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Spring Data Neo4J reference related nodes without overwriting them on save

I'm struggling to write this, so I may have to give an example to help explain the problem I'm experiencing.
Say we have nodes of three types (these nodes may have more relationships of their own, e.g. Product Family, has product manager):
Product
Product Family
Battery
With these relationships
A product can be be in 0 or more families
A product can have 0 or more batteries.
When using spring-data-neo4j and saving a new Product, I wish to include these relatiopnships, such as the batteries they require and the product family they belong to. However if I only supply say an ID rather then a fully populated object, it overwrites this object along with properties and relations accordingly.
This isn't great as it means that I have to end up sending a fully populated object, with all it's relations everytime I wish to save something, and some of these relations may go quite deep.
My domain is as follows:
#Node
public class Product {
#Id
#GeneratedValue(generatorClass = SnowflakeGenerator.class)
private Long productId;
private String name;
#Relationship(type = "REQUIRES_BATTERY", direction = OUTGOING)
private List<Battery> batteryList;
#Relationship(type = "IN_FAMILY", direction = OUTGOING)
private List<ProductFamily> productFamilyList;
}
#Node
public class Battery {
#Id
#GeneratedValue(generatorClass = SnowflakeGenerator.class)
private Long batteryId;
private String name;
}
#Node
public class ProductFamily {
#Id
#GeneratedValue(generatorClass = SnowflakeGenerator.class)
private Long familyId;
private String name;
}
This could very well by from coming from a Relational Database mindset and is a 'limitation' of using Neo4J.
TLDR When persisting somethign in Neo4J using spring-data how can I save just a relationship, rather than a whole related Node.

You can make use of projections in Spring Data Neo4j. (https://docs.spring.io/spring-data/neo4j/docs/current/reference/html/#projections)
This gives you the option to put a "mask" on the object tree, you want to persist (and what should stay untouched).
For example in your case:
interface ProductProjection {
// without defining e.g. String getName() here, SDN would not ever touch this property.
List<BatteryProjection> getBatteryList();
List<ProductFamilyProjection> getProductFamilyList();
}
interface BatteryProjection {
String getName();
}
interface ProductFamilyProjection {
String getName();
}

Elasticsearch: search for two different document in a single query

I have two entities Merchant and Customer:
public class Merchant{
private UUID id;
private String name;
//... other fields and getters/setters
}
public class Customer{
private UUID id;
private String name;
//... other fields and getters/setters
}
These two entities are sightly different from each-other.
What I'am trying to to do is when I search with the term "John" I want to get both a merchant named "John Market" and a customer called "John Smith".
To achieve this I indexed these entities to a single index.
#Document(indexName = "merchant_customer_index", type = "merchantorcustomer")
public class MerchantOrCustomer {
#Id
private UUID id;
private String name;
private int type;
//...
My query can return both Merchant and Customer:
List<MerchantOrCustomer> result = elasticsearchTemplate.queryForList(nativeSearchQuery, MerchantOrCustomer.class);
I distinguish them programmatic(if(result.get(i).getType() == 0 we received Merchant else Customer)
Then use their id to extract actual object from relational db.
I searched a lot, but couldn't find anything that can help to estimate if it is a good practice. Is it a good practice?
Please, give me a hint if there is a better way.

There doesn't seem to be anything wrong with what you did unless there is some collusion as mentioned by #Ivan in comments.
Here is another possible way to do if you were using elasticTemplate- Spring Data Elasticsearch: Multiple Index with same Document or if you are using queryBuilder - https://www.elastic.co/guide/en/elasticsearch/client/java-api/current/java-search.html

Criteria API createAlias on column containing null values kills the query

I do have a simple entity named "Address" that has a couple of properties and relations defined by itself and some inherited from some superclass.
public class Base {
private UUID id;
private Date createdAt;
#NotNull
private User createdBy; // User is a related Entity that cannot be null
private DataImport dataImport; // DataImport is another related entity that can be null
//getter & setter
}
public class Address extends Base {
private String street;
#NotNull
private Country country; // related entity that can't be null
//getter & setter
}
What I'm trying to achieve is with one query using the Criteria API, I want to get a list of Address objects containing all simple attributes like street and createdAt. At the same time I want only the IDs of the related entities if present: createdBy.id, dataImport.id and country.id.
I'm almost there using the following Criteria query:
entityManager.getDelegate().createCriteria(Address.class).add(criteriaExample)
.excludeZeroes()
.createAlias("createdBy", "subcreatedBy")
.createAlias("country", "subcountry")
.setProjection(Projections.projectionList().add(Projections.property("id").as("id"))
.add(Projections.property("createdAt").as("createdAt"))
.add(Projections.property("street").as("street")).
.add(Projections.property("subcreatedBy.id").as("createdBy.id"))
.add(Projections.property("subcountry.id").as("country.id")))
.setResultTransformer(new AliasToBeanNestedResultTransformer(Address.class));
List<Address> result = criteria.list();
This works just perfect!
Problem occurs when I only add the "alias" for the dataImport relation.
...createAlias("dataImport", "subdataImport")...
Even without adding the Projection for dataImport.id to the query, it returns an empty list, meaning list.size() = 0, as soon as I add this alias.
My current guess is, that I can't put an alias on a nullable property. Does anybody have an idea what the solution might be? So, when the related entity is not null, I want to get it's ID. And I want it to be simply null, when the relation is not set.
Thanks in advance.

Stupid me should have read the documentation and set the CriteriaSpecification.LEFT_JOIN.
...createAlias("dataImport", "subdataImport", CriteriaSpecification.LEFT_JOIN)...

How to return an entity by property if it already exists?

#Entity
public class Language {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#Column(length = 2)
private String code; //EN, DE, US
public Language(String code) {
this.code = code;
}
}
#Entity
public class ProductText {
#OneToOne(Cascade.ALL)
private Language lang;
}
ProductText text = new ProductText();
text.setLang(new Language("en")); //what if "en" exists?
dao.save(text);
Now, when I persist the ProductText, everytime a new Language object would be generated.
Can I prevent this, and in case a language table entry with code = 'en' exists this existing entity should be linked instead.
My initial goal is to not having to repeat the countryCodeString "EN" multiple times in my product-text table, but just reference the id. But does this really make sense? Should I rather just use the plain String without an extra table? (I later want to query a list of productTexts where lang = 'de').
Is the only change executing a select like dao.findByLang("en") before?
Or is there also some hibernate feature that would support this without explicit executing a query myself?

Do you process the value "en" further or do you display it directly? If only used for displaying purposes, I would just store the string, but if you want to reduce redundancy by using foreign key IDs you have to create an Entity containing the language string en which can be persisted via entity manager and which you have to obtain out of the entity manager before persisting to reuse it.

If there is only three different possible values for the language, you can also use an enum like this :
public enum Language {
EN("EN"),
DE("DE"),
US("US");
private String code; //EN, DE, US
public Language(String code) {
this.code = code;
}
// Getter...
}
#Entity
public class ProductText {
#Enumerated(EnumType.STRING)
// Or #Enumerated(EnumType.ORDINAL)
private Language lang;
}
EnumType.STRING will store the enum in the database as a String, while EnumType.ORDINAL will store it as an int. Int is maybe a little more efficient, but the mapping could change if you insert a new value in your enum. String is more flexible since it will use the names of your enum members.
In both case, you don't have to manage a separate entity and hibernate will not create an additional table, and it's more type-safe than using a plain string.

If the only value in Language is a 2 or 3 letter string, why not just have the string as a member? This will be quicker and more efficient.

How to create Many-One Mapping in hibernate?

