我正在使用 Spring for Android 并通过以下方式提取 JSON 数据:
课程活动.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);
...
}
课程.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 调用返回这种类型的数据:
[{"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。
如果我想有不同的名字怎么办?具体来说,如果我的 JSON 返回诸如“notify_on_follow”之类的键,我如何使用 Spring for Android 将其映射到不带下划线的“notifyOnFollow”?或者也许将“active”更改为“isActive”?