Develop Reference

Mapstruct: How to merge two fields into one

Using the MapStruct framework, how do you map multiple fields into a single one (based on custom logic) while still mapping the other fields one to one?
Here is a simple example to illustrate what I mean:
public class Source {
private String firstname;
private String surname;
// other fields eg:
private String address;
private int age;
private int favoriteNumber;
}
public class Target {
private String fullname; // Sould be firstname + surname
// other fields eg:
private String address;
private int age;
private int favoriteNumber;
}
I know it's possible using expressions:
#Mapping(target = "fullname", expression = "java(el.getFirstname() + el.getSurname())")
But in my special use case, not depicted in this example, I need to use some external library for the merging/mapping of the two fields, which isn't feasible with expressions.
Is there a way to achieve merging two fields without expressions?

One approach would be to add a custom mapping method from the Source object to the merged value, and then declare the source for the merged field to be the whole source object:
interface CustomMappingMethodMapper {
#Mapping(target = "fullname", source = ".")
Target map(Source source);
default String getFullName(Source s) {
return s.getFirstname() + " " + s.getSurname();
}
}

You can use #AfterMapping annotation
https://mapstruct.org/documentation/stable/reference/html/#customizing-mappings-with-before-and-after
You would like to replace your interface with abstract class and then
#AfterMapping
void customMapping(#MappingTarget Target target, Source source) {
// any custom logic
}

You claim that calling an external library in an expression isn't feasible. This may not be true, depending on the nature of the library you're calling and the frameworks being used.
Static method
If the method being called is a static method on a class, it can be called directly within the expression annotation element of #Mapping. To avoid having to fully qualify the class being called, the imports element of #Mapper can be used.
#Mapper(imports = ExternalLibrary.class)
public interface SourceToTargetMapper {
#Mapping(target = "fullname",
expression = "java(ExternalLibrary.toFullName(s.getFirstname(), s.getSurname()))")
Target map(Source s);
}
Spring Framework bean
If the library method is a method on a Spring bean, the mapper can be made into a bean using the Spring component model, and the Spring bean containing the library method can be injected into the mapper:
#Mapper(componentModel = "spring")
public static abstract class SourceToTargetMapper {
#Autowired
ExternalLibrary externalLibrary;
#Mapping(target = "fullname",
expression = "java(externalLibrary.toFullName(s.getFirstname(), s.getSurname()))")
abstract Target map(Source s);
}
To use this mapper, inject it as a Spring bean, and call the mapping method on the bean:
#Component
public class Example {
#Autowired
private SourceToTargetMapper mapper;
public void demonstrate(Source s) {
System.out.println(mapper.map(s));
}
}
I haven't tested it myself, but I imagine this injection approach would work with the other component models supported by MapStruct (cdi & jsr330).

I needed to compose 2 basic fields into am Wrapped Object. Here's how I did id
#Mapping(source = "quantity", target = "energyQuantity.quantity")
#Mapping(source = "quantityUnit", target = "energyQuantity.energyQuantityUnit", qualifiedByName = "tradeSearchEnergyQuantityUnitToEnergyQuantityUnit")
Trade tradeSearchToDomainDTO(api.client.generated.model.Trade trade);
where energyQuantity is a Wrapper class for the 2 fields

Spring boot elastic search repository

I have several indices where I save products:
product-example1
product-example2
product-example3
product-example4
product-example5
I have the a document in elastic search that has the same structure and it can used for different indices:
#Data
#org.springframework.data.elasticsearch.annotations.Document(indexName = "", type = "", createIndex = false)
public class ProductDocument {
#Id
private String id;
private String title;
private String seller;
private String releaseDate;
....
}
So basically I want to use same settings, same mappings same search service.
So I made indexName and type parametric in spring boot java, instead of creating 5 classes extending ProductDocument ?
#Autowired
private final ElasticsearchTemplate elasticsearchTemplate;
this.elasticsearchTemplate.createIndex("product-example1", loadFile("/files/settings.json"));
this.elasticsearchTemplate.putMapping("product-example1", "product-type1", loadFile("/files/mapping.json"));
this.elasticsearchTemplate.createIndex("product-example2", loadFile("/files/settings.json"));
this.elasticsearchTemplate.putMapping("product-example2", "product-type2", loadFile("/files/mapping.json"));
......
Now I want to create a ProductRepository but I don't have a class with defined index name. If I use generic class:
public interface DocumentRepository extends ElasticsearchRepository<ProductDocument, String> {
}
I get the error which is totally understable cause I created the index names in dynamic way:
lang.IllegalArgumentException: Unknown indexName. Make sure the indexName is defined. e.g #ProductDocument(indexName="foo")
So is it possible somehow to create repository for indexes that I created in dynamic way as described above, and pass the index name and type as parameter ?
Please help! I'm stuck.

AWS DynamoDBMapper save method keeps throwing `DynamoDBMappingException: not supported; requires #DynamoDBTyped or #DynamoDBTypeConverted`

I followed everything that is outlined here - https://github.com/derjust/spring-data-dynamodb/wiki/Use-Hash-Range-keys. But still no luck.
I have a DynamoDB table with a hash key and a sort key.
Here is my entity class RecentlyPlayed.class
#DynamoDBTable(tableName="some-table")
public class RecentlyPlayed {
#Id
private RecentlyPlayedId recentlyPlayedId;
// ----- Constructor methods -----
#DynamoDBHashKey(attributeName="keyA")
// Getter and setter
#DynamoDBRangeKey(attributeName="keyB")
// Getter and setter
}
Here is my key class RecentlyPlayedId.class
public class RecentlyPlayedId implements Serializable {
private static final long serialVersionUID = 1L;
private String keyA;
private String keyB;
public RecentlyPlayedId(String keyA, String keyB) {
this.keyA = keyA;
this.keyB = keyB;
}
#DynamoDBHashKey
// Getter and setter
#DynamoDBRangeKey
// Getter and setter
}
Here is my repository interface RecentlyPlayedRepository
#EnableScan
public interface RecentlyPlayedRepository extends CrudRepository<RecentlyPlayed, RecentlyPlayedId> {
List<RecentlyPlayed> findAllByKeyA(#Param("keyA") String keyA);
// Finding the entry for keyA with highest keyB
RecentlyPlayed findTop1ByKeyAOrderByKeyBDesc(#Param("keyA") String keyA);
}
I am trying to save an object like this
RecentlyPlayed rp = new RecentlyPlayed(...);
dynamoDBMapper.save(rp); // Throws that error
recentlyPlayedRepository.save(rp); // Also throws the same error
I am using Spring v2.0.1.RELEASE. The wiki in the original docs warns about this error and describes what to do to mitigate. I did exactly what they said. But still no luck.
The link to that wiki is here - https://github.com/derjust/spring-data-dynamodb/wiki/Use-Hash-Range-keys

DynamoDB only supports primitive data types, it does not know how to convert your complex field (recentlyPlayedId) into a primitive, such as a String.
To show that this is the case, you can add the annotation #DynamoDBIgnore to your recentlyPlayedId attribute like this:
#DynamoDBIgnore
private RecentlyPlayedId recentlyPlayedId;
You also need to remove the #id annotation.
Your save function will then work, but the recentlyPlayedId will not be stored in the item. If you do want to save this field, you need to use the #DynamoDBTypeConverted annotation and write a converter class. The converter class defines how to convert the complex field into a String, and then uncovert the String into the complex field.

Removing getters/setters for the #Id field fixed the problem for me. This is suggested in https://github.com/derjust/spring-data-dynamodb/wiki/Use-Hash-Range-keys

not supported; requires #DynamoDBTyped or #DynamoDBTypeConverted",
i was getting this error when i defined model class with field JsonNode,i converted it to MAP<String,String>,now it is working fine

Spring/JPA/persistence entity attribute field cannot be final?

I have a Spring MVC project using JPA which I have worked on for some time in the past without this issue. But now for some reason (likely an environmental issue as I have switch to a new laptop since I last worked on it) I am getting this weird error.
The project is essentially a tool for creating and performing surveys which are just a set of questions. There are multiple types of question such as "auto complete question", "multiple choice question", "integer question", etc which collect different types of data. Each of this question types is modeled by a subclass which extends an abstract class called DdmQuestion which looks something like this:
#Entity
#Table(name = "ddm_question")
#Inheritance(strategy=InheritanceType.SINGLE_TABLE)
#DiscriminatorColumn(discriminatorType = DiscriminatorType.STRING, name = "question_type")
#JsonIgnoreProperties({"dataType"})
#JsonTypeInfo(use = JsonTypeInfo.Id.MINIMAL_CLASS, include = JsonTypeInfo.As.PROPERTY, property = "#question_type")
#JsonSubTypes(value = { #Type(DdmTextQuestion.class),#Type(DdmDateQuestion.class),#Type(DdmTimeQuestion.class),#Type(DdmNumberIntegerQuestion.class),#Type(DdmChoiceMultiQuestion.class),#Type(DdmAutoCompleteQuestion.class) })
public abstract class DdmQuestion {
#Id
#GeneratedValue
#Column(name = "question_id")
private int questionId;
#Column(name = "name")
private String name;
public int getQuestionId() {
return questionId;
}
public void setQuestionId(int questionId) {
this.questionId = questionId;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#JsonIgnore
public abstract String getDataType();
}
Note the getDataType() method.
Then, for each question type, I have a subclass extending this which looks something like this:
#Entity
#DiscriminatorValue("ddm_question_date")
public class DdmDateQuestion extends DdmQuestion {
final private String DATA_TYPE = "Long"; // this is the line with the error
#Override
public String getDataType() {
return DATA_TYPE;
}
}
Now, I've never encountered this error before (that I can recall) but Eclipse is throwing up an error here that says:
"The Java field for attribute "DATA_TYPE" is final". That's all it
says.
If I remove the #Entity annotation from the class, this error disappears so evidently something in JPA doesn't like something about this but I never had this error before so I'm thinking something changed in a newer version. My POM is not particularly explicit with dependency versions so this would not be surprising.
Can anyone explain to me why this is happening and what the correct resolution is? I could just remove the "final" from the field declaration but this seems wrong to me as it is definitely a final value...
Thanks in advance.

If it is a field that should not be persisted in the database you usually should take advantage of the transient annotation which would tell the persistence provider to ommit that field in its processing.:
#Transient
final private String DATA_TYPE = "Long";
If Eclipse is smart enough, it should stop highlighting the error altogether.

in this linkshttp://docs.oracle.com/javaee/5/tutorial/doc/bnbqa.html#Entities;
An entity class must follow these requirements:
The class must be annotated with the javax.persistence.Entity annotation.
The class must have a public or protected, no-argument constructor. The class may have other constructors.
The class must not be declared final. No methods or persistent instance variables must be declared final.
If an entity instance be passed by value as a detached object, such as through a session bean’s remote business interface, the class must implement the Serializable interface.
Entities may extend both entity and non-entity classes, and non-entity classes may extend entity classes.
Persistent instance variables must be declared private, protected, or package-private, and can only be accessed directly by the entity class’s methods. Clients must access the entity’s state through accessor or business methods.

Serlaize case class with the variable inside it in scala with jackson

i am tiring to serialize a case class using jackson fasterxml, i can see the constructor parameters after deserialize (taskRequest and taskNameIn) but not the variables inside the class (jobsRequests is null for example):
//#JsonIgnoreProperties(ignoreUnknown = true) // tried to remove it with no luck
#JsonAutoDetect
case class Job(taskRequest: List[TaskRequest] = Nil,taskNameIn:String) {
{
this.jobsRequests = taskRequest
this.taskName= taskNameIn
}
#JsonProperty
#volatile private var jobsRequests: List[TaskRequest] = Nil
#JsonProperty
var task_name: String = ""
}
Any suggestions ?

Jackson uses Getter from the Java Beans standard to construct the json. Try adding #BeanProperty to your properties and constructor parameters to compile your class with Getter/Setters.
Example
Or you could use the Jackson Scala-Module. You can take a look at their tests to see how to use this module for serialization.

So there where some problems with the serialization, some where easy but i learn something that may help other people with this problem and the understanding of case class in general.
First, i used javap(.exe) to see the java code from the class files, to Job.scala with contained case class named Job, there are two class files: Job$.class and Job.class.
Job$.class:
public final class logic.Queue.Task.Job$ extends scala.runtime.AbstractFunction4<java.lang.Object, java.lang.String, scala.collection.immutable.List<logic.Job.TaskRequest>, org.Server.Job.TaskContainer, logic.Queue.Task.Job> implements scala.Serializable {
public static final logic.Queue.Task.Job$ MODULE$;
public static {};
public final java.lang.String toString();
.
.
.
}
Job.class:
public class logic.Queue.Task.Job implements scala.Product,scala.Serializable {
public static org.Server.Job.TaskContainer apply$default$4();
public static scala.collection.immutable.List<logic.Job.TaskRequest> apply$default$3();
.
.
.
}
Meaning that the scala case class is an anonymous inner class and when you try to serialize it (and you can since both implements scala.Serializable), i solved it with adding to the signature to:
#JsonAutoDetect
#JsonTypeInfo(use = JsonTypeInfo.Id.CLASS, include = JsonTypeInfo.As.PROPERTY, property = "#class")
#JsonCreator
case class Job(stat: Int = CmsTypeMsg.Pending, jobName: String = "", taskRequestIn: List[TaskRequest] = Nil, taskIn: TaskContainer = new TaskContainer())
For more help on this issue:
http://www.jpalomaki.fi/?p=527
Json deserialization into another class hierarchy using Jackson