I want to create Many-One Mapping between two tabels, Expense(ID, NAME, CATEGORY) and
Category(ID, NAME).
In my class i have created a field 'Category category' and its setters and getters.
I did them after seeing some stuff from internet. What are all the changes i have to do in my Category.java class. For now, its looks like,
public class Category{
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private int catId;
private String catName;
public Category() {
}
public int getCatId() {
return this.catId;
}
public void setCatId(int catId) {
this.catId = catId;
}
public String getCatName() {
return this.catName;
}
public void setCatName(String catName) {
this.catName = catName;
}
}
I dont want to do mappings with xml config. I think, annotations is good for a beginner like me.
And my Old! SQL query looks like,
SELECT EXPENSES.EXPNS_ID, EXPENSES.CAT_ID, EXPENSES.NAME, CATEGORY.CAT_NAME FROM EXPENSES INNER JOIN CATEGORY ON EXPENSES.CAT_ID = CATEGORY.CAT_ID WHERE USER_NAME="+currentUserName
How to use inner join in Hibernate?
Any Suggestions!!
Thanks!
Update
Thanks for all answerers,
I tried what you told and it returns a empty list.
To, test i set the 'userName=Tamil' which is in the table.
The query generated by Hibernate is looks like below,
select expens0_.expnsId as expnsId1_, expens0_.catId as catId1_, expens0_.category_catId as category7_1_, expens0_.userName as userName1_ from Expens expens0_ inner join Category category1_ on expens0_.category_catId=category1_.catId where expens0_.userName=?
As a beginner, i have some doubts in JPQL, I want catName from Category[catId, catName] table. And the catId is also available in Expens[expnsId, catId, userName].
By adding the below lines in Expens.java class, how it will give me catName along with the other variables in the Expens table.
#ManyToOne
private Category category
// getters, setters
I cant able to understand it. Without understanding this i cant move further, i have to give more mappings in my project. If clear with this mapping, i can move to the rest with confidence.
The query i used is pascal's version: Query query = hSession.createQuery("SELECT e FROM Expens e JOIN e.category c WHERE e.userName = :userName").setParameter("userName", userName);
For me, the query generated by hibernate is looks like same as my Old SQl query. I cant able to find problem here.

Actually, a big part of the documentation that would be useful in your case is located in the Hibernate Annotations Reference Guides (links provided below). Reading it would be very worth it.
That being said, regarding your specific question, the simplest possible mapping would be:
#Entity
public class Expense {
#Id #GeneratedValue
private Long;
#ManyToOne
private Category category
// getters, setters
...
}
That's all.
If you want to make it bi-directional, you'll have to add a OneToMany on the other side (and don't forget the mappedBy element since the association is bidirectional):
#Entity
public class Category {
#Id #GeneratedValue
private Long id;
#OneToMany(mappedBy="category")
private Set<Expense> expenses = new HashSet<Expense>();
....
}
And a possible JPQL query would be:
SELECT e FROM Expense e JOIN e.category c WHERE e.username = :username
Update: Hibernate and JDBC are different. With Hibernate, you need to think objects and the above HQL query (which was more an example) will actually return a List<Expense>. To get a category name, iterate over the results and navigate through the association. For example:
List<Expense> expenses = ... // some code to retrieve a list by username
for (Expense expense : expenses) {
System.out.println(expense.getCategory().getName());
}
References
2.2. Mapping with JPA (Java Persistence Annotations)
2.2.5.2. Many-to-one

As Bozho suggested,
#ManyToOne(fetch=FetchType.EAGER) // Gonna be eager by default anyway
#JoinColumn(name="CATEGORY_ID")
private Category category;
Plus this in your Category class to make it bidirectional,
#OneToMany(mappedBy="category")
private List<Expense> expense;
You need not do an inner join like that. When you query the expense, the related category will automatically get loaded eagerly, most likely using join.

In your Expense class have:
#ManyToOne
#JoinColumn(name="CATEGORY_ID")
private Category category
As pointed in the comments, if you need to access all expenses in a given category, i.e. have the one-to-many relationship, you can have:
#OneToMany
private List<Expense> expenses;
I, for example, prefer to use as little #OneToMany mappings as possible - you'd have to manager eager/lazy loading, at some point limiting the number of results, etc. For them I tend to use HQL queries that fetch the subset of objects (expenses in your case) that I need.