Question

我把一些带有JSP表单的代码插入到JDBC中。我在编码方面相对较新，但经过长时间的互联网搜索，我仍然无法解决问题。我需要插入＆＃39;数据人员在我的jsp表单中输入的表格。

JSP：

<form action="CampaignToJDBC" method="post">
        Enter company name: <input type="text" name="pname" /><br/>
        Enter company email: <input type="email" name="pemail" /><br/>
        <input type="submit" value="Apply" />
</form>

Servlet：

        Connection con = null;
        PreparedStatement ps = null;

        // ____________________________________________

        String pname = request.getParameter("pname");
        String pemail = request.getParameter("pemail");

        Class.forName(driver);
        con = DriverManager.getConnection(URL);
        String Query = "insert into campaigns(pname, pemail) values(?, ?)";
        ps = con.prepareStatement(Query);
        ps.setString(1, pname);
        ps.setString(2, pemail);

        ps.executeUpdate();

SQL：

DROP TABLE campaigns; 

CREATE TABLE campaigns(
Pname varchar2(32),
Pemail varchar2(32)
);

我的问题是它没有将它插入数据库，并且凭借我新获得的经验，我不知道为什么。我查了几个例子，但它们看起来都像我的。

亚历山大

FULL SERVLET：

@WebServlet(name = "CampaignToJDBC", urlPatterns = {"/CampaignToJDBC"})
public class CampaignToJDBC extends HttpServlet {
public static String driver = "oracle.jdbc.driver.OracleDriver";
public static String URL = "jdbc:oracle:thin:@datdb.cphbusiness.dk:1521:dat";
public static String ID = "";
public static String PW = "";
private int id = 0;
Connection con = null;
PreparedStatement ps = null;

protected void processRequest(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    String pname = request.getParameter("pname");
    String pemail = request.getParameter("pemail");

    try {
        Class.forName(driver);
        con = DriverManager.getConnection(URL, ID, PW);
        String Query = "insert into campaigns(Pname, Pemail) values(?, ?)";
        ps = con.prepareStatement(Query);
        ps.setString(1, pname);
        ps.setString(2, pemail);

        ps.executeUpdate();

        id = id + 1;
    } catch (Exception e) {

    } finally {
        try {
            ps.close();
        } catch (SQLException ex) {
            Logger.getLogger(CampaignToJDBC.class.getName()).log(Level.SEVERE, null, ex);
        }
        try {
            con.close();
        } catch (SQLException ex) {
            Logger.getLogger(CampaignToJDBC.class.getName()).log(Level.SEVERE, null, ex);
        }
    }

    RequestDispatcher rd = request.getRequestDispatcher("display.jsp");
    rd.forward(request, response);
}

// <editor-fold defaultstate="collapsed" desc="HttpServlet methods. Click on the + sign on the left to edit the code.">
/**
 * Handles the HTTP <code>GET</code> method.
 *
 * @param request servlet request
 * @param response servlet response
 * @throws ServletException if a servlet-specific error occurs
 * @throws IOException if an I/O error occurs
 */
@Override
protected void doGet(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    processRequest(request, response);
}

/**
 * Handles the HTTP <code>POST</code> method.
 *
 * @param request servlet request
 * @param response servlet response
 * @throws ServletException if a servlet-specific error occurs
 * @throws IOException if an I/O error occurs
 */
@Override
protected void doPost(HttpServletRequest request, HttpServletResponse response)
        throws ServletException, IOException {
    processRequest(request, response);
}

/**
 * Returns a short description of the servlet.
 *
 * @return a String containing servlet description
 */
@Override
public String getServletInfo() {
    return "Short description";
}// </editor-fold>
}

Answer 1

请将您的业务逻辑与JSP和Servlet端分开。为逻辑创建一个单独的类，让我们说经理/控制器或其他什么。在那里实现逻辑并从控制台或UnitTest尝试。

如果工作正常，请从Servlet调用该方法。首先，尝试使用常量参数。请参阅日志文件，生成日志信息或使用System.out.println（）;看看发生了什么。在日志中搜索例外。

SELECT查询是否适用于Servet？意味着连接很好。试着靠近。

Answer 2

问题已解决。

代码是正确的，但我忘了添加我需要的驱动程序。所以我所做的只是添加带驱动程序的库，它工作正常。对不起，我的坏！

感谢大家的帮助：）

使用servlet将JSP表单转换为JDBC

2 个答案: