I'm using Spring Boot 2, Spring JPA, Spring Data.
I'm trying to get the first result from a table in which I want to sort results by a int property asc.
This is my bean:
#Entity
public class DatabaseInstance extends AbstractEntity {
#NotNull
#Enumerated(EnumType.STRING)
#Column(nullable = false)
private Supplier supplier;
NotNull
#Column(nullable = false, columnDefinition = "INT DEFAULT 0.0")
private int databaseCount = 0;
and this is my repository:
#Transactional
public interface DatabaseInstanceRepository extends JpaRepository<DatabaseInstance, Long> {
public Optional<DatabaseInstance> findFirstOrderByDatabaseCountAsc();
}
According to Spring manual I can order by property name + asc word. At the same time I want the first result (so i get the instance with the lower databaseCount value).
I get this error:
Caused by: org.springframework.data.mapping.PropertyReferenceException: No property asc found for type int! Traversed path: DatabaseInstance.databaseCount.
I am not able to figure out what's wrong with my method name. Some advice?
Related
i'm using Spring Boot 2.4.2 and Data module for JPA implementation.
Now, i'm using an Oracle View, mapped by this JPA Entity:
#Entity
#Immutable
#Table(name = "ORDER_EXPORT_V")
#ToString
#Data
#NoArgsConstructor
#EqualsAndHashCode(onlyExplicitlyIncluded = true)
public class OrderExportView implements Serializable {
private static final long serialVersionUID = -4417678438840201704L;
#Id
#Column(name = "ID", nullable = false)
#EqualsAndHashCode.Include
private Long id;
....
The view uses an UNION which allows me to obtain two different attributes of the same parent entity, so for one same parent entity (A) with this UNION I get the attribute B in row 1 and attribute C in row 2: this means that the rows will be different from each other.
If I run the query with an Oracle client, I get the result set I expect: same parent entity with 2 different rows containing the different attributes.
Now the issue: when I run the query with Spring Data (JPA), I get the wrong result set: two lines but duplicate.
In debug, I check the query that perform Spring Data and it's correct; if I run the same query, the result set is correct, but from Java/Spring Data not. Why??
Thanks for your support!
I got it! I was wrong in the ID field.
The two rows have the same parent id, which is not good for JPA, which instead expects a unique value for each line.
So, now I introduced a UUID field into the view:
sys_guid() AS uuid
and in JPA Entity:
#Id
#Column(name = "UUID", nullable = false)
#EqualsAndHashCode.Include
private UUID uuid;
#Column(name = "ID")
private Long id;
and now everything works fine, as the new field has a unique value for each row.
Using javers 5.11.2 I get the following exception although the id is set to be ignored. Why is that?
JaversException ENTITY_INSTANCE_WITH_NULL_ID: Found Entity instance 'my.package.javers.Leaf' with null Id-property 'id'
Update: I learned that
JaVers matches only objects with the same GlobalId
The id is specified using javax.persistence.Id. However, with each ORM it is possible to have an entity with a collection, then add a new element without id to that entity and then save it (CascadeType.Persist).
Is there any way to compare objects with javers in such a case?
Example (used lombok for boiler plate code).
The leaf:
#AllArgsConstructor
#NoArgsConstructor
#Builder
#Data
#Entity
public class Leaf {
#DiffIgnore <============ id is ignored
#Id
private Long id;
private String color;
}
The tree:
#Builder
#Data
#Entity
public class Tree {
#Id
private Long id;
private String name;
#OneToMany
private Set<Leaf> leafs;
}
Test adds a leaf to the oakSecond without an id set. The diff cannot be made. An Exception is thrown.
#Test
public void testCompare_AddLeafToTree() {
Leaf leaf = Leaf.builder().id(1L).color("11").build();
Set<Leaf> leafsOfOakFirst = new HashSet<>();
leafsOfOakFirst.add(leaf);
Tree oakFirst = Tree.builder().id(1L).name("oakFirst").build();
oakFirst.setLeafs(leafsOfOakFirst);
Set<Leaf> leafsOfOakSecond = new HashSet<>();
leafsOfOakSecond.add(leaf);
leafsOfOakSecond.add(Leaf.builder().color("12").build());
Tree oakSecond = Tree.builder().id(1L).name("oakFirst").build();
oakSecond.setLeafs(leafsOfOakSecond);
Javers javers = JaversBuilder.javers().build();
Changes changes = javers.compare(oakFirst, oakSecond).getChanges();
assertThat(changes).isNotEmpty();
}
Same with the following definition of the Javers instance:
EntityDefinition leafEntityDefinition = EntityDefinitionBuilder.entityDefinition(Leaf.class).withIgnoredProperties("id").build();
Javers javers = JaversBuilder.javers().registerEntity(leafEntityDefinition).build();
You can't pass an Entity to Javers with null Id because it would be non-identifiable. If you use Hibernate to generate your Ids, make sure that you pass your object to javers.commit() after hibernate are done with its job. That's how the #JaversSpringDataAuditable aspect works.
