jpa

How to mock EntityManager?

让人想犯罪 __ 提交于 2020-12-30 04:50:07
问题 I need to mock entity-manager to make testing service layer (in my case a session facade) to be independent of the underlying layer (which in my case is the entity-manager). So how I can accomplish this? Should I use dbunit? Do I need easy/j(Mock)? 回答1: I suggest to use Mockito Framework it is very easy to use and understand. @Mock private EntityManager entityManager; If you want to use any method that belongs to entityManager, you should call. Mockito.when(METHOD_EXPECTED_TO_BE_CALLED)

How to mock EntityManager?

你。 提交于 2020-12-30 04:49:43
问题 I need to mock entity-manager to make testing service layer (in my case a session facade) to be independent of the underlying layer (which in my case is the entity-manager). So how I can accomplish this? Should I use dbunit? Do I need easy/j(Mock)? 回答1: I suggest to use Mockito Framework it is very easy to use and understand. @Mock private EntityManager entityManager; If you want to use any method that belongs to entityManager, you should call. Mockito.when(METHOD_EXPECTED_TO_BE_CALLED)

How to save parent and child in one shot (JPA & Hibernate)

本小妞迷上赌 提交于 2020-12-29 11:57:45
问题 I start showing you my scenario. This is my parent object: @Entity @Table(name="cart") public class Cart implements Serializable{ @GeneratedValue(strategy=GenerationType.IDENTITY) @Id @Column(name="id") private Integer id; @OneToMany(mappedBy="cart", fetch = FetchType.EAGER, cascade = CascadeType.ALL) private List<CartItem> cartItems; ... } This is my child object: @Entity @Table(name="cart_item") public class CartItem implements Serializable{ @GeneratedValue(strategy=GenerationType.IDENTITY)

Spring JPA :: No converter found capable of converting from type

心不动则不痛 提交于 2020-12-28 06:43:26
问题 I am using spring with JPA and trying to execute query using @query with SQL query.and trying to map the result to an object. I have different entity class and mapping to other dto as I do not want all the columns. getting below error No converter found capable of converting from type [org.springframework.data.jpa.repository.query.AbstractJpaQuery$TupleConverter$TupleBackedMap] to type [webapi.dto.StatsDTO]", Below is my repository interface public interface UserRepository extends

Spring JPA :: No converter found capable of converting from type

余生长醉 提交于 2020-12-28 06:43:25
问题 I am using spring with JPA and trying to execute query using @query with SQL query.and trying to map the result to an object. I have different entity class and mapping to other dto as I do not want all the columns. getting below error No converter found capable of converting from type [org.springframework.data.jpa.repository.query.AbstractJpaQuery$TupleConverter$TupleBackedMap] to type [webapi.dto.StatsDTO]", Below is my repository interface public interface UserRepository extends

Spring JPA :: No converter found capable of converting from type

心不动则不痛 提交于 2020-12-28 06:42:44
问题 I am using spring with JPA and trying to execute query using @query with SQL query.and trying to map the result to an object. I have different entity class and mapping to other dto as I do not want all the columns. getting below error No converter found capable of converting from type [org.springframework.data.jpa.repository.query.AbstractJpaQuery$TupleConverter$TupleBackedMap] to type [webapi.dto.StatsDTO]", Below is my repository interface public interface UserRepository extends

How to use listagg with querydsl?

怎甘沉沦 提交于 2020-12-27 05:38:31
问题 I have a query as shown below; SELECT . . LISTAGG(DISTINCT CC.POPULATION, ', ') WITHIN GROUP (ORDER BY CC.POPULATION ASC), . . FROM COUNTRY C JOIN CITY CC ON C.ID = CC.COUNTRY_ID --WHERE GROUP BY C.ID; I should be implement with querydsl for custom filtering and sorting operations but I got "No pattern found for LISTAGG" error JPAQuery<Tuple> jpaQuery = jpaQueryFactory.select( SQLExpressions.listagg(QCity.city.POPULATION, ",") .withinGroup() .orderBy(QCity.city.POPULATION.asc()) ) .from

foreign key must have same number of columns as the referenced primary key for manytoone mapping

两盒软妹~` 提交于 2020-12-27 04:05:21
问题 Hi below is my entities with manytoone association between them student.java @Entity @Table(name = "student") public class student{ @Id @Column(name = "UserID") private String userid; @ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.LAZY) @JoinColumns({ @JoinColumn(name = "userrole", referencedColumnName = "VALUE"), @JoinColumn(name = "userrole", referencedColumnName = "DESCRIPTION") }) private studentdetails userrole; //setters and getters //constructor } studentdetails.java @Data

foreign key must have same number of columns as the referenced primary key for manytoone mapping

*爱你&永不变心* 提交于 2020-12-27 04:01:31
问题 Hi below is my entities with manytoone association between them student.java @Entity @Table(name = "student") public class student{ @Id @Column(name = "UserID") private String userid; @ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.LAZY) @JoinColumns({ @JoinColumn(name = "userrole", referencedColumnName = "VALUE"), @JoinColumn(name = "userrole", referencedColumnName = "DESCRIPTION") }) private studentdetails userrole; //setters and getters //constructor } studentdetails.java @Data

foreign key must have same number of columns as the referenced primary key for manytoone mapping

无人久伴 提交于 2020-12-27 03:59:42
问题 Hi below is my entities with manytoone association between them student.java @Entity @Table(name = "student") public class student{ @Id @Column(name = "UserID") private String userid; @ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.LAZY) @JoinColumns({ @JoinColumn(name = "userrole", referencedColumnName = "VALUE"), @JoinColumn(name = "userrole", referencedColumnName = "DESCRIPTION") }) private studentdetails userrole; //setters and getters //constructor } studentdetails.java @Data