Question

从数据库中检索项目时出现问题，以下是我的代码：

UserDAO.java

public class UserDAO {

    public static Connection getConnection() {
        Connection con = null;
        try {
            Class.forName("org.hsqldb.jdbcDriver");
            con = DriverManager.getConnection(
                "jdbc:hsqldb:hsql://localhost/oneDB", "sa", "");
        } catch(Exception e) {
            e.printStackTrace();
        }
        return con;
    }

    public static List<User> list() {
        try{
            Connection connection = getConnection();

            List<User> users = new ArrayList<User>();

            try {
                PreparedStatement statement = connection.prepareStatement("SELECT * FROM user");
                ResultSet rs = statement.executeQuery();

                while (rs.next()) {
                    int id = rs.getInt("id");
                    String name = rs.getString("name");
                    String address = rs.getString("address");
                    User user = new User(id, name, address);
                    System.out.println("user:" + user.getTheName());
                    users.add(user);
                }
                statement.close();
                connection.close();
            } catch (SQLException e) {
                e.printStackTrace();
            }

            return users;
        } catch (Exception e) {
            e.printStackTrace();
           return null;
       }
    }
}

在我的servlet中执行：

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        List<User> users = UserDAO.list();
        request.setAttribute("users", users);
        request.getRequestDispatcher("showUser.jsp").forward(request, response);
}

showUser.jsp

<c:forEach items="${users}" var="user">
        Name: <c:out value="${users.theName}" /> <br>
        Address: <c:out value="${users.theAddress}" /><br>
</c:forEach>

堆栈追踪：

root cause
java.lang.NumberFormatException: For input string: "name"
    java.lang.NumberFormatException.forInputString(Unknown Source)
    java.lang.Integer.parseInt(Unknown Source)
    java.lang.Integer.parseInt(Unknown Source)
    javax.el.ListELResolver.coerce(ListELResolver.java:157)
    javax.el.ListELResolver.getValue(ListELResolver.java:70)
    org.apache.jasper.el.JasperELResolver.getValue(JasperELResolver.java:110)
    org.apache.el.parser.AstValue.getValue(AstValue.java:169)
    org.apache.el.ValueExpressionImpl.getValue(ValueExpressionImpl.java:184)
    org.apache.jasper.runtime.PageContextImpl.proprietaryEvaluate(PageContextImpl.java:943)
    org.apache.jsp.showUser_jsp._jspx_meth_c_005fout_005f0(showUser_jsp.java:217)
    org.apache.jsp.showUser_jsp._jspx_meth_c_005fforEach_005f0(showUser_jsp.java:178)
    org.apache.jsp.showUser_jsp._jspService(showUser_jsp.java:134)
    org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
    org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:438)
    org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:396)
    org.apache.jasper.servlet.JspServlet.service(JspServlet.java:340)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
    org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
    controllers.GetUserServlet.doGet(GetUserServlet.java:37)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:622)
    javax.servlet.http.HttpServlet.service(HttpServlet.java:729)
    org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)

我是jsp和servlet的新手，所以我不确定我哪里出错了。任何帮助将不胜感激。

Answer 1

在foreach片段中，您应该将元素称为user，这是var属性中提供的名称。同样，User类的字段很可能是name和address，或theName和theAddress。

<c:forEach items="${users}" var="user">
        Name: <c:out value="${user.name}" /> <br>
        Address: <c:out value="${user.address}" /><br>
</c:forEach>

使用jsp和servlet从hsqldb检索数据

1 个答案: