JPA : Select parent and child with oneToMany relatinships

Question

Implemented one to many relationship and select parent and child I have a onetomany relationship and I want to do a query (select *)

Folders.java

@Entity
@Table(name = "FOLDERS")
public class Folders implements Serializable {
    @Id
    @Column(name = "id")
    Long id;
    @ManyToOne
    @JoinColumn(name = "Folder_Author", insertable=false, updatable=false)
    private Authors author;
    // Getter + setter
}

Authors.java

@Entity
@Table(name = "AUTHORS")
public class Authors implements Serializable {

    @Id
    @Column(name = "id")
    Long id;
    @OneToMany(fetch=FetchType.EAGER, cascade=CascadeType.PERSIST)
    @JoinColumn(name = "Folder_Author", referencedColumnName = "id", nullable=false)
    private List<Folders> folders = new ArrayList<Folders>();
}

My request :

Query query = em.createQuery("SELECT f FROM Folders f");
return query.getResultList();

I got this result :

[
    {   
        "id":29,
        "noFolder":"0017",
        "author":{
            "id":9,
            "name":"Alpha",
            "service":null,
            "folders":[
                {"id":29,
                "noFolder":"0017",
                "author":{
                    "id":9,
                    "name":"Alpha",
                    "service":null,
                    "folders":[
                    {
                        "id":29,
                        "noFolder":"0017",
                        "author":{
                        "id":9,
                        "name":"Alpha",
                        "service":null,
                        "folders":[
                            ..
                            ..
            }
]

What's wrong in my code ? What is the problem when I execute query I got this result .. .. I would like to get this result

[
    {   
        "id":29,
        "noFolder":"0017",
        "author":{
            "id":9,
            "name":"Alpha",
            "service":null,
            }
    }
]

you mean you want the 1-N field to be LAZY loading, and you set it to EAGER? — Neil Stockton, Sep 04 '17 at 09:27
Yes you right, but when I tried lazy I got this error : `Caused by: org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: business.entities.folders.Authors.folders, could not initialize proxy - no Session at rg.hibernate.collection.internal.AbstractPersistentCollection.throwLazyInitializationException(AbstractPersistentCollection.java:587) at ` .. — user1814879, Sep 04 '17 at 09:33
also the `OneToMany` needs a mappedBy specifying to link both sides. — Neil Stockton, Sep 04 '17 at 09:33
I updated it and now it is like : `@OneToMany(mappedBy="author", targetEntity=Folders.class, fetch=FetchType.LAZY, cascade=CascadeType.PERSIST) @Fetch(FetchMode.SELECT) private List folders = new ArrayList(); ` — user1814879, Sep 04 '17 at 09:36
wtf is field "initiator" ? Should be "author" from the code you post — Neil Stockton, Sep 04 '17 at 09:38

score 3 · Accepted Answer · answered Sep 04 '17 at 12:33

3

Just use the @JsonIgnore on your collection field like this:

@OneToMany(fetch=FetchType.EAGER, cascade=CascadeType.PERSIST)
    @JoinColumn(name = "Folder_Author", referencedColumnName = "id", nullable=false)
    @JsonIgnore
    private List<Folders> folders = new ArrayList<Folders>();

answered Sep 04 '17 at 12:33

Chaitanya

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yes, but if I do it, how can I get values of the parent (Author) ? Like for each folders I would like to get his Author ... – user1814879 Sep 04 '17 at 12:34
@user1814879 You have to break the infinite loop using `@JsonIgnore` or else it never works. So you have to put on one-to-many side or many-to-one side. – Chaitanya Sep 04 '17 at 12:36

score 2 · Answer 2 · answered Sep 04 '17 at 12:40

2

Use @JsonBackReference on the list of folders in the author class to prevent recursive serialization.

JPA : Select parent and child with oneToMany relatinships

2 Answers2