Mixing
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Dynamically Query from List or two different entities
1
I have an entity called Person, inside that basic metadata, then inside that Tag and Language. I want to get all rows that contain specific tag name and language.
I came to know about Criteria Query about. How can we interlink two different entities together?
Example: Get all rows having the tag as Model and language as English.
@Entity
public Person {
@Id
private String id;
private BasicMetadata basicMetadata;
-----------
}
Basic Metadata table
@Entity
public BasicMetadata {
@Id
private String id;
private List<Tag> tags;
private List<Language> language;
-------------
}
Tag Table
@Entity
public Tag {
@Id
private String id;
private String name;
-------------
}
Language Table
@Entity
public Language{
@Id
private String id;
private String name;
-------------
}
I created a simple method for specification Query is that correct
private Specification<Person> containsText(String keyword) {
return (root,query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Predicate genreExp = builder.like(builder.lower(root.get("basicMetadata").get("tags")), finalText);
return builder.or(genreExp);
};
java spring-boot jpa hibernate-criteria criteria-api
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
add a comment |
1
I have an entity called Person, inside that basic metadata, then inside that Tag and Language. I want to get all rows that contain specific tag name and language.
I came to know about Criteria Query about. How can we interlink two different entities together?
Example: Get all rows having the tag as Model and language as English.
@Entity
public Person {
@Id
private String id;
private BasicMetadata basicMetadata;
-----------
}
Basic Metadata table
@Entity
public BasicMetadata {
@Id
private String id;
private List<Tag> tags;
private List<Language> language;
-------------
}
Tag Table
@Entity
public Tag {
@Id
private String id;
private String name;
-------------
}
Language Table
@Entity
public Language{
@Id
private String id;
private String name;
-------------
}
I created a simple method for specification Query is that correct
private Specification<Person> containsText(String keyword) {
return (root,query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Predicate genreExp = builder.like(builder.lower(root.get("basicMetadata").get("tags")), finalText);
return builder.or(genreExp);
};
java spring-boot jpa hibernate-criteria criteria-api
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
add a comment |
1
1
1
1
I have an entity called Person, inside that basic metadata, then inside that Tag and Language. I want to get all rows that contain specific tag name and language.
I came to know about Criteria Query about. How can we interlink two different entities together?
Example: Get all rows having the tag as Model and language as English.
@Entity
public Person {
@Id
private String id;
private BasicMetadata basicMetadata;
-----------
}
Basic Metadata table
@Entity
public BasicMetadata {
@Id
private String id;
private List<Tag> tags;
private List<Language> language;
-------------
}
Tag Table
@Entity
public Tag {
@Id
private String id;
private String name;
-------------
}
Language Table
@Entity
public Language{
@Id
private String id;
private String name;
-------------
}
I created a simple method for specification Query is that correct
private Specification<Person> containsText(String keyword) {
return (root,query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Predicate genreExp = builder.like(builder.lower(root.get("basicMetadata").get("tags")), finalText);
return builder.or(genreExp);
};
java spring-boot jpa hibernate-criteria criteria-api
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
I have an entity called Person, inside that basic metadata, then inside that Tag and Language. I want to get all rows that contain specific tag name and language.
I came to know about Criteria Query about. How can we interlink two different entities together?
Example: Get all rows having the tag as Model and language as English.
@Entity
public Person {
@Id
private String id;
private BasicMetadata basicMetadata;
-----------
}
Basic Metadata table
@Entity
public BasicMetadata {
@Id
private String id;
private List<Tag> tags;
private List<Language> language;
-------------
}
Tag Table
@Entity
public Tag {
@Id
private String id;
private String name;
-------------
}
Language Table
@Entity
public Language{
@Id
private String id;
private String name;
-------------
}
I created a simple method for specification Query is that correct
private Specification<Person> containsText(String keyword) {
return (root,query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Predicate genreExp = builder.like(builder.lower(root.get("basicMetadata").get("tags")), finalText);
return builder.or(genreExp);
};
java spring-boot jpa hibernate-criteria criteria-api
java spring-boot jpa hibernate-criteria criteria-api
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
edited Jan 2 at 14:05
Nandakumar Kadavannoore
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
asked Jan 2 at 13:55
Nandakumar KadavannooreNandakumar Kadavannoore
249
249
add a comment |
add a comment |
1 Answer
1
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oldest
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you can write your specification like this
public class PersonSpecifications {
public static Specification<Person> hasTag(String keyword) {
return (root, query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Join<Person, BasicMetaData> md = root.join("basicMetaData");
return builder.like(builder.lower(md.join("tags").get("name")), finalText);
}
}
}
and you can use this specification to get the filtered results like this
repository.findAll(PersonSpecifications. hasTag("abc"),PageRequest,of(0,10));
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
But while using Person repository asPage<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured asCaused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!
