Spring MVC：如何为以下 JSON 响应正确创建实体

Question

我在为 Spring MVC RestTemplate.exchange 请求正确创建模型/实体时遇到问题，该请求将对应于此 JSON 响应：

{
    "resourceLookup": [
        {
            "version": 0,
            "permissionMask": 1,
            "creationDate": "2017-08-16T12:52:30",
            "updateDate": "2017-08-16T12:52:30",
            "label": "",
            "description": "",
            "uri": "",
            "resourceType": ""
        },
        {
            "version": 5,
            "permissionMask": 1,
            "creationDate": "2017-08-16T11:34:04",
            "updateDate": "2017-08-17T08:27:02",
            "label": "",
            "description": "",
            "uri": "",
            "resourceType": ""
        }
    ]
}

我创建了以下模型：

报告类（getters/setters 省略）：

public class Report implements Serializable {
    private Long version;
    private String permissionMask;
    private Date creationDate;
    private Date updateDate;
    private String label;
    private String description;
    private String url;
    private String resourceType;

    public Report() {
    }

ResourceLookup 类：

public class ResourceLookup implements Serializable {
    List<Report> reports;

    public ResourceLookup() {
    }

    public List<Report> getReports() {
        return reports;
    }

    public void setReports(List<Report> reports) {
        this.reports = reports;
    }
}

最后回复是reports=null。如果我用这段代码返回String，一切都很好：

restTemplate.exchange(serverUrl, HttpMethod.GET, httpEntity, String.class).getBody();

我需要将它正确映射到模型，以便我可以将格式正确的 JSON 返回给我的其他 API。

Answer 1

试试这个方法：

public class ResourceLookup implements Serializable {
    List<Report> resourceLookup;

    public ResourceLookup() {
    }

    public List<Report> getResourceLookup() {
        return resourceLookup;
    }

    public void setResourceLookup(List<Report> resourceLookup) {
        this.resourceLookup = resourceLookup;
    }
}