Alternatively, you can model those objects with unstable IDs as Value Object in Javers.
I'd like to find all Offer documents by Offer.ProductProperties.brand:
#Document(collection = "offers")
public class Offer {
#Id
private String id;
#NotNull
#DBRef
private ProductProperties properties;
ProductProperties:
#Document(collection = "product_properties")
public class ProductProperties {
#Id
private String id;
#NotNull
#NotEmpty
private String brand;
Service:
Flux<ProductProperties> all = productPropertiesRepository.findAllByBrand(brand);
List<String> productPropIds = all.toStream()
.map(ProductProperties::getId)
.collect(Collectors.toList());
Flux<Offer> byProperties = offerRepository.findAllByProperties_Id(productPropIds);
But unfortunately byProperties is empty. Why?
My repository:
public interface OfferRepository extends ReactiveMongoRepository<Offer, String> {
Flux<Offer> findAllByProperties_Id(List<String> productPropertiesIds);
}
How to find all Offers by ProductProperties.brand?
Thanks!
After reading some documentation found out that You cannot query with #DBRef. Hence the message
Invalid path reference properties.brand! Associations can only be
pointed to directly or via their id property
If you remove DBRef from the field, you should be able to query by findAllByProperties_BrandAndProperties_Capacity.
So the only ways is how you are doing. i.e. Fetch id's and query by id.
As I said in the comment, the reason it is not working is because return type of findAllByProperties_Id is a Flux. So unless u execute a terminal operation, you wont have any result. Try
byProperties.collectList().block()
As Spring jpa Provides some usefull features to find Items from a repository by defining it in the method name. e .x findByTitle(String title) then Spring is automatically searching the Title Colum for the given String. If i have an int column named numberOfCopies and i want only to find the datasets with >0 greater then null how would define such a method ?
to filter out those books with the numberOfCopies equals 0 = zero
#Entity
public class Book {
#Id
private int id;
private String title;
private int numberOfCopies;
}
can i use the Repomethod
public List findBooksByNumberOfCopies.greater then 0 ?To Use this Spring Feature without some if or for loops
First, you should use Integer, since it is better, in my opinion, to use wrapper classes than to primitives, and enforce not null requirement through annotations, e.g. #Column(nullable = false)
#Entity
public class Book {
#Id
private Integer id;
private String title;
private Integer numberOfCopies;
}
Then you can add the following two methods in your BookRepository;
List<Book> findByNumberOfCopiesGreaterThan(Integer numberOfCopies);
default List<Book> findAllAvailableBooks() {
return findByNumberOfCopiesGreaterThan(0);
}
and use the default findAllAvailableBooks method, with hardcoded 0 value which is your requirement.
you can easily use
List<Book> findByNumberOfCopiesGreaterThanEqual(int numberOfCopies);
Pretty sure this would work:
public interface BookRepo extends JpaRepository<Book, Integer> {
#Query("SELECT b FROM Book b WHERE b.numberOfCopies >= 0")
public Optional<List<Book>> getTheBooksWithMultCopies();
}
// back in your component class:
{
...
Optional<List<Book>> optionalBookList = myBookRepo.getTheBooksWithMultCopies();
if (optionalBookList.isPresent()){
List<Book> bookList = optionalBookList.get();
}
}
Note that the language within the query is called HQL, which is what is used by Hibernate internally (which is used by JPA internally). It's really not very intimidating - just, know that you the objects in your POJO, which map to your database table, rather than your database table directly.
Also, I'd recommend using Integer over int in entity classes, at least if your value is nullable. Otherwise, numberOfCopies will always default to 0, which may not be desirable and may cause exceptions that are difficult to decipher.
