java 如何使用 jpa spring 找出电子邮件是否已经存在并向前端发送一些错误消息

How to find out if an email already exist with jpa spring and sending some error messag to the front end

提问by Greg

So i have a simple UsersDao

所以我有一个简单的 UsersDao

public interface UserDao extends JpaRepository<User, Long> {

}

And inside my user controller i want to do something like this :

在我的用户控制器中，我想做这样的事情：

@RequestMapping(value = "/register",method = RequestMethod.POST)
public void addUser(@RequestBody User user) {

    //How do i check if user already exist with email instead of id
    // i managed to do this but can i search on something else than the id
    User user1 = userDao.findOne(1);

    if (user.getEmail().equals(user1.getEmail()))
    {

        // And how should i give one error to the front end if the email 
       //already exist I'm using angular js

    }
    else {

        userDao.save(user);

    }

}

I also have some extra questions on this topic:

我还有一些关于这个主题的额外问题：

Somethings that are not clear are following. I have done a small tutorial on jpa but there they use:

不清楚的东西还在后面。我在 jpa 上做了一个小教程，但他们在那里使用：

EntityManager, EntityTransaction

实体管理器、实体交易

Note : when using EntityManagerFactory it goes as follow :

注意：使用 EntityManagerFactory 时，如下所示：

    EntityManagerFactory emf = null,

    //Then they use EntityManagerFactory

        emf = Persistence.createEntityManagerFactory("SomeValue")
    //where can i get "someValue" When using application .properties 

//because in the example they use xml but can't find the right properties in application.properties

Or do i not need to use these in springboot

或者我不需要在springboot中使用这些

Sorry for all these question. I really want to get into spring but somethings are still a bit unclear at this point ;)

很抱歉所有这些问题。我真的很想进入春天，但在这一点上还有点不清楚；)

回答by ufuoma

You can do the following:

您可以执行以下操作：

Assuming Userhas an attribute email, define a method in the interface like this to generate a dynamic query:

假设User有一个属性email，在接口中定义一个方法是这样生成动态查询的：

public interface UserDao extends JpaRepository<User, Long> {
    public User findByEmail(String email);
}

Then you can find a user by email. If nullis returned, no user with the given email exists. Also, within the Userentity class, you can define an annotation to ensure that emailis unique like this:

然后您可以通过电子邮件找到用户。如果null返回，则不存在具有给定电子邮件的用户。此外，在User实体类中，您可以定义一个注释以确保它email是唯一的，如下所示：

public class User {

    ....

    @Column(unique=true)
    String email;

}

回答by asm0dey

You have 2 options:

您有 2 个选择：

1. Use method User findByEmail(String email);in repository interface.
2. Use method like @Query("SELECT COUNT(u.id) FROM User u WHERE u.email=:email) Long countUsersWithEmail(String email);Than it's obvious how to use rusults of these queries. I would use 2nd choice because of smaller overhead.

1. User findByEmail(String email);在存储库界面中使用方法。
2. 使用像@Query("SELECT COUNT(u.id) FROM User u WHERE u.email=:email) Long countUsersWithEmail(String email);这样的方法 很明显如何使用这些查询的结果。由于开销较小，我会使用第二个选择。