问题描述
我将确认码链接发送到注册用户的电子邮件中，当点击链接时，我需要激活状态以便会员被批准，但我收到标题中显示的错误。一切都是完全真实的，但我无法理解为什么会发生这种情况。这是我的代码：
1)在此代码部分，用户成为会员，并会向其电子邮件发送确认代码。
@WebServlet(name = "RegisterUserSaveController",urlPatterns = {"/register-user-save"})
public class RegisterUserSaveController extends HttpServlet {
protected void processRequest(HttpServletRequest request,HttpServletResponse response)
throws servletexception,IOException {
try {
String name = request.getParameter("name");
String surname = request.getParameter("surname");
String email = request.getParameter("email");
String password = request.getParameter("password");
String repeatPassword = request.getParameter("password-repeat");
if (!password.equals(repeatPassword)) {
request.setAttribute("error","Password repeat is not same!");
request.getRequestdispatcher("register").forward(request,response);
} else {
User user = new User();
user.setName(name);
user.setSurname(surname);
user.setEmail(email);
user.setPassword(PasswordHasher.hashPassword(password));
String activationCode = UUID.randomUUID().toString();
user.setActivationCode(MD5.hashedMd5(activationCode));
LocalDateTime expiredDate = LocalDateTime.Now().plusHours(1);
user.setExpiredDate(expiredDate);
UserDaoService userDaoService = new UserDaoManager();
userDaoService.save(user);
String subject = "Confirm Registration";
String link = "http://localhost:8084/employee/registerconfirm?code=" + activationCode;
String title = "Your confirmation link:\n " + link;
SendEmail.sendAsync(email,title,subject);
request.setAttribute("info2","Your Registration was successfully! Pls check your email.");
request.getRequestdispatcher("success-info").forward(request,response);
}
} catch (GeneralSecurityException e) {
}
2)在此代码部分，执行用户单击发送到电子邮件的链接后的步骤。所以如果代码正确，我会验证数据库中用户的状态列。
@WebServlet(name = "RegisterConfirmController",urlPatterns = {"/registerconfirm"})
public class RegisterConfirmController extends HttpServlet {
protected void processRequest(HttpServletRequest request,IOException {
String code = request.getParameter("code");
UserDaoService userDaoService = new UserDaoManager();
if (code == null) {
request.setAttribute("info1","Activation code is not correct!");
request.getRequestdispatcher("error-info").forward(request,response);
} else {
code = MD5.hashedMd5(code);
User user = userDaoService.findByActivationCode(code);
if (user == null) {
request.setAttribute("info1","Activation code is not correct!");
request.getRequestdispatcher("error-info").forward(request,response);
} else {
LocalDateTime expiredDate = user.getExpiredDate();
LocalDateTime currentDate = LocalDateTime.Now();
if (expiredDate.isBefore(currentDate)) {
request.setAttribute("info1","Activation code is expired!");
request.setAttribute("info2","resend?id=" + user.getId() + "");
request.getRequestdispatcher("error-info").forward(request,response);
} else if (user.getStatus() == UserStatusEnum.CONFIRMED.getValue()) {
request.setAttribute("info1","Your account already confirmed!");
request.getRequestdispatcher("error-info").forward(request,response);
} else {
userDaoService.updateStatusById(user.getId(),UserStatusEnum.CONFIRMED);
request.setAttribute("info2","Your account is confirmed!");
request.getRequestdispatcher("success-info").forward(request,response);
}
}
}
}
3)但问题是，当我点击发送到电子邮件的链接时，出现以下错误。 错误 java.time.format.DateTimeParseException: 无法解析文本 '2021-07-24 18:11:33.0'，在索引 19 处找到未解析的文本
@Override
public User findByActivationCode(String activationCode) {
try (Connection connection = DbConnection.getConnection()) {
PreparedStatement preparedStatement = connection.prepareStatement("select id,expired_date,status from users where activation_code=?");
preparedStatement.setString(1,activationCode);
ResultSet resultSet = preparedStatement.executeQuery();
if (resultSet.next()) {
User user = new User();
user.setId(resultSet.getInt(1));
user.setExpiredDate(LocalDateTime.parse(resultSet.getString(2),DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss")));
user.setStatus(resultSet.getInt(3));
return user;
}
} catch (Exception e) {
System.out.println("findByActivationCode error " + e);
}
return null;
}
解决方法
您必须决定如何处理 resultSet.getString(2) 返回的日期字符串中的“.0”。你可以做一个子字符串来删除它。或者您可以更改格式字符串以解决它。我为纳秒选择了“.n”，但您必须研究该值的含义。
再想一想，您甚至可能不想使用 resultSet.getString(2)。如果这是来自数据库的内容，您可能需要考虑返回时间戳的 resultSet.getTimestamp(2)。 可以在此处找到有关如何将其转换为 LocalDateTime 的信息 How to convert java.sql.timestamp to LocalDate (java8) java.time?
public static void main(String[] args)
{
String text = "2021-07-24 18:11:33.0";
//
String text1 = text.substring(0,19);
LocalDateTime ldt = LocalDateTime.parse(text1,DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss"));
//
LocalDateTime ldt2 = LocalDateTime.parse(text,DateTimeFormatter.ofPattern("yyyy-MM-dd HH:mm:ss.n"));
int ai = 0;
}
您遇到问题是因为您错过了秒的模式。您可以使用以下模式：
u-M-d H:m:s[.[SSSSSSSSS][SSSSSSSS][SSSSSSS][SSSSSS][SSSSS][SSSS][SSS][SS][S]]
其中使用方括号将 fraction-of-second 设为可选。
演示：
import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;
import java.util.Locale;
public class Main {
public static void main(String[] args) {
DateTimeFormatter dtf = DateTimeFormatter.ofPattern(
"u-M-d H:m:s[.[SSSSSSSSS][SSSSSSSS][SSSSSSS][SSSSSS][SSSSS][SSSS][SSS][SS][S]]",Locale.ENGLISH);
String strDateTime = "2021-07-24 18:11:33.0";
LocalDateTime ldt = LocalDateTime.parse(strDateTime,dtf);
System.out.println(ldt);
}
}
输出：
2021-07-24T18:11:33
我还建议您将列类型更改为 Date-Time 类型，这将使您能够直接使用 java.time 类型而不是处理 String。查看 this answer 和 this answer 以了解如何将 java.time API 与 JDBC 结合使用。
从 Trail: Date Time 了解有关现代 Date-Time API 的更多信息。