GreaterThanEqual takes an Integer not int
List<Book> findByNumberOfCopiesGreaterThanEqual(Integer numberOfCopies);
I have a database service using Spring Boot 1.5.1 and Spring Data Rest. I am storing my entities in a MySQL database, and accessing them over REST using Spring's PagingAndSortingRepository. I found this which states that sorting by nested parameters is supported, but I cannot find a way to sort by nested fields.
I have these classes:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
#ManyToOne
protected Address address;
#ManyToOne(targetEntity = Name.class, cascade = {
CascadeType.ALL
})
#JoinColumn(name = "NAME_PERSON_ID")
protected Name name;
#Id
protected Long id;
// Setter, getters, etc.
}
#Entity(name = "Name")
#Table(name = "NAME")
public class Name{
protected String firstName;
protected String lastName;
#Id
protected Long id;
// Setter, getters, etc.
}
For example, when using the method:
Page<Person> findByAddress_Id(#Param("id") String id, Pageable pageable);
And calling the URI http://localhost:8080/people/search/findByAddress_Id?id=1&sort=name_lastName,desc, the sort parameter is completely ignored by Spring.
The parameters sort=name.lastName and sort=nameLastName did not work either.
Am I forming the Rest request wrong, or missing some configuration?
Thank you!
The workaround I found is to create an extra read-only property for sorting purposes only. Building on the example above:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
// read only, for sorting purposes only
// #JsonIgnore // we can hide it from the clients, if needed
#RestResource(exported=false) // read only so we can map 2 fields to the same database column
#ManyToOne
#JoinColumn(name = "address_id", insertable = false, updatable = false)
private Address address;
// We still want the linkable association created to work as before so we manually override the relation and path
#RestResource(exported=true, rel="address", path="address")
#ManyToOne
private Address addressLink;
...
}
The drawback for the proposed workaround is that we now have to explicitly duplicate all the properties for which we want to support nested sorting.
LATER EDIT: another drawback is that we cannot hide the embedded property from the clients. In my original answer, I was suggesting we can add #JsonIgnore, but apparently that breaks the sort.
I debugged through that and it looks like the issue that Alan mentioned.
I found workaround that could help:
Create own controller, inject your repo and optionally projection factory (if you need projections). Implement get method to delegate call to your repository
#RestController
#RequestMapping("/people")
public class PeopleController {
#Autowired
PersonRepository repository;
//#Autowired
//PagedResourcesAssembler<MyDTO> resourceAssembler;
#GetMapping("/by-address/{addressId}")
public Page<Person> getByAddress(#PathVariable("addressId") Long addressId, Pageable page) {
// spring doesn't spoil your sort here ...
Page<Person> page = repository.findByAddress_Id(addressId, page)
// optionally, apply projection
// to return DTO/specifically loaded Entity objects ...
// return type would be then PagedResources<Resource<MyDTO>>
// return resourceAssembler.toResource(page.map(...))
return page;
}
}
This works for me with 2.6.8.RELEASE; the issue seems to be in all versions.
From Spring Data REST documentation:
Sorting by linkable associations (that is, links to top-level resources) is not supported.
https://docs.spring.io/spring-data/rest/docs/current/reference/html/#paging-and-sorting.sorting
An alternative that I found was use #ResResource(exported=false).
This is not valid (expecially for legacy Spring Data REST projects) because avoid that the resource/entity will be loaded HTTP links:
JacksonBinder
BeanDeserializerBuilder updateBuilder throws
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of ' com...' no String-argument constructor/factory method to deserialize from String value
I tried activate sort by linkable associations with help of annotations but without success because we need always need override the mappPropertyPath method of JacksonMappingAwareSortTranslator.SortTranslator detect the annotation:
if (associations.isLinkableAssociation(persistentProperty)) {
if(!persistentProperty.isAnnotationPresent(SortByLinkableAssociation.class)) {
return Collections.emptyList();
}
}
Annotation
#Retention(RetentionPolicy.RUNTIME)
#Target(ElementType.FIELD)
public #interface SortByLinkableAssociation {
}
At project mark association as #SortByLinkableAssociation:
#ManyToOne
#SortByLinkableAssociation
private Name name;
Really I didn't find a clear and success solution to this issue but decide to expose it to let think about it or even Spring team take in consideration to include at nexts releases.
Please see https://stackoverflow.com/a/66135148/6673169 for possible workaround/hack, when we wanted sorting by linked entity.