0
我是一个begginer,我试图在数据库中显示数据。我有一个控制器servlet,DBhelper,table.jsp和索引...在servlet中显示jsp的数据
这是控制器servlet
import java.io.IOException;
import java.sql.ResultSet;
import javax.annotation.Resource;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import javax.sql.DataSource;
import demo2.DBhelper;
/**
* Servlet implementation class servletJsp
*/
@WebServlet("/servletJsp")
public class controller extends HttpServlet {
private static final long serialVersionUID = 1L;
@Resource(name="")
private DataSource dataSource;
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String page;
if(request.getParameterMap().containsKey("page")){
page =request.getParameter("page");
}else{
page="index";
}
if(page.equals("table")){
ResultSet rs =new DBhelper().getUsers(dataSource);
request.getRequestDispatcher("table.jsp").forward(request, response);
request.setAttribute("users", rs);
}
}
}
这是DBhelper ..
package demo2;
import java.sql.Connection;
import java.sql.ResultSet;
import java.sql.Statement;
import javax.sql.DataSource;
public class DBhelper {
private ResultSet users =null;
public ResultSet getUsers(DataSource dataSource) {
Connection con;
Statement st;
try{
con =dataSource.getConnection();
st=con.createStatement();
String query="select * from authorities";
users= st.executeQuery(query);
}catch(Exception e){
e.printStackTrace();
}
return users;
}
}
这是表.JSP
<%@ page import="java.sql.ResultSet" %>
<%
ResultSet rs = (ResultSet)request.getAttribute("users");
while(rs.next()){
String username = rs.getString("username");
out.println(username);``
}
%>
我在table.jsp得到一个错误,在该行同时(rs.next)
'out.println(username);'......你在这条线上有反射。 –
我刚刚删除它,并得到错误 – oma
你会得到哪个错误?你为什么强迫我们猜测而不是阅读并发布它? –