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TA贡献1806条经验 获得超8个赞
创建一个custom object并deserialize放入json string该对象的一个实例中。我将在下面展示一个使用gson库的解决方案
Maven 依赖
<dependency>
<groupId>com.google.code.gson</groupId>
<artifactId>gson</artifactId>
<version>2.8.5</version>
</dependency>
模型类
public class Model {
private boolean archived;
private String releaseDate;
private String name;
private String self;
private String userReleaseDate;
private long id;
private long projectId;
private boolean released;
//getters && setters ommitted
public Model(){
}
然后你可以deserialize the JSON string into the Model class like this
String json = "{\"archived\":false,\"releaseDate\":\"2019-07-16\",\"name\":\"test 1.0\",\"self\":\"https://test/rest/api/latest/test/10000\",\"userReleaseDate\":\"16/Jul/19\",\"id\":\"10000\",\"projectId\":10000,\"released\":true}";
Gson gson = new GsonBuilder().create();
Model model = gson.fromJson(json, Model.class);
这样,您就不必每次都分别检查每个参数以将其分配给相应的成员变量。
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