlet-class>
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>java.config.ApplicationConfig</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>
In the xml the issues lies with the param-value tag
Here is the ApplicationConfig file
package config;
import java.util.HashSet;
import java.util.Set;
import javax.ws.rs.ApplicationPath;
import javax.ws.rs.core.Application;
import service.TestService;
#ApplicationPath("/rest")
public class ApplicationConfig extends Application{
#Override
public Set<Class<?>> getClasses() {
Set<Class<?>> s = new HashSet<Class<?>>();
s.add(TestService.class);
return s;
}
}
and my directories look like this
error in console
java.lang.ClassNotFoundException: java.config.ApplicationConfig

I think the problem lies in your web.xml. Fully qualified name of ApplicationConfig class is wrong.
Change it to config.ApplicationConfig -
<init-param>
<param-name>javax.ws.rs.Application</param-name>
<param-value>config.ApplicationConfig</param-value>
</init-param>

Java - Accessing Tomcat/Jersey url gives 404 Not Found

I've been struggling with this for the past two hours. I've looked at many different StackOverflow answers (a lot of questions appear like this one) but not one of them worked for me.
I have a simple 'CountryResources' class, that uses the #Path("/countries"), but when I navigate to localhost:8080/restservices/countries it just gives me a 404 error.
CountryResource class, located in a worldmap.services package:
package worldmap.services;
import model.Country;
import model.CountryService;
import model.ServiceProvider;
import javax.json.Json;
import javax.json.JsonArray;
import javax.json.JsonArrayBuilder;
import javax.json.JsonObjectBuilder;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
#Path("/countries")
public class CountryResource {
#GET
#Produces("application/json")
public String getCountries() {
System.out.println("Opgehaald");
CountryService cs = ServiceProvider.getWorldService();
JsonArrayBuilder jab = Json.createArrayBuilder();
for(Country c : cs.getAllCountries()) {
JsonObjectBuilder job = Json.createObjectBuilder();
job.add("code", c.getCode());
job.add("name", c.getName());
job.add("continent", c.getContinent());
job.add("region", c.getRegion());
job.add("surface", c.getSurface());
job.add("population", c.getPopulation());
job.add("government", c.getGovernment());
jab.add(job);
}
JsonArray array = jab.build();
return array.toString();
}
}
Web XML:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="3.0"
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"
metadata-complete="false">
<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>worldmap.services</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/restservices/*</url-pattern>
</servlet-mapping>
</web-app>

The problem has been found: Maven is not copying the dependencies of the program to the target's lib folder. Will have to try and figure out how to fix that.

REST Jersey HTTP Status 404 - The requested resource is not available

I tries looking on stackoverflow to solve this issue, but I can't find anything. Code seems to be OK, but I get a HTTP 404. Could you help me, please?
package rest;
import java.util.ArrayList;
import java.util.List;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import javax.ws.rs.core.Response;
import domain.Employee;
#Path("/employees")
public class EmployeeResource {
#GET
#Produces(MediaType.TEXT_PLAIN)
public Response getAllEmployees(){
List<Employee> employees = new ArrayList<Employee>();
employees.add(new Employee("Keith", "Brown", "Official", 27000));
employees.add(new Employee("Jenny", "Pitman", "Trainer", 31000));
String output = employees.toString();
return Response.status(200).entity(output).build();
}
}
and the 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" 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>VPP JavaEE 18 Rest DWP eu</display-name>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
<!-- configuration for jax rs -->
<servlet>
<servlet-name>Jersey Web Application</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 Web Application</servlet-name>
<url-pattern>/webservice/*</url-pattern>
</servlet-mapping>
</web-app>
I go the browser and write:
http://localhost:8080/VPP JavaEE 18 Rest DWP eu/webservice/employees
But I get A HTTP 404 The requested resource is not available with no clues.
However, Tomcat REST servlet IS working, because if I write:
http://localhost:8080/VPP JavaEE 18 Rest DWP eu/blablalba
I receive a different message:
HTTP Status 404 - /project-name/blablalba
I must also mention that it works if I write:
http://localhost:8080/VPP JavaEE 18 Rest DWP eu/index.html

Getting 404 error with TOMCAT-Jersey Rest

I am using JERSEY2.15:-
java class:-
package packages.newJersey;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
#Path("/rest")
public class SimpleWebService {
private static String versions = "4.1";
#GET
#Produces(MediaType.TEXT_HTML)
public String simpleMessage() {
return "<p>This is a simple REST</p>";
}
#Path("/version")
#GET
#Produces(MediaType.TEXT_HTML)
public String version() {
return "<p>Version Number:</p> " + versions;
}
}
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>LatestJersey</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.package</param-name>
<param-value>packages.newJersey</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey REST Service</servlet-name>
<url-pattern>/hello/*</url-pattern>
</servlet-mapping>
</web-app>
even though i use display name as :- LatestJersey
by default tomcat is opening:-
http://localhost:8080/RESTFULServiceWithLatestJersey/
and when i hit:-
http://localhost:8080/RESTFULServiceWithLatestJersey/hello/rest
I AM GETTING 404 ERROR
Could someone please help me here?

Everything looks good, except for this
jersey.config.server.provider.package
It should be
jersey.config.server.provider.packages
You're missing the s