How to access the String json payload and mapping it to an Object in Spring rest controller?

Question

I'm building a rest API using Spring Boot rest services. I have a Java class:

class Person{
 int id;
 @notNull
 String name;
 @notNull
 String password;
 }

And I want to make an API to create a Person object. I will recieve a POST request with json body like:

{
"name":"Ahmad",
"password":"myPass",
"shouldSendEmail":1
}

As you can see there are an extra field "shouldSendEmail" that I have to use it to know if should I send an email or not after I create the Person Object. I am using the following API:

@RequestMapping(value = "/AddPerson", method = RequestMethod.POST)
public String savePerson(
        @Valid @RequestBody Person person) {

     personRepository.insert(person);

    // Here I want to know if I should send an email or Not

    return "success";
}

Is there a method to access the value of "shouldSendEmail" while I using the autoMapping in this way?

Sure. Since what you're receiving is not actually a person, but a command which contains some partial information about a person, and also other stuff like shouldSendEmail, just design a class matching the JSON structure you receive, and use that as an argument of savePerson(). — JB Nizet, Sep 19 '18 at 21:46
if you dont what to persist this field, and it's ok to keep in the entity, declared it with @Transient — Deividi Cavarzan, Sep 19 '18 at 22:02

score 0 · Answer 1 · answered Sep 19 '18 at 21:47

You will need an intermediary DTO, or you will otherwise have to modify person to include a field for shouldSendEmail. If that is not possible, the only other alternative is to use JsonNode and manually select the properties from the tree.

For example,

@Getter
public class PersonDTO {
    private final String name;
    private final String password;
    private final Integer shouldSendEmail;

    @JsonCreator
    public PersonDTO(
        @JsonProperty("name") final String name,
        @JsonProperty("password") final String password,
        @JsonProperty("shouldSendEmail") final Integer shouldSendEmail
    ) {
        this.name = name;
        this.password = password;
        this.shouldSendEmail = shouldSendEmail;
    }
}

score 0 · Answer 2 · answered Sep 19 '18 at 21:55

0

You can use @RequestBody and @RequestParam together as following

.../addPerson?sendEmail=true

So send the “sendEmail” value as request param and person as request body

Spring MVC - Why not able to use @RequestBody and @RequestParam together

answered Sep 19 '18 at 21:55

Amr Alaa

545
3
7

Samir Ghoneim · Answer 3 · 2018-09-19T22:26:24.753

0

You have mutli solutions

1 - You can put @Column(insertable=false, updatable=false) above this property

2 - send it as request param @RequestParam @RequestMapping(value = "/AddPerson", method = RequestMethod.POST) public String savePerson( @Valid @RequestBody Person person, @RequestParam boolean sendMail) {}

3- use DTO lets say PersonModel and map it to Person before save

edited Sep 19 '18 at 22:26

answered Sep 19 '18 at 22:04

Samir Ghoneim

591
1
6
14

score 0 · Accepted Answer · answered Sep 20 '18 at 13:37

There's many options for you solve. Since you don't want to persist the shouldSendEmail flag and it's ok to add into you domain class, you can use the @Transient annotation to tell JPA to skip the persistence.

@Entity
public class Person {

    @Id 
    private Integer id;

    @NotNull
    private String name;

    @NotNull
    private String password;

    @Transient
    private Boolean shouldSendEmail;

}

If you want more flexible entity personalizations, I recommend using DTO`s.

MapStruct is a good library to handle DTO`s

How to access the String json payload and mapping it to an Object in Spring rest controller?

4 Answers4