Spring Boot REST - How to implement a search that requires a Java object?

Question

I have the following JPA models:

Issue

@Entity
public class Issue {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private long id;
private String title;
private String text;

@ManyToOne
@JoinColumn(name="user_id")
private User user;

public Issue() {}

public Issue(String title, String text) {
    this.title = title;
    this.text  = text;
}

public long getId() {
    return id;
}

public String getTitle() {
    return title;
}

public void setTitle(String title) {
    this.title = title;
}

public String getText() {
    return text;
}

public void setText(String text) {
    this.text = text;
}

public User getUser() {
    return user;
}

public void setUser(User user) {
    this.user = user;
}

@Override
public String toString() {
    return "Issue [id=" + id + ", title=" + title + ", text=" + text + ", user=" + user + "]";
}
}

User

@Entity
public class User {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private long id;
private String username;
private String firstname;
private String lastname;

public User() {}

public User(String username) {
    this.username = username;
}

public String getUsername() {
    return username;
}

public void setUsername(String username) {
    this.username = username;
}

public String getFirstname() {
    return firstname;
}

public void setFirstname(String firstname) {
    this.firstname = firstname;
}

public String getLastname() {
    return lastname;
}

public void setLastname(String lastname) {
    this.lastname = lastname;
}

@Override
public String toString() {
    return "User [id=" + id + ", username=" + username + ", firstname=" + firstname + ", lastname=" + lastname
            + "]";
}
}

And an Issue repository that extends PagingAndSortingRepository and contains the method List<Issue> findByUser(User user); See below:

public interface IssueRepository extends PagingAndSortingRepository<Issue,Long> {
    List<Issue> findByUser(User user);
}

I'm trying to find a way to navigate these relationships with HTTP calls, namely how do I call findByUser(User user) and get all the issues for that user?

Using the following call I can execute that particular query:

GET http://localhost:8080/issues/search/findByUser

But I'm unclear what I should be providing as the User? Do I send the id as a query param? Do I construct an object and send that as a query param? Am I just modeling this the wrong way?

I'd like to get back a JSON list containing all the Issues for this particular User.

Thanks in advance for any help or guidance.

I guess it might be useful to read docs.spring.io/spring-boot/docs/current/reference/htmlsingle/… and docs.spring.io/spring/docs/4.3.8.RELEASE/… — Jeremy Grand
– Jeremy Grand, Commented May 29, 2017 at 16:51

jerney · Accepted Answer · 2017-05-29 17:01:53Z

1

Changing the repository to this solved the issue. The key is to do the lookup based on a field of the User, not the User itself.

public interface IssueRepository extends PagingAndSortingRepository<Issue,Long> {
    List<Issue> findByUserUsername(@Param("username") String username);
}

GET http://localhost:8080/issues/search/findByUserUsername?username=jerney

This returns a list of issues.

answered May 29, 2017 at 17:01

jerney

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RANJITH VALLAMDASU · Accepted Answer · 2017-05-29 16:57:59Z

0

put another read only column like @Column(name = "user_id", insertable = false, updatable = false) private Long userId; in Issue entity and use findByUserId(Long userId) repo method to find it and pass userId parameter(i.e path varible) to controller to do this using http calls.

answered May 29, 2017 at 16:57

RANJITH VALLAMDASU

1

Comments

Zeromus · Accepted Answer · 2017-05-29 21:34:56Z

0

You can use a simple findOne(Long userId) if you need only one record as its probably faster than query by string field

answered May 29, 2017 at 21:34

Zeromus

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