Question

在表单提交时，请求被传递给httpservlet请求，其中我试图将值插入用户模型（使用setter方法），如下所示：

的Servlet

protected void doPost(HttpServletRequest request, HttpServletResponse response)
            throws ServletException, IOException {

    System.out.println("Request Parameters from Registraion Servlet: ");
    System.out.println("=========================================== ");
    System.out.println(request.getParameter("uname")); //Doesn't print anything in console
    System.out.println(request.getParameter("lname")); //Doesn't print anything in console
    System.out.println(request.getParameter("email")); //Doesn't print anything in console
    System.out.println(request.getParameter("pwd"));   //Doesn't print anything in console

    try {
        System.out.println("Registration Servlet INSIDE try block:"); //Doesn't print anything in console
        User user = new User();
        user.setFirstName(request.getParameter("uname"));
        user.setLastName(request.getParameter("lname"));
        user.setEmail(request.getParameter("email"));
        user.setPassword(request.getParameter("pwd"));
        RegistrationDAO rsDao = new RegistrationDAO();
        rsDao.insert(user);
    } catch (Exception e) {
        System.out.println("DB related Error");
    }

    System.out.println("Registration Servlet OUTSIDE try block:"); //Doesn't print anything in console
    RequestDispatcher rd = request.getRequestDispatcher("EmailConfirmation.jsp"); //Successfully Redirects
    rd.forward(request, response);
}

它成功重定向到我提到的JSP页面，但没有打印request.getparameters(xxx)中的任何值。我想知道什么时候从表单的动作方法成功调用servlet，为什么不打印值？

抓挠我的头但却无法弄清楚出了什么问题？

这是我的HTML页面：

<form name="myForm" action="RegistrationServlet.do" method="POST" novalidate>

    <input type="text" class="form-control input-sm" name="uname" ng-model="uname" ng-pattern="/^[a-zA-Z]{3,20}/"  unique-username="" required placeholder="First Name" /><br>

    <input type="text" class="form-control input-sm" name="lname" ng-model="lname" ng-pattern="/^[a-zA-Z]{3,20}/" required placeholder="Last Name"/><br>

    <input type="email" class="form-control input-sm" name="email" ng-model="email" ng-pattern="/^[a-zA-Z]{3,20}/" required placeholder="Email"/><br>

    <input type="password" class="form-control input-sm glyphicon glyphicon-ok" name="pwd" ng-model="pwd" required placeholder="Password"/><br>

    <input type="password" class="form-control input-sm glyphicon glyphicon-ok" name="pwd2" ng-model="pwd2" value-matches="pwd" required placeholder="Confirm password"/><br>

    <input type="checkbox" class="" name="agreement" value=""><small> I agree to RTH agreement</small>

    <!--<button class="form-control btn btn-success" type="submit" ng-click="submit($event)">Connect now</button>-->

    <input type="submit" class="form-control btn btn-success" value="Connect now" />
</form>

DAO课程：

public class RegistrationDAO {

    public void insert(User user) {

        try {
            System.out.println("Printing USER Values");
            System.out.println("====================");
            System.out.println(user.getFirstName());
            System.out.println(user.getLastName());
            System.out.println(user.getEmail());
            System.out.println(user.getPassword());

            Connection con = DBConnection.getConnection();
            String query = "insert into TBL_USER(USR_FIRST_NAME,USR_LST_NAME,USR_PRIMARY_EMAIL,USR_PASSWORD) values(?,?,?,?)";
            PreparedStatement pst = con.prepareStatement(query);
            pst.setString(1, user.getFirstName());
            pst.setString(2, user.getLastName());
            pst.setString(3, user.getEmail());
            pst.setString(4, user.getPassword());
            pst.executeUpdate();
        } catch (Exception e) {
            System.out.println("@@@@Record insertion error in Registration DAO@@@@");
            System.out.println(e);
        }
    }
}

模特课：

public class User implements Serializable{

    private String firstName;
    private String lastName;
    private String password;
    private String email; 
    //setters and getters here
}

