Question

我正在使用Spring for Android并通过执行以下操作来提取JSON数据：

CourseActivity.java

private void getCourses()
{
  RestTemplate restTemplate = new RestTemplate();
  restTemplate.getMessageConverters().add(new MappingJacksonHttpMessageConverter());
  String url = "http://192.168.1.74:3000/andy/courses.json";
  Course[] coursesArray = restTemplate.getForObject(url, Course[].class);

  ...
}

Course.java

@JsonIgnoreProperties(ignoreUnknown = true)
public class Course implements Parcelable, DBObject
{
    @JsonProperty
    private String number = null;
    @JsonProperty
    private String title = null;
    @JsonProperty
    private int school_id = -1;
    @JsonProperty
    private String department = null;

    ...
}

JSON调用返回此类数据：

[{"id":1,"number":"CE-490","title":"Senior Design","school_id":1,"department":"Computer Engineering"},{"id":2,"number":"CE-491","title":"Introduction to Mobile App Development","school_id":1,"department":"Computer Engineering"}]

现在您可以看到JSON数据键直接映射到我的课程属性名称。

如果我想要不同的名字怎么办？具体来说，如果我的JSON返回一个键，例如“notify_on_follow”，我如何使用Spring for Android将其映射到没有下划线的“notifyOnFollow”？或者将“活跃”改为“isActive”？

Answer 1

我找不到文档，但我很确定这会有效：

@JsonProperty("notify_on_follow")
private SomeType notifyOnFollow;

适用于Android的Spring - 自定义休息模板

1 个答案: