【发布时间】:2015-06-19 07:17:21
【问题描述】:
我不能用这个插入或编辑记录。
我的 SQLException 有一个错误“无法访问 SQLException 的 catch 块。这个异常永远不会从 try 语句体中抛出”
刚接触java所以一切都是一个学习的过程
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
String forward = "";
String action = request.getParameter("action");
if (action.equalsIgnoreCase("delete")) {
int surveyid = Integer.parseInt(request.getParameter("survey_id"));
firstdao.removeFirst(surveyid);
forward = list_first;
try {
request.setAttribute("firsts", firstdao.getFirst());
} catch (SQLException e) {
e.printStackTrace();
}
} else if (action.equalsIgnoreCase("edit")) {
forward = insert_or_edit;
int surveyid = Integer.parseInt(request.getParameter("survey_id"));
try {
First first = firstdao.getFirstById(surveyid);
request.setAttribute("first", first);
} catch (SQLException e) {
e.printStackTrace();
}
} else if (action.equalsIgnoreCase("listFirst")) {
forward = list_first;
try {
request.setAttribute("firsts", firstdao.getFirst());
} catch (SQLException e) {
e.printStackTrace();
}
} else {
forward = insert_or_edit;
}
RequestDispatcher view = request.getRequestDispatcher(forward);
view.forward(request, response);
}
FirstController.java
private static final long serialVersionUID = 1L;
private static String insert_or_edit = "/first.jsp";
private static String list_first = "/listfirst.jsp";
private FirstDAO firstdao;
public FirstController() {
super();
firstdao = new FirstDAO();
}
FirstDAO.java
public void addFirst(First first) {
try {
String query = "insert into survey_data_27 (uname, p1q1, p1q2, p1q3, p1q4) values('"+ first.getuname() +"', '"+ first.getp1q1() +"', '"+ first.getp1q2() +"', '"+ first.getp1q3() +"', '"+ first.getp1q4() +"')";
Statement stmt = connection.createStatement();
stmt.executeUpdate(query);
} catch (SQLException e) {
e.printStackTrace();
}
}
【问题讨论】:
-
你能展示一下
firstdao的实现吗? -
我已经添加了 FirstController 和 FirstDAO
-
你为什么不使用 PreparedStatement ?