Spring MVC + 杰克逊 - JsonView答案

【问题标题】：Spring MVC + Jackson - JsonViewSpring MVC + 杰克逊 - JsonView
【发布时间】：2016-10-12 11:06:35
【问题描述】：

我需要从我的模型中公开两组不同的值，所以我实现了 2 个视图

public class Views {

    public static class Small{ }

    public static class Large extends Small { }

}

然后，在我的模型中（所有其他字段都用 JSONIgnore 注释

@JsonView(Views.Small.class)
    @Id
    @GeneratedValue(strategy = IDENTITY)
    @Column(name = "id_posto", unique = true, nullable = false)
    public int getIdPosto() {
        return this.idPosto;
    }

    public void setIdPosto(int idPosto) {
        this.idPosto = idPosto;
    }

@JsonView(Views.Large.class)
    @NotNull
    @Column(name = "nome_posto_park")
    public String getNomePosto() {
        return this.nomePosto;
    }
public void setNomePosto(String nomePosto) {
        this.nomePosto = nomePosto;
    }

在我的控制器上，我有 2 种方法：

@RequestMapping(value = "/spots", method = RequestMethod.GET)
    public ResponseEntity<Posto> getSpotStatus(@RequestParam(value = "idPosto") int idPosto,
            @RequestParam(value = "occupied") boolean occupied) {
        Posto posto = postoService.findByIdPosto(idPosto);
        ObjectMapper mapper = new ObjectMapper();
        mapper.disable(MapperFeature.DEFAULT_VIEW_INCLUSION);
        mapper.setConfig(mapper.getSerializationConfig()
                .withView(Views.Small.class));
        mapper.convertValue(posto, JsonNode.class);

return new ResponseEntity<Posto>(posto, HttpStatus.OK);

和

@RequestMapping(value="/spot", method = RequestMethod.GET)
    public ResponseEntity<List<Posto>> getSpotList(@RequestParam (value = "idPiano") int idPiano){
        Piano piano = pianoService.findById(idPiano);

    List<Posto> posti = postoService.showSpotsByFloor(-1, piano);
     ObjectMapper mapper = new ObjectMapper();
        mapper.disable(MapperFeature.DEFAULT_VIEW_INCLUSION);
        mapper.setConfig(mapper.getSerializationConfig()
                .withView(Views.Large.class));
    mapper.convertValue(posti, JsonNode.class);

    return new ResponseEntity<List<Posto>>(posti, HttpStatus.OK);
}

Che 结果是一样的...（显然第一个是单个 Posto，第二个是 List，但是模型中的所有字段都被序列化了....

使用视图时我做错了什么？

【问题讨论】：

标签： java json spring

【解决方案1】：

您需要使用同意视图和 @ResponseBody 注释定义生产和消费

示例：更改您的需要

@Produces(value = { MediaType.APPLICATION_JSON_VALUE })
@Consumes(value = { MediaType.APPLICATION_JSON_VALUE })
public @ResponseBody   public ResponseEntity<List<Posto>> getSpotList(...

//when request, put your client agree request view
protected HttpEntity<T> headers()
{
    final HttpHeaders headers = new HttpHeaders();
    headers.set("Accept", MediaType.APPLICATION_JSON_VALUE);
    headers.set("application", MediaType.APPLICATION_JSON_VALUE);
    // T define your type here
    return new HttpEntity<T>(headers);
}

【讨论】：

【解决方案2】：

您的问题是 Spring 在 ApplicationContext 启动时实例化了它自己的 Jackson ObjectMapper。

您必须自动装配 Spring 管理的 ObjectMapper 并配置该实例，而不是使用 new 创建您自己的实例。

private final ObjectMapper mapper;

public MyController(ObjectMapper mapper) {
    this.mapper = mapper;
}

【讨论】：