Question

    try {
    int c=1;
    pstmt = conn.prepareStatement(NHMQueryConstant.INSERT_INTO_USER.toString());

       pstmt.setInt(c++, userID);
       pstmt.setInt(2, userTypeID);
       pstmt.setString(c++,userName);
        pstmt.setString(c++,NameInHindi);
        pstmt.setString(c++,loginID);
        pstmt.setString(c++,password);

       int  result = Integer.parseInt(mobile);
       System.out.println(result);
      pstmt.setInt(c++,result);
       pstmt.setString(c++,email);
        System.out.println(email);
        /*
        java.util.Date date = new SimpleDateFormat("yyyy-MM-dd").parse(dob);
        java.sql.Date sqlDate = new java.sql.Date(date.getTime());
        pstmt.setDate(c++, sqlDate);
        System.out.println(sqlDate);
        */
        //pstmt.setDate(++c,new Date(new java.util.Date().getTime()));
        pstmt.executeUpdate();
}



 public static final String INSERT_INTO_USER = new StringBuilder("")
    .append("INSERT INTO m_user_master (M_User_Id,User_Type_ID,M_User_Name,M__User_Name_Hindi,Login_ID,Password,Mobile,Email)")
        .append("VALUES(USER_ID_SEQ.NEXTVAL,'2',?,?,?,?,?,?)").toString();

我收到错误User_ID_seq.nextval unknown column.how我可以插入获取参数索引超出范围。

Answer 1

如果你有自动生成的ID（例如MySQL语法中的AUTOINCREMENT）并且不想关心id你可能会喜欢这个

public static final String INSERT_INTO_USER = "INSERT INTO " +
            "m_user_master (M_User_Name,M__User_Name_Hindi,Login_ID,Password,Mobile,Email)" +
            " VALUES('2',?,?,?,?,?,?)";

你必须删除

pstmt.setInt(c++, userID);
pstmt.setInt(2, userTypeID);

如果你想在Java中处理ID生成，那么你可以这样做

public static final String INSERT_INTO_USER = "INSERT INTO " +
            "m_user_master (M_User_Id,M_User_Name,M__User_Name_Hindi,Login_ID,Password,Mobile,Email)" +
            " VALUES(?,'2',?,?,?,?,?,?)";

并仅删除

pstmt.setInt(2, userTypeID);

我收到错误User_ID_seq.nextval未知列

1 个答案: