Map an Object Property in a POJO given an ID from a JSON
First of all, sorry, I cannot explain myself in any better way.
I am programming an example API, I have a POJO (with JPA) called Movie, so in my controller, I want to give it a JSON to insert a Movie.
Movie has a @ManyToOne(optional = false) property, relating another POJO called Genre (idGenre, Name)
I want to give in a JSON not an object with every property but an id, so:
CONTROLLER
@RequestMapping(value = "/sendMovie", method = RequestMethod.POST)
public void setMovie(@RequestBody Movie movie) {
mRepo.save(movie);
}
Movie
@Entity
public class Movie {
@Id
@Column(name = "ID_MOVIE", nullable = false)
@GeneratedValue(strategy = GenerationType.AUTO)
private Long idMovie;
@Column(name = "NAME")
private String name;
@Column(name = "SYNOPSIS")
private String synopsis;
@Column(name = "POSTER")
private String poster;
@Column(name = "DIRECTOR")
private String director;
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
private Genre genre;
JSON I want to use
{
"name": "MATRIX",
"idGenre": 3,
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS"
}
Is there possibility to achieve that?
spring jpa jackson
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
add a comment |
First of all, sorry, I cannot explain myself in any better way.
I am programming an example API, I have a POJO (with JPA) called Movie, so in my controller, I want to give it a JSON to insert a Movie.
Movie has a @ManyToOne(optional = false) property, relating another POJO called Genre (idGenre, Name)
I want to give in a JSON not an object with every property but an id, so:
CONTROLLER
@RequestMapping(value = "/sendMovie", method = RequestMethod.POST)
public void setMovie(@RequestBody Movie movie) {
mRepo.save(movie);
}
Movie
@Entity
public class Movie {
@Id
@Column(name = "ID_MOVIE", nullable = false)
@GeneratedValue(strategy = GenerationType.AUTO)
private Long idMovie;
@Column(name = "NAME")
private String name;
@Column(name = "SYNOPSIS")
private String synopsis;
@Column(name = "POSTER")
private String poster;
@Column(name = "DIRECTOR")
private String director;
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
private Genre genre;
JSON I want to use
{
"name": "MATRIX",
"idGenre": 3,
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS"
}
Is there possibility to achieve that?
spring jpa jackson
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
1
Sure. Design a class that matches with the JSON you want to send and receive, different from the Movie class, Let's call it a MovieCreationCommand. When you receive a MovieCreationCommand in your controller, create a Movie, initializing all its properties by getting them out of the command, and getting the genre from the database using the genre ID found in the command, then insert that movie.
– JB Nizet
Nov 14 '18 at 17:35
add a comment |
First of all, sorry, I cannot explain myself in any better way.
I am programming an example API, I have a POJO (with JPA) called Movie, so in my controller, I want to give it a JSON to insert a Movie.
Movie has a @ManyToOne(optional = false) property, relating another POJO called Genre (idGenre, Name)
I want to give in a JSON not an object with every property but an id, so:
CONTROLLER
@RequestMapping(value = "/sendMovie", method = RequestMethod.POST)
public void setMovie(@RequestBody Movie movie) {
mRepo.save(movie);
}
Movie
@Entity
public class Movie {
@Id
@Column(name = "ID_MOVIE", nullable = false)
@GeneratedValue(strategy = GenerationType.AUTO)
private Long idMovie;
@Column(name = "NAME")
private String name;
@Column(name = "SYNOPSIS")
private String synopsis;
@Column(name = "POSTER")
private String poster;
@Column(name = "DIRECTOR")
private String director;
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
private Genre genre;
JSON I want to use
{
"name": "MATRIX",
"idGenre": 3,
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS"
}
Is there possibility to achieve that?
spring jpa jackson
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
First of all, sorry, I cannot explain myself in any better way.
I am programming an example API, I have a POJO (with JPA) called Movie, so in my controller, I want to give it a JSON to insert a Movie.
Movie has a @ManyToOne(optional = false) property, relating another POJO called Genre (idGenre, Name)
I want to give in a JSON not an object with every property but an id, so:
CONTROLLER
@RequestMapping(value = "/sendMovie", method = RequestMethod.POST)
public void setMovie(@RequestBody Movie movie) {
mRepo.save(movie);
}
Movie
@Entity
public class Movie {
@Id
@Column(name = "ID_MOVIE", nullable = false)
@GeneratedValue(strategy = GenerationType.AUTO)
private Long idMovie;
@Column(name = "NAME")
private String name;
@Column(name = "SYNOPSIS")
private String synopsis;
@Column(name = "POSTER")
private String poster;
@Column(name = "DIRECTOR")
private String director;
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
private Genre genre;
JSON I want to use
{
"name": "MATRIX",
"idGenre": 3,
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS"
}
Is there possibility to achieve that?
spring jpa jackson
spring jpa jackson
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
edited Nov 14 '18 at 17:35
Thomas Fritsch
5,351122133
5,351122133
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
asked Nov 14 '18 at 17:29
mariotepromariotepro
335
335
1
Sure. Design a class that matches with the JSON you want to send and receive, different from the Movie class, Let's call it a MovieCreationCommand. When you receive a MovieCreationCommand in your controller, create a Movie, initializing all its properties by getting them out of the command, and getting the genre from the database using the genre ID found in the command, then insert that movie.
– JB Nizet
Nov 14 '18 at 17:35
add a comment |
1
Sure. Design a class that matches with the JSON you want to send and receive, different from the Movie class, Let's call it a MovieCreationCommand. When you receive a MovieCreationCommand in your controller, create a Movie, initializing all its properties by getting them out of the command, and getting the genre from the database using the genre ID found in the command, then insert that movie.
– JB Nizet
Nov 14 '18 at 17:35
1
1
Sure. Design a class that matches with the JSON you want to send and receive, different from the Movie class, Let's call it a MovieCreationCommand. When you receive a MovieCreationCommand in your controller, create a Movie, initializing all its properties by getting them out of the command, and getting the genre from the database using the genre ID found in the command, then insert that movie.
– JB Nizet
Nov 14 '18 at 17:35
Sure. Design a class that matches with the JSON you want to send and receive, different from the Movie class, Let's call it a MovieCreationCommand. When you receive a MovieCreationCommand in your controller, create a Movie, initializing all its properties by getting them out of the command, and getting the genre from the database using the genre ID found in the command, then insert that movie.
– JB Nizet
Nov 14 '18 at 17:35
add a comment |
1 Answer
1
active
oldest
votes
You did put the JsonIdentity... annotations at the wrong place.
You need to put these annotations on your @Genre class:
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
public class Genre {
@Id
@Column(...)
private Long idGenre;
//....
}
and remove these annotations from the property Genre genre in your Movie class.
You also need to tell Jackson with @JsonProperty("idGenre") that you want this property
serialized with name "idGenre".
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
@JsonProperty("idGenre")
private Genre genre;
Then the JSON output will be something like this:
{
"name": "MATRIX",
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS",
"idGenre": 3
}
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using thiscom.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:
– mariotepro
Nov 15 '18 at 11:31
add a comment |
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You did put the JsonIdentity... annotations at the wrong place.
You need to put these annotations on your @Genre class:
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
public class Genre {
@Id
@Column(...)
private Long idGenre;
//....
}
and remove these annotations from the property Genre genre in your Movie class.
You also need to tell Jackson with @JsonProperty("idGenre") that you want this property
serialized with name "idGenre".
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
@JsonProperty("idGenre")
private Genre genre;
Then the JSON output will be something like this:
{
"name": "MATRIX",
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS",
"idGenre": 3
}
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using thiscom.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:
– mariotepro
Nov 15 '18 at 11:31
add a comment |
You did put the JsonIdentity... annotations at the wrong place.
You need to put these annotations on your @Genre class:
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
public class Genre {
@Id
@Column(...)
private Long idGenre;
//....
}
and remove these annotations from the property Genre genre in your Movie class.
You also need to tell Jackson with @JsonProperty("idGenre") that you want this property
serialized with name "idGenre".
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
@JsonProperty("idGenre")
private Genre genre;
Then the JSON output will be something like this:
{
"name": "MATRIX",
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS",
"idGenre": 3
}
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using thiscom.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:
– mariotepro
Nov 15 '18 at 11:31
add a comment |
You did put the JsonIdentity... annotations at the wrong place.
You need to put these annotations on your @Genre class:
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
public class Genre {
@Id
@Column(...)
private Long idGenre;
//....
}
and remove these annotations from the property Genre genre in your Movie class.
You also need to tell Jackson with @JsonProperty("idGenre") that you want this property
serialized with name "idGenre".
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
@JsonProperty("idGenre")
private Genre genre;
Then the JSON output will be something like this:
{
"name": "MATRIX",
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS",
"idGenre": 3
}
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
You did put the JsonIdentity... annotations at the wrong place.
You need to put these annotations on your @Genre class:
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "idGenre")
@JsonIdentityReference(alwaysAsId = true)
public class Genre {
@Id
@Column(...)
private Long idGenre;
//....
}
and remove these annotations from the property Genre genre in your Movie class.
You also need to tell Jackson with @JsonProperty("idGenre") that you want this property
serialized with name "idGenre".
@ManyToOne(optional = false)
@JoinColumn(name = "ID_GENRE", referencedColumnName = "ID_GENRE")
@JsonProperty("idGenre")
private Genre genre;
Then the JSON output will be something like this:
{
"name": "MATRIX",
"synopsis": "NEO DOING THINGS",
"poster": "matrix.jpg",
"director": "WACHOWSKIS",
"idGenre": 3
}
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
answered Nov 14 '18 at 18:06
Thomas FritschThomas Fritsch
5,351122133
5,351122133
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using thiscom.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:
– mariotepro
Nov 15 '18 at 11:31
add a comment |
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using thiscom.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:
– mariotepro
Nov 15 '18 at 11:31
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using this
com.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:– mariotepro
Nov 15 '18 at 11:31
Thank you very much, Thomas, but I want to give that JSON as input to add a new movie into the DB, and i get a JSON parse error when using this
com.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for:– mariotepro
Nov 15 '18 at 11:31
add a comment |
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1
Sure. Design a class that matches with the JSON you want to send and receive, different from the Movie class, Let's call it a MovieCreationCommand. When you receive a MovieCreationCommand in your controller, create a Movie, initializing all its properties by getting them out of the command, and getting the genre from the database using the genre ID found in the command, then insert that movie.
– JB Nizet
Nov 14 '18 at 17:35