Question

我是JSP Servlet的新手，我想要做的任务就是在JSP页面中使用servlet的ResultSet数据...当我运行代码时它会给我一些错误，比如

org.apache.jasper.JasperException: PWC6033: Error in Javac compilation for JSP

PWC6199: Generated servlet error:
source value 1.5 is obsolete and will be removed in a future release

PWC6199: Generated servlet error:
target value 1.5 is obsolete and will be removed in a future release

PWC6199: Generated servlet error:
To suppress warnings about obsolete options, use -Xlint:-options.

PWC6197: An error occurred at line: 148 in the jsp file: /products.jsp
PWC6199: Generated servlet error:
cannot find symbol
symbol:   class ResultSet
location: class org.apache.jsp.products_jsp

PWC6197: An error occurred at line: 148 in the jsp file: /products.jsp
PWC6199: Generated servlet error:
method processRequest in class jdbc.units cannot be applied to given types;
required: javax.servlet.http.HttpServletRequest,javax.servlet.http.HttpServletResponse
found: no arguments
reason: actual and formal argument lists differ in length

这是我的代码..

units.java

package jdbc;

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;
import java.sql.*;
import java.util.*;
import java.util.logging.*;
import org.json.*;
import javax.servlet.http.HttpSession;

public class units extends HttpServlet {
     protected ResultSet processRequest(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {
        ResultSet rs = null;
        response.setContentType("text/html;charset=UTF-8");
        try (PrintWriter out = response.getWriter()) {
            try {
                Class.forName("com.mysql.jdbc.Driver");
            } catch (ClassNotFoundException ex) {
                Logger.getLogger(units.class.getName()).log(Level.SEVERE, null, ex);
            }
            try {
                Connection con = DriverManager.getConnection("jdbc:mysql://localhost:3306/gst_application", "root", "");
                String sql = "SELECT * FROM `product_units` ORDER BY `product_unit_id` ASC";
                PreparedStatement ps = null;
                ps = con.prepareStatement(sql);
                rs = ps.executeQuery();

            } catch (SQLException ex) {
                Logger.getLogger(units.class.getName()).log(Level.SEVERE, null, ex);
                out.print(ex);
            }

        }
        return rs;
    }
}

product.jsp

<%@page import="jdbc.units"%>
<select class="form-control" name="product_unit" required="required">
    <%
        units obj = new units();
        ResultSet rs = obj.processRequest();
        while(rs.next())
        {
            out.print("<option>"+rs.getString("unit_val")+"</option>"); 
        }
    %>
 </select>

Answer 1

首先警告：您不应直接在页面中使用ResultSet。您应该将结果放在一个中间对象中，然后关闭ResultSet，Statement以及必要时的连接。

话虽如此，据我所知，您在JSP页面中缺少import语句，因此JSP编译器无法找到ResultSet类。尝试在JSP页面中导入jdbc包。

Answer 2

您可以将ResultSet对象放在List中并将其传递给您的页面，并在JSP中提取此列表。在这里，我假设Demo是您的班级名称。您可以根据您的班级名称更改代码。不要忘记关闭连接。

在Servlet中

Demo d = null;
List <Demo> myList = new ArrayList<>();
while (rs.next()) {
        myList.add(d);
}
rs.close();
ps.close();
con.close();
request.setAttribute("List", myList);

将您的响应转发给JSP文件，如下所示：

RequestDispatcher requestDispatcher = request.getRequestDispatcher("/pageName.jsp");
requestDispatcher.forward(request, response);

在JSP中

使用此标记：

<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>

使用forEach

访问列表

<c:forEach items="${List}" var="theList">
            <tr>
                <td></td>
                <td>${theList.attributeName}</td>
                ...
            </tr>
</c:forEach>

如何在jsp页面中使用Servlet ResultSet？

2 个答案: