✅ The verified answer to this question is available below. Our community-reviewed solutions help you understand the material better.
Нехай в додатку servletapp визначено сервлет Exam_servlet:import java.io.PrintWriter;import java.io.IOException;import javax.servlet.ServletException;import javax.servlet.annotation.WebServlet;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse; @WebServlet("/example")public class HelloServlet extends HttpServlet { protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { response.setContentType("text/html"); PrintWriter writer = response.getWriter(); String name = request.getParameter("subject"); String age = request.getParameter("rating"); try { writer.println("<h2>Subject: " + subject + "; Rating: " + rating + "</h2>"); } finally { writer.close(); } }}Як можна передати значення Java та 95 відповідних параметрів subject та rating сервлету через адресний рядок?