I was already searching for a solution but I cannot find any.
My problem is that I cannot my Postgres database, called IncomeOutgo, via Hibernate.
I always get this error message when calling my Thymeleaf/HTML website.
So does my application.properties look like
spring.jpa.properties.hibernate.dialect = org.hibernate.dialect.PostgreSQLDialect
spring.jpa.hibernate.ddl-auto=none
spring.jpa.hibernate.show-sql=true
spring.datasource.url=jdbc:postgresql://192.168.0.227:5432/IncomeOutgo
spring.datasource.username=postgres
spring.datasource.password=XXX
#spring.jpa.properties.hibernate.format_sql=true
So does my table look like in Postgres (DBeaver):
And last but not least so does my Domain/Entity look like:
#Data
#NoArgsConstructor
#AllArgsConstructor
#Builder
#Entity
#Table
public class IncomeOutgo extends AbstractPersistable<Long> {
#Version
#Column(name ="id")
private Long id;
#Column(name="dayofweek")
private Date dayofweek;
#Size(min = 5, max = 50)
#Column(name ="person")
private String person;
#Min(0)
#Column(name ="version")
private Integer version;
#Column(name="income")
private int income;
#Column(name="outgo")
private int outgo;
}
Maybe, someone can tell me what my error is?
Thanks in advance!
I was able to make it work now.
LesiakĀ“s hint with the schema name made me do a workaround.
I created a new database: incomeoutgo (lower letter case), and under schema: public, a new table: incomeoutgo with lower letters, too.
And so does my Domain definition look like:
#Data
#NoArgsConstructor
#AllArgsConstructor
#Builder
#Entity
#Table(name="incomeoutgo", schema = "public")
public class IncomeOutgo extends AbstractPersistable<Long> {
#Version
#NotNull
#Column(name ="id")
private Long id;
#Column(name="dayofweek")
private Date dayofweek;
#Size(min = 5, max = 50)
#Column(name ="person")
private String person;
#Min(0)
#Column(name ="version")
private Integer version;
#Column(name="income")
private int income;
#Column(name="outgo")
private int outgo;
}
Related
The full error message is:
Unable to create unique key constraint (aircraft_series_id, service_enum) on table aircraft_service: database column 'service_enum' not found. Make sure that you use the correct column name which depends on the naming strategy in use (it may not be the same as the property name in the entity, especially for relational types)
My entity is specified as:
#Entity
#Table(uniqueConstraints = { #UniqueConstraint(columnNames = { "aircraft_series_id", "service_enum" }) })
#Getter
#Setter
#NoArgsConstructor
#ToString
public class AircraftService {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#NotNull
private Integer minimumQuantity;
#NotNull
private Integer maximumQuantity;
#NotNull
private Integer defaultQuantity;
#NotNull
#ManyToOne(optional = false)
#JsonIgnore
private AircraftSeries aircraftSeries;
#NotNull
#Enumerated(EnumType.STRING)
private ServiceEnum serviceEnum;
}
If I comment out the #Table(uniqueConstraints = { #UniqueConstraint(columnNames = { "aircraft_series_id", "service_enum" }) }) annotation then the columns are created and I can see the field names when opening the table under the SQL client.
service_enum
aircraft_series_id
For now I'm running the application against the H2 database.
I could have the application running not throw an exception if the class is boasting the column annotations, as in:
#Column(name = "service_enum")
#ManyToOne(optional = false)
#JoinColumn(name = "service_profile_id")
I don't see why this is the case, as by default, the column names are exactly the same, when attributed by the application itself.
I have the following code. The number of rows in the One table will be about 100 and the number of rows in the Many table will be several hundred thousand (millions). I'm going to need to paginate the Many rows per single row in One when I query.
I'm not sure how to accomplish this.
My first thought is to get only the rows (and NOT the manyList content) in One but not even sure of that in JPA OneManyRepository. And then for each row use it's id primary key to query the Many with that one_id value and do pagination that way.
Any help will be apprecited.
Thanks,Jim
public interface OneManyRepository extends JpaRepository<One, Long> {
#Query(value = "select one.id,one.columnOne,one.columnTwo from one", nativeQuery = true)
Optional<List<One2>> findAllOne();
#Query(value = "select many.someTimeValue,many.value from many where many.one_id = :id", nativeQuery = true)
Optional<List<Many>> findAllBy(#Param("id") long id);
}
#Builder
#AllArgsConstructor
#NoArgsConstructor
public class One2 {
private Long id;
private String columnOne;
private String columnTwo;
}
#Entity
#Table(name = "one")
#Data
#Builder
#AllArgsConstructor
#NoArgsConstructor
public class One {
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Id
private Long id;
private String columnOne;
private String columnTwo;
#OneToMany(cascade = CascadeType.ALL, mappedBy = "one")
private List<Many> manyList;
}
#Entity
#Table(name = "many")
#Data
#Builder
#AllArgsConstructor
#NoArgsConstructor
public class Many {
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Id
private Long id;
#ManyToOne
private One one;
private long someTimeValue;
#Column(length = 10)
private String value;
}
I am using a view in my entity class. Although the view-column mappings are exactly the same as in entity class, I am receiving the exception "ERROR: relation "location_view" does not exist - Position: 15 - underlying exception is SQLGrammarException could not extract ResultSet at org.springframework.orm.jpa.vendor.HibernateJpaDialect.convertHibernateAccessException(HibernateJpaDialect.java:261)"
When I replace the view to the underlying SQL query, it works in SpringBoot. Can you please let me know where I am going wrong?
#Getter
#Setter
#NoArgsConstructor
#AllArgsConstructor
#Data
#ToString
#Entity
#Table(name = "location_view")
public class Location{
#Id
#Column(name = "id")
long id;
#Column(name = "name")
String name;
#Column(name = "total")
long total;
#Column(name = "created_by")
long createdBy;
#Column(name = "created_on")
Timestamp createdOn;
}
public interface LocationDAO extends JpaRepository<Location, Long> {
#Query(value = "select * from location_view", nativeQuery = true)
List<Location> getLocations();
}
In my controller, I am making call to locationDAO.getLocations();
Probably the table/view is in a different schema? Or maybe the user for the JDBC connection has no permission to read from that table/view?
Here's my Reader :
private static final String SELECT_ALL_BOOKS = "SELECT * FROM BOOKS WHERE COLOR = 'yellow'";
#Bean
public JdbcCursorItemReader<BookEntity> itemReader(final DataSource dataSource) {
return new JdbcCursorItemReaderBuilder<BookEntity>()
.name("book_reader")
.sql(SELECT_ALL_BOOKS)
.dataSource(dataSource)
.rowMapper(new BeanPropertyRowMapper<>(BookEntity.class))
.build();
}
And my entity :
#Entity
#Getter
#Setter
#Table(name = "book")
#AllArgsConstructor
#NoArgsConstructor
public class BookEntity implements java.io.Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id_book")
private Integer idBook;
#Column(name = "color")
private String color;
#Column(name = "page_number")
private Integer pageNumber;
#Column(name = "read")
private Boolean read;
#ManyToOne(fetch = FetchType.EAGER, cascade = {CascadeType.ALL})
#JoinColumn(name = "id_author")
private Author author;
...
...
}
Problem is : with my job executing my step, I'm having the books but with an author = null. So the foreign key is not mapped correctly, all the other String/Integer/Boolean fields are retrieved correctly.
I'm new with Spring batch so maybe I'm using a wrong kind of reader.
Any one has an idea about the problem ? Thanks
Please refer this- You are using JDBC Item reader that is mapped to native columns and properties are binded by BeanPropertyRowMapper
https://docs.spring.io/spring-batch/docs/current/reference/html/readersAndWriters.html#JdbcCursorItemReader
Change the query join with AUTHOR tables as a native query or use JPA readers to support ORM
Below reference will give repository reader
https://docs.spring.io/spring-batch/docs/current/api/org/springframework/batch/item/data/RepositoryItemReader.html
Below example give some idea to implement and its SO references
https://github.com/gpassini/Spring-Batch-RepositoryItemReader-NativeQuery-Example/tree/master/src/main
I followed by tutorial : http://www.codejava.net/frameworks/hibernate/hibernate-one-to-one-mapping-with-foreign-key-annotations-example
I have following code:
#Entity
#Table(name = DomainConstant.TABLE_USER)
public class User{
#Id
#Column(name = DomainConstant.DOMAIN_USER_ID)
#GeneratedValue
private Long userId;
private UserActivationCode userActivationCode;
///////////////////// CONSTRUCTOR....
/// STANDARD GET AND SET....
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = DomainConstant.DOMAIN_ACTIVATION_LINK_ID)
public UserActivationCode getUserActivationCode() {
return userActivationCode;
}
}
#Entity
#Table(name = DomainConstant.TABLE_USER_ACTIVATON_LINK)
public class UserActivationCode {
#Id
#Column(name = DomainConstant.DOMAIN_ACTIVATION_LINK_ID)
#GeneratedValue
private Long userActivationCodeId;
#Column(name = DomainConstant.DOMAIN_ACTIVATION_DATE)
#Temporal(javax.persistence.TemporalType.DATE)
private Date date;
#Column(name = DomainConstant.DOMAIN_ACTIVATION_CODE)
private String code;
///////////////////// CONSTRUCTOR....
/// STANDARD GET AND SET....
}
When I save the User object it does not make record in UserActivationCode, why?
Like this:
User newUser = new User();
newUser.setUserActivationCode(new UserActivationCode("this is example"));
userDao.save(newUser);
I have record only in user table.
Can you tell me why?
Your problem is that you are mixing access types. In the User entity you have specified #Id on a field (private Long userId) whereas you have defined the join mapping on a property (the getter to UserActivationCode). If you specify the join mapping on the field, it should work as is.
#Entity
#Table(name = DomainConstant.TABLE_USER)
public class User{
#Id
#Column(name = DomainConstant.DOMAIN_USER_ID)
#GeneratedValue
private Long userId;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = DomainConstant.DOMAIN_ACTIVATION_LINK_ID)
private UserActivationCode userActivationCode;
///////////////////// CONSTRUCTOR....
/// STANDARD GET AND SET....
public UserActivationCode getUserActivationCode() {
return userActivationCode;
}
}
For more information on access and access types, see Access, Java EE 7