SQL表格式：

CREATE TABLE "RTH"."TBL_USER" 
(
    "USR_ID" VARCHAR2(1 BYTE) NOT NULL ENABLE, 
    "USR_IS_ANONYMOUS" RAW(1), 
    "USR_FIRST_NAME" CLOB, 
    "USR_MID_NAME" CLOB, 
    "USR_LST_NAME" CLOB, 
    "USR_PRIMARY_EMAIL" CLOB, 
    "USR_ALT_EMAIL" CLOB, 
    "USR_PRIMARY_CNCT_NMBR" CLOB, 
    "USR_SECONDARY_CNCT_NMBR" CLOB, 
    "USR_TYPE" CLOB, 
    "USR_CATEGORY" CLOB, 
    "USR_DOB" TIMESTAMP (0), 
    "USR_CREATE_DT" TIMESTAMP (0), 
    "CUPN_ID" VARCHAR2(1 BYTE), 
    "USR_MARITIAL_STUTUS" VARCHAR2(20 BYTE), 
    "USR_GENDER" VARCHAR2(1 BYTE), 
    "USR_IMAGE" RAW(1), 
    "USR_PASSWORD" VARCHAR2(20 BYTE), 
     CONSTRAINT "XPKTBL_USER" PRIMARY KEY ("USR_ID")
)

更新

问题是我没有调用dao类的insert(User user)方法。现在它的工作，但我有插入语句String query = "insert into TBL_USER(USR_FIRST_NAME,USR_LST_NAME,USR_PRIMARY_EMAIL,USR_PASSWORD) values(?,?,?,?)";

的问题

我收到错误：java.sql.SQLIntegrityConstraintViolationException: ORA-01400: cannot insert NULL into ("RTH"."TBL_USER"."USR_ID")

UPDATE2：

很少研究我将sql查询更改为以下内容：

`String query = "update TBL_USER set USR_FIRST_NAME = ?, USR_LST_NAME = ?, USR_PRIMARY_EMAIL=?, USR_PASSWORD = ?";`

现在我没有得到任何错误，但表中没有插入

Answer 1

您确定要用需要插入的值替换？你没有向我们展示你是如何做到的，我怀疑这是出错的地方。

Answer 2

你也可以发布用户实体类吗？

您获得的第一个异常意味着您在表的列

上传递空值

Answer 3

创建序列并触发自动更新ID列修复了问题。这是解决方案：

CREATE TABLE "RTH"."TBL_USER" 
(   
    "USR_ID" VARCHAR2(1 BYTE) NOT NULL ENABLE, 
    "USR_IS_ANONYMOUS" RAW(1), 
    "USR_FIRST_NAME" CLOB, 
    "USR_MID_NAME" CLOB, 
    "USR_LST_NAME" CLOB, 
    "USR_PRIMARY_EMAIL" CLOB, 
    "USR_ALT_EMAIL" CLOB, 
    "USR_PRIMARY_CNCT_NMBR" CLOB, 
    "USR_SECONDARY_CNCT_NMBR" CLOB, 
    "USR_TYPE" CLOB, 
    "USR_CATEGORY" CLOB, 
    "USR_DOB" TIMESTAMP (0), 
    "USR_CREATE_DT" TIMESTAMP (0), 
    "CUPN_ID" VARCHAR2(1 BYTE), 
    "USR_MARITIAL_STUTUS" VARCHAR2(20 BYTE), 
    "USR_GENDER" VARCHAR2(1 BYTE), 
    "USR_IMAGE" RAW(1), 
    "USR_PASSWORD" VARCHAR2(20 BYTE), 
     CONSTRAINT "XPKTBL_USER" PRIMARY KEY ("USR_ID")
);

自动递增ID值的序列和触发器：

CREATE SEQUENCE TBL_USER_SEQ
  START WITH 1
  INCREMENT BY 1
  CACHE 100;

 CREATE OR REPLACE TRIGGER "RTH"."TBL_USER_TRIGGER"
  BEFORE INSERT ON TBL_USER
  FOR EACH ROW
BEGIN
  SELECT TBL_USER_SEQ.nextval
    INTO :new.USR_ID
    FROM dual;
END;

无法使用dao类将值插入DB

3 个答案: