I have an entity:
#Entity
public class MailingList extends PanacheEntity{
public String name;
public String email;
public Organization organization;
.
.
.
}
So, I am trying to write a query to simply return me a list having the id and email only but not a list of the MailingList entities. I am using MySQL and I want something as shown here. I have read through different documentation but I haven't found something to solve my problem. Please help out.
Yes,
With panache it is possible to use DTO projections like in hibernate or jpa.
Here you have an example from the quarkus panache documentation on how to achieve a projection of only certain fields.
Related
required format image
I want to object data into MongoDB using spring and I have hardcoded it.
please how to write a schema for that and I have taken it as an example only.
I have a different type of categories in it I have taken only clothes.
please tell me how to write one schema for a different type of categories and query too.
please find the attachment for your reference
I would recommend going though Spring Data MongoDB documentation for specifics on mapping java objects to MongoDB documents. Your case would look similar to:
#Document
public class Clothes {
#Id
private ObjectId id;
private Men men;
private Women women;
// getters & setters
}
You would need to define each sub class but this should be the gist of it.
What you can do is create a simple POJO (Plain Old Java Object) and with that you can insert that object into the data base. The the following example:
#Document
public class OAuthModel implements Serializable {
#Id
String username;
#Indexed
String oAuthID;
#Indexed
String type;
// Getter and Setters and Construct.
}
When I insert this object in the DB by calling:
OAuthModel authModel = new OAuthModel(username,firebaseToken.getUid(), OAuthHosts.GOOGLE.getType());
oAuthRepo.insert(authModel);
It will then be seen as this in the Database:
Keep in mind this will work no matter what your object looks like, you can have hashmaps etc. The should be a built in serialization.
I'm trying to get a page of a partial entity (NetworkSimple) using the new feature of spring data, projections
I've checked the documentation and if I request only:
Collection<NetworkSimple> findAllProjectedBy();
It works, but if I'm using pageable:
Page<NetworkSimple> findAllProjectedBy(Pageable pageable);
It throws an error:
org.hibernate.jpa.criteria.expression.function.AggregationFunction$COUNT cannot be cast to org.hibernate.jpa.criteria.expression.CompoundSelectionImpl
Any one has already work with this ?
My NetworkSimple class is the following:
public interface NetworkSimple {
Long getId();
String getNetworkName();
Boolean getIsActive();
}
Note: This feature should work in the way described by the original poster but due to this bug it didn't. The bug has been fixed for the Hopper SR2 release, if you're stuck on an earlier version then the workaround below will work.
It is possible to use Pageable with the new query projection features introduced in Spring Data JPA 1.10 (Hopper). You will need to use the #Query annotation and manually write a query for the fields you require, they must also be aliased using AS to allow Spring Data to figure out how to project the results. There is a good example in spring-boot-samples part of the spring boot repository.
In your example it would be quite simple:
#Query("SELECT n.id AS id, n.name AS networkName, n.active AS isActive FROM Network n")
Page<NetworkSimple> findAllProjectedBy(Pageable pageable);
I have made the assumption that your entity looks something like this:
#Entity
public class Network
{
#Id
#GeneratedValue
private Long id;
#Column
private String name;
#Column
private boolean active;
...
}
Spring Data will derive a count query automatically for the paging information. It is also possible to make use of joins in the query to fetch associations and then summarise them in the projection.
I think you need create findAllProjectedBy() as specification.Then you can use findAll() method like this.
example :findAll(findAllProjectedBy(),pageable)
Following link may be help to find how to create specification in spring.
https://spring.io/blog/2011/04/26/advanced-spring-data-jpa-specifications-and-querydsl/
The issue may come from the method name. The by keyword means that you ae filterig data by a specific property: findByName for example. Its called query creation from method name:
http://docs.spring.io/spring-data/jpa/docs/1.10.1.RELEASE/reference/html/#repositories.query-methods.query-creation
So try with Page<NetworkSimple> findAll(Pageable pageable);
Even with spring-data-jpa 1.11.4, something like
public interface NetworkRepository extends JpaRepository<Network, String> {
Page<NetworkSimple> findAll(Pageable pageable);
}
would not compile; reporting
findAll(org.springframework.data.domain.Pageable) in NetworkRepository clashes with findAll(org.springframework.data.domain.Pageable) in org.springframework.data.repository.PagingAndSortingRepository
return type org.springframework.data.domain.Page<NetworkSimple> is not compatible with org.springframework.data.domain.Page<Network>
The workaround we found was to rename findAll to findAllBy, e.g.
public interface NetworkRepository extends JpaRepository<Network, String> {
Page<NetworkSimple> findAllBy(Pageable pageable);
}
You can use interface projection with Pageable like this :
Page<NetworkSimple> findPagedProjectedBy(Pageable pageable);
with some parameter :
Page<NetworkSimple> findPagedProjectedByName(String name, Pageable pageable);
Implementing interface projection with pagination
1. Our ResourceEntity.java class
#Getter
#Setter
#NoArgsConstructor
#Entity
public class ResourceEntity{
private Long id;
private String name;
}
2. Creating projection Interface name ProjectedResource.java, which maps data collected by the SQL query from repository layer method
public interface ProjectedResource {
Long getId();
String getName();
String getAnotherProperty();
}
3. Creating Repository layer method: getProjectedResources()
We are considering the database table name is resource.
We are only fetching id and name here.
#Query(name="select id, name, anotherProperty from resource", countQuery="select count(*) from resource", nativeQuery=true)
Page<ProjectedResource> getProjectedResources(Pageable page);
Hope the issue will be resolved!
You can use:
#Query("SELECT n FROM Network n")
Page<? extends NetworkSimple> findAllProjectedBy(Pageable pageable);
Is it possible to publish two different repositories for the same JPA entity with Spring Data Rest?
I gave the two repositories different paths and rel-names, but only one of the two is available as REST endpoint.
The point why I'm having two repositories is, that one of them is an excerpt, showing only the basic fields of an entity.
The terrible part is not only that you can only have 1 spring data rest repository (#RepositoryRestResource) per Entity but also that if you have a regular JPA #Repository (like CrudRepository or PagingAndSorting) it will also interact with the spring data rest one (as the key in the map is the Entity itself).
Lost quite a few hours debugging random load of one or the other. I guess that if this is a hard limitation of spring data rest at least an Exception could be thrown if the key of the map is already there when trying to override the value.
The answer seems to be: There is only one repository possible per entity.
I ended up using the #Subselect to create a second immutable entity and bound that to the second JpaRepsotory and setting it to #RestResource(exported = false), that also encourages a separation of concerns.
Employee Example
#Entity
#Table(name = "employee")
public class Employee {
#Id
Long id
String name
...
}
#RestResource
public interface EmployeeRepository extends PagingAndSortingRepository<Employee, Long> {
}
#Entity
#Immutable
#Subselect(value = 'select id, name, salary from employee')
public class VEmployeeSummary {
#Id
Long id
...
}
#RestResource(exported = false)
public interface VEmployeeRepository extends JpaRepository<VEmployeeSummary, Long> {
}
Context
Two packages in the monolithic application had different requirements. One needed to expose the entities for the UI in a PagingAndSortingRepository including CRUD functions. The other was for an aggregating backend report component without paging but with sorting.
I know I could have filtered the results from the PagingAndSorting Repository after requesting Pageable.unpaged() but I just wanted a Basic JPA repository which returned List for some filters.
So, this does not directly answer the question, but may help solve the underlying issue.
You can only have one repository per entity... however, you can have multiple entities per table; thus, having multiple repositories per table.
In a bit of code I wrote, I had to create two entities... one with an auto-generated id and another with a preset id, but both pointing to the same table:
#Entity
#Table("line_item")
public class LineItemWithAutoId {
#Id
#GeneratedValue(generator = "system-uuid")
#GenericGenerator(name = "system-uuid", strategy = "uuid")
private String id;
...
}
#Entity
#Table("line_item")
public class LineItemWithPredefinedId {
#Id
private String id;
...
}
Then, I had a repository for each:
public interface LineItemWithoutId extends Repository<LineItemWithAutoId,String> {
...
}
public interface LineItemWithId extends Repository<LineItemWithPredefinedId,String> {
...
}
For the posted issue, you could have two entities. One would be the full entity, with getters and setters for everything. The other, would be the entity, where there are setters for everything, but only getters for the fields you want to make public. Does this make sense?
I got a problem with simple Spring Data issue. Let's assume we got two entities.
public class Request {
// all normal stuff
#ManyToOne
private Document doc;
}
public class Document {
private Long id;
private String name;
}
Simple relation. My question is - is it possible to retrieve Request entities using Spring Data Method-DSL and sorting by Document? So what I want to achieve is to create repository method like:
public List<Request> findAllOrderByDoc()
or similar:
public List<Request> findAllOrderByDocId()
Unfortunately when I try that I am given error message saying that there is no Doc field or it cannot be mapped to long. I assume it is possible to be done using QueryDSL and predicates but I am wondering if this pretty obvious and simple thing can be done by plain Spring Data?
Yes, sure.
you need to provide the direction:
public List<Request> findAllOrderByDocAsc()
public List<Request> findAllOrderByDocDesc()
I am trying to relate two tables with spring / hibernate in MYSQL like this
#Table (name = candidatresumeinfo)
public class CandidateResumeInfo implements Serializable
{
List<SelectedResumes> selectedResumes;
.............
..............
#JoinColumn(name = "selectedresumeid")
#OneToMany
public List<SelectedResumes> getSelectedResumes() {
return selectedResumes;
}
public void setSelectedResumes(List<SelectedResumes> selectedResumes) {
this.selectedResumes = selectedResumes;
}
Now ,i got the data in my list correctly( i checked in debug)but the call from server is getting failed which is saying cause:Nullpointer exception .
thanks
You can use OneToMany annotation only on Collections, so you should change the field to Set or List, because hibernate will return multiple result if you use OneToMany. I think you'd like to use ManyToOne annotation here.
ManyToOne means here that you have multiple CandidateResumeInfo for one SelectedResumes.
OneToMany means here that you have multiple SelectedResumes for one CandidateResumeInfo.
This annotation naming can be a bit strange for first time. Hope I helped.
Answer for your comment:
The best way is you declare the relationship both side.
Here is the example:
CandidateResumeInfo.java:
#OneToMany(mappedBy="candidateResumeInfo")
List<SelectedResumes> selectedResumes;
SelectedResumes.java:
#ManyToOne
#JoinColumn(name="candidate_resume_info_id")
CandidateResumeInfo candidateResumeInfo;