언어/JSP

[JSP] web.xml 에서 servlet 등록하기

Sime 2016. 10. 31. 12:45


web.xml에서 servlet이란 태그를 이용해서 서블릿을 등록 할 수 있다.



   1. 등록하고자 하는 서블릿의 이름과 등록하고자 하는 서블릿 구현 클래스를 지정 해줘야함..



   2. servlet-mapping을 통해 name과  url-pattern을 지정해줘야함..



<예제>


HelloServlet.java

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package servlet;
 
import java.io.IOException;
import java.io.PrintWriter;
 
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
 
public class HelloServlet extends HttpServlet {
    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        // TODO Auto-generated method stub
        // super.doGet(req, resp);
        resp.setContentType("text/html; charset=UTF-8");
        PrintWriter pw = resp.getWriter();
        pw.println("<html>");
        pw.println("<title>");
        pw.println("제목입니다.");
        pw.println("</title>");
        pw.println("<body>");
        pw.println("Hello World kk <br> Welcome to hell");     
        pw.println("</body>");
        pw.println("</html>");
        
    }
}
 
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web.xml

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<?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>JSP_1031</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>Hello</servlet-name> //servlet 이름 명시
  <servlet-class>servlet.HelloServlet</servlet-class> //패키지.클래스 이름을 등록 해줌
  </servlet>
  
  <servlet-mapping>
  <servlet-name>Hello</servlet-name> //servlet의 name을 매핑
  <url-pattern>/hello</url-pattern>  / 가 반드시 필요!! (주소 뒤에 값이라서)
  </servlet-mapping>
</web-app>
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http://localhost:8080/프로젝트이름/hello

run 실행 




자주 쓰는 방법으로는..


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<?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>JSP_1031</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>Hello</servlet-name>
  <servlet-class>servlet.HelloServlet</servlet-class>
  </servlet>
  
  <servlet-mapping>
  <servlet-name>Hello</servlet-name>
  <url-pattern>*.do</url-pattern>
  </servlet-mapping>
</web-app>
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  url-pattern에 *.do 방법이 있는데 특정 url 패턴을 지정할수 있다.

<예제 화면>




그외....


<포워딩 예제>


  

   req.getRequestURI()  -> url 경로를 return ex) /JSP_1031/hello.do


   req.getContextPath() -> content까지 경로 ex) /JSP_1031


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    if (req.getRequestURI().equals(req.getContextPath() + "/hello.do")) {
            req.getRequestDispatcher("hello.jsp").forward(req, resp);
        } else if (req.getRequestURI().equals(req.getContextPath() + "/whatTime.do")) {
            req.setAttribute("time", new Date());
            req.getRequestDispatcher("whatTime.jsp").forward(req, resp);
        }
    
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