标签: spring-data

我有一个带有一个简单实体和一个 JPA 存储库的 Spring 项目。该实体具有三个变量：Name (String)、active (boolean)和date (java.util.Date)。部署后，Hibernate 在我的 MySQL 数据库中使用varchar,tinyint和datetime. 一切似乎都正确，但是当我创建/修改实体的实例并调用save()存储库的方法时，除日期外的所有字段都被保存。使用 JPA 存储库存储日期是否有任何问题，或者我做错了什么？

我没有在这里放任何代码，因为它只是一个简单的类和存储库的接口。此外，实体正在被保存。我的问题仅与日期字段（以及我可以定义的任何其他日期字段）有关。话虽如此，如果有什么需要，尽管问。

编辑：

根上下文.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:jpa="http://www.springframework.org/schema/data/jpa"
    xmlns:task="http://www.springframework.org/schema/task"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
    http://www.springframework.org/schema/data/jpa http://www.springframework.org/schema/data/jpa/spring-jpa.xsd
    http://www.springframework.org/schema/task http://www.springframework.org/schema/task/spring-task-3.0.xsd">

    <!-- Root Context: defines shared resources visible to all other web components -->

    <import resource="infraestructure.xml" />

    <jpa:repositories base-package="com.smarttabletv.repository" />

</beans>

servlet-context.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/mvc"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xmlns:beans="http://www.springframework.org/schema/beans"
    xmlns:context="http://www.springframework.org/schema/context"
    xmlns:task="http://www.springframework.org/schema/task"
    xsi:schemaLocation="http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc.xsd
        http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/context …

我尝试并阅读有关此问题的其他问题,但我无法将逻辑应用于我的案例.我想从这个表中选择:

@Entity
public class LabelStatistics {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private int ID;

    @Enumerated(EnumType.STRING)
    private AnalysisType type;

    private String labelId;
    private String hexLabelId;
    private Timestamp timestamp;

    @OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER)
    private List<LabelStatisticsItem> results;

我正在尝试执行以下语句:

@Query(value = "SELECT s1.labelId, s1.type, s1.timestamp "
        + "FROM LabelStatistics s1 "
        + "INNER JOIN LabelStatistics s2 on s1.labelId = s2.labelId  and s1.type = s2.type and s1.timestamp < s2.timestamp")
List<Object[]> findLatestStatisticsEntries();

我一直收到这个错误:

org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! 

有人可以解释一下如何解决这个问题吗？最好的祝福

是否可以@Value从另一个变量设置

例如.

System properties : firstvariable=hello ,secondvariable=world

@Value(#{systemProperties['firstvariable'])
String var1;

现在我想要var2与var1它连接并依赖它,就像这样

    @Value( var1 + #{systemProperties['secondvariable']
    String var2;

public void message(){ System.out.printlng("Message : " + var2 );

我需要将Spring依赖项注入到JPA实体侦听器中。我知道我可以使用@Configurable和Spring的AspectJ weaver作为javaagent来解决此问题，但这似乎是一个棘手的解决方案。还有其他方法可以完成我想做的事情吗？

我对这个独特的约束功能感到非常惊讶.我正在使用H2数据库开发一个弹簧启动应用程序进行单元测试.

我的一个实体有一个独特的约束.在User类的标题下面.

@Entity
@Table(name = "user", uniqueConstraints = { @UniqueConstraint(columnNames = { "identifier" }) })
@SequenceGenerator(name = "user_pkey", sequenceName = "user_id_seq")

我写了一个小单元测试来检查唯一约束.

测试前功能

@Before
public void before() {
        User user1 = new User();
        user1.setBlackListed(false);
        user1.setIdentifier("test@tby.com");
        user1.setRefreshToken("azerty");          
        this.userDao.save(user1);
}

我的考试

@Test(expected = DataIntegrityViolationException.class)
public void user_identifierUnicityTest() {
    User user4 = new User();
    user4.setBlackListed(false);
    user4.setIdentifier("test@tby.com");
    user4.setRefreshToken("azerty4");
    User response = this.userDao.save(user4);
    this.userDao.findByIdentifier("test@tby.com");
}

我很惊讶,我认为在保存请求期间会抛出异常.或者在查找请求期间抛出异常.

在选择请求期间检查约束完整性而不是插入.我不明白结果.

这是唯一约束的正常行为吗？

谢谢你的回答:)

在我的控制器中我使用CrudRepository方法findAll()来查找我的数据库中的所有用户,如下所示:

userRepository.findAll()

问题是,这样做需要1.3分钟才能加载1.500个用户.从那里我用Thymeleaf在模型中加载数据我在一个html表中显示它:名称,时间创建,电子邮件,id,数据包和每个用户的状态.有没有办法提高性能或解决我的问题？任何帮助都会非常准确.

