不要使用 jdbc 和 jsp 在 mysql 中插入数据答案

【问题标题】：Dont work insert data in mysql with jdbc and jsp不要使用 jdbc 和 jsp 在 mysql 中插入数据
【发布时间】：2015-12-10 12:22:57
【问题描述】：

我在 MySQL 中有一个表，其中包含 applicant_id、profession_id、last_name、first_name 和 entrance_year 列。

这是我用来插入或更新记录的代码：

public void saveApplicant(Applicant applicant) throws Exception {
    PreparedStatement preparedStatement = null;

    try {
        if (applicant.getId() == -1) {
            preparedStatement = connection.prepareStatement("INSERT INTO applicant (first_name, last_name, profession_id, entrance_year) VALUES (?,?,?,?)");

            preparedStatement.setString(1, applicant.getFirstName());
            preparedStatement.setString(2, applicant.getLastName());
            preparedStatement.setLong(3, (long) applicant.getProfessionId());
            preparedStatement.setInt(4, (int) applicant.getEntranceYear());

        } else {
            preparedStatement = connection.prepareStatement("UPDATE applicant SET first_name=?, last_name=?, profession_id=?, entrace_year=?  WHERE applicant_id=?");

            preparedStatement.setString(1, applicant.getFirstName());
            preparedStatement.setString(2, applicant.getLastName());
            preparedStatement.setLong(3, (long) applicant.getProfessionId());
            preparedStatement.setInt(4, (int) applicant.getEntranceYear());
            preparedStatement.setInt(5, (int) applicant.getId());
        }
        preparedStatement.executeUpdate();
    } catch (SQLException e) {
        throw new Exception(e);
    } finally {
        if (preparedStatement != null) {
            preparedStatement.close();
        }
    }
}

这是实现上述代码的类：

public class SaveApplicantCommand implements ICommand {

    private ApplicantDBProvider provider = ApplicantDBProvider.INSTANCE;

    @Override
    public String execute(HttpServletRequest request, HttpServletResponse resp) {

        Applicant applicant = new Applicant();

        applicant.setFirstName(request.getParameter("first_name"));
        applicant.setLastName(request.getParameter("last_name"));
        applicant.setProfessionId(Long.parseLong(request.getParameter("profession_id")));
        applicant.setEntranceYear(Integer.parseInt(request.getParameter("entrance_year")));

        if (request.getParameter("applicant_id") != null) {
            applicant.setId(Long.parseLong(request.getParameter("applicant_id")));
        }

        try {
            provider.saveApplicant(applicant);
        } catch (Exception e) {
            request.setAttribute("error", e);
            return "pages/error.jsp";
        }

        return "controller?command=applicants";
    }
}

最后，我的jsp：

<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core" %>
<html>
<head>
  <title></title>
</head>
<body>
<h1>Add application</h1>

<form method="post" action="controller?command=saveApplication">

  <table>
    <tr>
      <th>First Name</th>
      <th>Last Name</th>
      <th>Profession ID</th>
      <th>Entrance Year</th>
    </tr>
  </table>

  <c:choose>
    <c:when test="${application ne null}">
      <input type="text" name="first_name" value="${application.getFirstName()}"/>
      <input type="text" name="last_name" value="${application.getLastName()}"/>
      <input type="text" name="profession_id" value="${application.getProfessionId()}"/>
      <input type="text" name="entrance_year" value="${application.getEntranceYear()}"/>
      <input type="hidden" name="application_id" value="${application.getId()}"/>
    </c:when>
    <c:otherwise>
      <input type="text" name="first_name" value=""/>
      <input type="text" name="last_name" value=""/>
      <input type="text" name="profession_id" value=""/>
      <input type="text" name="entrance_year" value=""/>
    </c:otherwise>
  </c:choose>
  <input type=submit value=submit>
</form>
</body>
</html>

此表格允许我输入/编辑 4 个参数。但是，在提交表单后，会引发异常：

java.lang.NullPointerException
    org.sourceit.main.Controller.processRequest(Controller.java:28)
    org.sourceit.main.Controller.doPost(Controller.java:23)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:650)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:731)
    org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)

note The full stack trace of the root cause is available in the Apache Tomcat/7.0.64 logs.

我的控制器类是：

public class Controller extends HttpServlet {

    private Chooser chooser = Chooser.INSTANCE;

    @Override
    protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        processRequest(req, resp);
    }

    @Override
    protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
        processRequest(req, resp);
    }

    private void processRequest(HttpServletRequest req, HttpServletResponse resp) {
        try {
            String page = chooser.chooseCommand(req.getParameter("command")).execute(req, resp);
            req.getRequestDispatcher(page).forward(req, resp);
        } catch (ServletException | IOException e) {
            e.printStackTrace();
        }
    }
}

对代码量感到抱歉，但我认为需要展示整个问题。

感谢您的关注。

【问题讨论】：

您将 command 参数作为表单中的 GET 方法发送，但表单方法是 POST。
这里String page = chooser.chooseCommand(req.getParameter("command")).execute(req, resp); 或这里req.getRequestDispatcher(page).forward(req, resp); 有问题。是哪一个？

标签： java mysql jsp tomcat jdbc

【解决方案1】：

你的代码有问题，你的jsp没有包含“applicant_id”命名字段，但是你在SaveApplicantCommand类中调用参数名称为“applicant_id”，可能是“application_id”吧？

  if (request.getParameter("applicant_id") != null) {
        applicant.setId(Long.parseLong(request.getParameter("applicant_id")));
    }

然后把上面的代码改成

if (request.getParameter("application_id") != null) {
        applicant.setId(Long.parseLong(request.getParameter("application_id")));
    }

【讨论】：