将JSON作为响应Spring Boot返回

Question

I am trying to get rest response as json instead I am getting as string.

Controller

@RestController
@RequestMapping("/api/")
public class someController{

  @Autowired
  private SomeService someService;

  @GetMapping("/getsome")
  public Iterable<SomeModel> getData(){
    return someService.getData();
  }
}

Service

@Autowired
private SomeRepo someRepo;

public Iterable<someModel> getData(){
  return someRepo.findAll();
}

Repository

public interface SomeRepo extends CrudRepository<SomeModel,Integer>{

}

Models

@Entity
@Table(name="some_table")
public class SomeModel{

  @Id
  @Column(name="p_col", nullable=false)
  private Integer id;
  @Column(name="s_col")
  private String name
  @Column(name="t_col")
  private String json;   // this column contains json data

  //constructors, getters and setters
}

when I run localhost:8080/api/getsome I am getting:

[
 {
    "p_col":1,
    "s_col":"someName",
    "t_col":" 
{\r\n\t"school_name\":\"someSchool\",\t\r\n\t"grade\":"A\",\r\n\t\"class\": 
 [{\"course\":"abc",\t"course_name\":\"def" }]}"
  }
]

Field t_col is returning string instead of json. How do I get json objects in response?

As for the database, the three columns are int, varchar and varchar.

Any help would be appreciated. Thanks !!

Answer 1

In your controller add json response:

 @RequestMapping(value = "/getsome", method = RequestMethod.GET, produces = "application/json"

And wrap the getData string as response

    public class StringResponse {

        private String response;

        public StringResponse(String s) { 
           this.response = s;
        }

}

Answer 2

Change your some model class like this

@Entity
@Table(name="some_table")
public class SomeModel {

    @Id
    @Column(name="p_col", nullable=false)
    private Integer id;
    @Column(name="s_col")
    private String name
    @Column(name="t_col")
    private String json;   // this column contains json data

    @Column(name = "t_col", columnDefinition = "json")
    @Convert(attributeName = "data", converter = JsonConverter.class)
    private Map<String, Object> json = new HashMap<>();

    //constructors
    //getters and setters
}

Write a json converter class.

@Converter
public class JsonConverter
                    implements AttributeConverter<String, Map<String, Object>> 
{


    @Override
    public Map<String, Object> convertToDatabaseColumn(String attribute)
    {
        if (attribute == null) {
           return new HashMap<>();
        }
        try
        {
            ObjectMapper objectMapper = new ObjectMapper();
            return objectMapper.readValue(attribute, HashMap.class);
        }
        catch (IOException e) {
        }
        return new HashMap<>();
    }

    @Override
    public String convertToEntityAttribute(Map<String, Object> dbData)
    {
        try
        {
            ObjectMapper objectMapper = new ObjectMapper();
            return objectMapper.writeValueAsString(dbData);
        }
        catch (JsonProcessingException e)
        {
            return null;
        }
    }
}

It will convert database json attribute to your desire results. Thanks

Answer 3

You need to define your json attribute as JsonNode so jackson can read it back and forrward, but mark is as @Transient so JPA does not try to store it on database.

Then you can code getter/setter for JPA, where you translate from JsonNode to String back and forward. You define a getter getJsonString that translate JsonNode json to String . That one can be mapped to a table column, like 'json_string', then you define a setter where you receive the String from JPA and parse it to JsonNode that will be avaialable for jackson, jackson then will translate it to a json object not a string as you mention.

@Entity
@Table(name = "model")
public class SomeModel {

  private Long id;
  private String col1;

  //  Attribute for Jackson 
  @Transient
  private JsonNode json;

  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  public Long getId() {
    return id;
  }

  @Column(name ="col1")
  public String getCol1() {
    return col1;
  }

  // Getter and setter for name

  @Transient
  public JsonNode getJson() {
    return json;
  }

  public void setJson(JsonNode json) {
    this.json = json;
  }

  // Getter and Setter for JPA use
  @Column(name ="jsonString")
  public String getJsonString() {
    return this.json.toString();
  }

  public void setJsonString(String jsonString) {
    // parse from String to JsonNode object
    ObjectMapper mapper = new ObjectMapper();
    try {
      this.json = mapper.readTree(jsonString);
    } catch (Exception e) {
      e.printStackTrace();
    }
  }
}

Notice, @Column are defined at gettters because we need to indicate JPA to use getJsonString and JPA requires consistency so all column's getters must be mark with @Columns .

Answer 4

  @RequestMapping(value = "/getsome", method = RequestMethod.POST, consumes = 
  "application/json;")
 public @ResponseBody
  ModelAndView getSome(@RequestBody List<Map<String, String>> request) {
 request.stream().forEach(mapsData->{
    mapsData.entrySet().forEach(mapData -> {
      System.Out.Println("key :"+mapData.getKey() + " " + 
        " value : " +mapData.getValue());
    });
  }
});
return new ModelAndView("redirect:/home");

}

Answer 5

Cast your DB answer from findAll to a list of object type, like (List of Sometype), then return data. This will be resolve your problem.