这是我的用户实体

    @Id
    @SequenceGenerator(name = "user_id_generator", sequenceName = "user_id_seq", allocationSize = 1)
    @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "user_id_generator")
    private Long id;
    @Column(nullable = false, unique = true)
    private String email;
    @Column(name = "uuid", nullable = false, unique = true)
    private String uuid;
    @Column(name = "reset_pwd_uuid", unique = true)
    private String resetPwdUuid;
    @Column(nullable = false)
    private String password;
    @Enumerated(EnumType.STRING)
    @Column(nullable = false)
    private Status status;
    @Enumerated(EnumType.STRING)
    @Column(nullable = false)
    private Packet packet = Packet.BASE;
    @Enumerated(EnumType.STRING)
    @Column
    private Situation situation; …

我正在尝试发布一个用户组：

public UserGroup createUserGroup(UserGroup userGroup) {
    ResponseEntity<UserGroup> userGroupResponseEntity = oauthRestTemplate
            .postForEntity(GROUPS_ENDPOINT, userGroup, UserGroup.class);
    return userGroupResponseEntity.getBody();
}

由于我正在使用@RepositoryRestResource，因此必须配置对象映射器：

ObjectMapper objectMapper = new ObjectMapper();
objectMapper.registerModule(new Jackson2HalModule());
objectMapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
objectMapper.configure(SerializationFeature.WRITE_DATES_AS_TIMESTAMPS, false);

MappingJackson2HttpMessageConverter messageConverter = new MappingJackson2HttpMessageConverter();
messageConverter.setObjectMapper(objectMapper);
messageConverter.setSupportedMediaTypes(Collections.singletonList(MediaTypes.HAL_JSON));

this.oauthRestTemplate.setMessageConverters(Collections.singletonList(messageConverter));

但是，从上面运行POST会抛出我

Caused by: com.fasterxml.jackson.databind.exc.MismatchedInputException: Unexpected token (START_OBJECT), expected VALUE_STRING: Expected array or string.
 at [Source: (org.apache.catalina.connector.CoyoteInputStream); line: 1, column: 87] (through reference chain: mahlzeit.api.hibernate.model.UserGroup["voteUntil"])
    at com.fasterxml.jackson.databind.exc.MismatchedInputException.from(MismatchedInputException.java:63) ~[jackson-databind-2.9.4.jar:2.9.4]
    at com.fasterxml.jackson.databind.DeserializationContext.wrongTokenException(DeserializationContext.java:1507) ~[jackson-databind-2.9.4.jar:2.9.4]
    at com.fasterxml.jackson.datatype.jsr310.deser.LocalDateTimeDeserializer.deserialize(LocalDateTimeDeserializer.java:138) ~[jackson-datatype-jsr310-2.9.2.jar:2.9.2]
    at com.fasterxml.jackson.datatype.jsr310.deser.LocalDateTimeDeserializer.deserialize(LocalDateTimeDeserializer.java:39) ~[jackson-datatype-jsr310-2.9.2.jar:2.9.2]
    at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:127) ~[jackson-databind-2.9.4.jar:2.9.4]
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:369) ~[jackson-databind-2.9.4.jar:2.9.4] …

是否可以在@Query批注内调用参数的方法？

例：

@Query("SELECT user " +
    "FROM User user " +
    "WHERE (?1.getFilter() = '*' OR user.name = ?1)");
List<User> getUsers(UserNameFilter userNameFilter);

我知道我可以做这样的事情：

@Query("SELECT user " +
    "FROM User user " +
    "WHERE (?1 = '*' OR user.name = ?1)");
List<User> getUsers(String userName);

但是，当过滤器数量增加时，这意味着我需要更改许多参数。

我正在尝试检查用户名在spring-boot中是否唯一。我想将结果作为JSON对象发送。这是REST控制器

@RequestMapping(value="/checkEmailUnique",method=RequestMethod.POST)
public String checkEmailUnique(@RequestBody String username){

    AppUser app = userRepo.findByUsername(username);
    if(app!=null){
        // I want to return somthing like emailNotTaken: true
    }
    else{
        // and here : emailNotTaken: false  
    }
}

我想获得角度结果，以便在组件中显示错误消息。我怎样才能做到这一点？

角边

客户服务

checkEmailNotTaken(email:string){
    if(this.authService.getToken()==null) {
      this.authService.loadToken();
    }
    return this.http.post(this.host+
      "/checkEmailUnique/",{email},{headers:new HttpHeaders({'Authorization':this.authService.getToken()})});
  }

在client.component.ts中

ngOnInit() {

    this.form = this.formBuilder.group({
      prenom: ['', Validators.required],
      nom: ['', Validators.required],
      tel: ['', Validators.required],
      cin: ['', Validators.required],
      username: ['', Validators.required , Validators.email  , this.validateEmailNotTaken.bind(this)],
      passwordG: this.formBuilder.group({
        password: ['',[Validators.required,Validators.minLength(9)]],
        Confirmationpassword : ['',[Validators.required,Validators.minLength(9)]]

      }, {validator: passwordMatch})

    }); …

我基于Spring Initializr（渐变风味）创建了一个Spring Boot应用程序。

我还加了

compile('org.springframework.boot:spring-boot-starter-data-mongodb')

使用MongoDB进行持久化。我还添加了一个可以正常工作的简单集成测试：

@RunWith(SpringRunner.class)
@SpringBootTest
@AutoConfigureMockMvc
public class TileServiceApplicationTests {

    @Autowired
    private MockMvc mvc;

    @Autowired
    private UserSettingRepository userSettingRepository;

    @Test
    public void contextLoads() throws Exception {
        Folder folder = random( Folder.class, "color", "elements" );
        EserviceTile eserviceTile1 = random( EserviceTile.class , "color");
        EserviceTile eserviceTile2 = random( EserviceTile.class, "color" );
        folder.setElements( Arrays.asList(eserviceTile1) );
        TileList usersTiles = new TileList( Arrays.asList( folder, eserviceTile2 ) );

        userSettingRepository.save( new UserSetting( "user1", usersTiles ));


        String string = mvc.perform( get( "/user1" ) ).andExpect( status().isOk() …