– Nandakumar Kadavannoore
Jan 3 at 11:05
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
Yes extending bothPagingAndSortingRepository<Person, String>andJpaSpecificationExecutor<Person>
– Nandakumar Kadavannoore
Jan 4 at 7:04
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }
– Nandakumar Kadavannoore
Jan 7 at 6:42
|
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
you can write your specification like this
public class PersonSpecifications {
public static Specification<Person> hasTag(String keyword) {
return (root, query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Join<Person, BasicMetaData> md = root.join("basicMetaData");
return builder.like(builder.lower(md.join("tags").get("name")), finalText);
}
}
}
and you can use this specification to get the filtered results like this
repository.findAll(PersonSpecifications. hasTag("abc"),PageRequest,of(0,10));
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
But while using Person repository asPage<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured asCaused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!
– Nandakumar Kadavannoore
Jan 3 at 11:05
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
Yes extending bothPagingAndSortingRepository<Person, String>andJpaSpecificationExecutor<Person>
– Nandakumar Kadavannoore
Jan 4 at 7:04
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }
– Nandakumar Kadavannoore
Jan 7 at 6:42
|
show 2 more comments
0
you can write your specification like this
public class PersonSpecifications {
public static Specification<Person> hasTag(String keyword) {
return (root, query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Join<Person, BasicMetaData> md = root.join("basicMetaData");
return builder.like(builder.lower(md.join("tags").get("name")), finalText);
}
}
}
and you can use this specification to get the filtered results like this
repository.findAll(PersonSpecifications. hasTag("abc"),PageRequest,of(0,10));
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
But while using Person repository asPage<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured asCaused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!
– Nandakumar Kadavannoore
Jan 3 at 11:05
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
Yes extending bothPagingAndSortingRepository<Person, String>andJpaSpecificationExecutor<Person>
– Nandakumar Kadavannoore
Jan 4 at 7:04
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }
– Nandakumar Kadavannoore
Jan 7 at 6:42
|
show 2 more comments
0
0
0
you can write your specification like this
public class PersonSpecifications {
public static Specification<Person> hasTag(String keyword) {
return (root, query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Join<Person, BasicMetaData> md = root.join("basicMetaData");
return builder.like(builder.lower(md.join("tags").get("name")), finalText);
}
}
}
and you can use this specification to get the filtered results like this
repository.findAll(PersonSpecifications. hasTag("abc"),PageRequest,of(0,10));
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
you can write your specification like this
public class PersonSpecifications {
public static Specification<Person> hasTag(String keyword) {
return (root, query, builder) -> {
String finalText = keyword.toLowerCase();
if (!finalText.contains("%")) {
finalText = "%" + finalText + "%";
}
Join<Person, BasicMetaData> md = root.join("basicMetaData");
return builder.like(builder.lower(md.join("tags").get("name")), finalText);
}
}
}
and you can use this specification to get the filtered results like this
repository.findAll(PersonSpecifications. hasTag("abc"),PageRequest,of(0,10));
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
edited Jan 7 at 12:53
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
answered Jan 2 at 14:36
Reza NasiriReza Nasiri
6211116
6211116
But while using Person repository asPage<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured asCaused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!
– Nandakumar Kadavannoore
Jan 3 at 11:05
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
Yes extending bothPagingAndSortingRepository<Person, String>andJpaSpecificationExecutor<Person>
– Nandakumar Kadavannoore
Jan 4 at 7:04
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }
– Nandakumar Kadavannoore
Jan 7 at 6:42
|
show 2 more comments
But while using Person repository asPage<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured asCaused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!
– Nandakumar Kadavannoore
Jan 3 at 11:05
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
Yes extending bothPagingAndSortingRepository<Person, String>andJpaSpecificationExecutor<Person>
– Nandakumar Kadavannoore
Jan 4 at 7:04
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }
– Nandakumar Kadavannoore
Jan 7 at 6:42
But while using Person repository as
Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured as Caused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!– Nandakumar Kadavannoore
Jan 3 at 11:05
But while using Person repository as
Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest);. After running the application error occured as Caused by: org.springframework.data.mapping.PropertyReferenceException: No property findAll found for type Person!– Nandakumar Kadavannoore
Jan 3 at 11:05
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
is your repository extending JpaSpecificationExecutor?
– Reza Nasiri
Jan 3 at 12:40
Yes extending both
PagingAndSortingRepository<Person, String> andJpaSpecificationExecutor<Person>– Nandakumar Kadavannoore
Jan 4 at 7:04
Yes extending both
PagingAndSortingRepository<Person, String> andJpaSpecificationExecutor<Person>– Nandakumar Kadavannoore
Jan 4 at 7:04
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
that doesn't seem right. can you share your repository code as well as the code that causing the exception?
– Reza Nasiri
Jan 4 at 14:32
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }– Nandakumar Kadavannoore
Jan 7 at 6:42
public interface PersonRepository extends CrudRepository<Person, String>, JpaSpecificationExecutor<Person> { Page<Person> findAll(Specification<Person> containsText, Pageable pageRequest); Page<Person> findAll(Pageable pageable); }– Nandakumar Kadavannoore
Jan 7 at 6:42
|
show 2 more comments
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