在Android中返回null的改进响应

Question

我想显示一个JSON值的文本.我使用Retrofit,使API呼叫我不知道如果我这样做是正确的.无论如何这是我的代码.

这是网址链接:http://api.icndb.com/jokes/random.

该网站每次都会显示一个随机笑话.以下是该网站输出的示例:

{ "type": "success", "value": { "id": 175, "joke": "When Chuck Norris was a baby, he didn't suck his mother's breast. His mother served him whiskey, straight out of the bottle.", "categories": [] } } 

---

fragment.java

    String url = "http://api.icndb.com/";

    Retrofit.Builder builder = new Retrofit.Builder()
            .baseUrl(url)
            .addConverterFactory(GsonConverterFactory.create());

    Retrofit retrofit = builder.build();

    RetroFitHelper client = retrofit.create(RetroFitHelper.class);
    Call<Api> call = client.findJoke("random");

    call.enqueue(new Callback<Api>() {
        @Override
        public void onResponse(Response<Api> response, Retrofit retrofit) {

            String result = response.body().getJoke();

            Toast.makeText(getContext()," The word is: " + result ,Toast.LENGTH_LONG).show();
        }

        @Override
        public void onFailure(Throwable t) {
            Toast.makeText(getContext()," Error..." ,Toast.LENGTH_LONG).show();

        }
    });

---

RetroFitHelper.java

public interface RetroFitHelper {

    @GET("/jokes/{random}")
    Call<Api> findJoke(@Path("random") String random);

}

---

模型类

public class Api {

    @SerializedName("joke")
    private String joke;

    public String getJoke() {
        return joke;
    }

    public void setJoke(String joke){
        this.joke = joke;
    }
}

Answer 1

提供的json响应有一个名为value另一个json对象内部的json对象(它的形式{..,"value":{}}).所以我们需要两个模型类 - 一个用于outer json object另一个用于inner json object(值).

你需要有两个这样的模型类

public class Api {

@SerializedName("type")
@Expose
private String type;
@SerializedName("value")
@Expose
private Value value;

public String getType() {
return type;
}

public void setType(String type) {
this.type = type;
}

public Value getValue() {
return value;
}

public void setValue(Value value) {
this.value = value;
}

以及价值对象的下一个

public class Value {

@SerializedName("id")
@Expose
private Integer id;
@SerializedName("joke")
@Expose
private String joke;
@SerializedName("categories")
@Expose
private List<Object> categories = null;

public Integer getId() {
return id;
}

public void setId(Integer id) {
this.id = id;
}

public String getJoke() {
return joke;
}

public void setJoke(String joke) {
this.joke = joke;
}

public List<Object> getCategories() {
return categories;
}

public void setCategories(List<Object> categories) {
this.categories = categories;
}

}

现在,response.body()将有结果outer json object(Api),并response.body().getValue()将会有结果inner json object(Value).

现在在你的响应回调中,得到这样的响应

String result = response.body().getValue().getJoke();

还要确保您在清单中声明了必要的互联网权限

<uses-permission android:name="android.permission.INTERNET" />

确保在文件中设置了最新的依赖项app level build.gradle

implementation 'com.squareup.retrofit2:retrofit:2.4.0'

implementation 'com.squareup.retrofit2:converter-gson:2.4.0'

Answer 2

正如@Aswin 建议和@Navneet 回答的那样，您的 POJO 课程有问题。我建议您使用jsonschema2pojo或RoboPOJOGenerator以便下次避免陷入此类错误。

脚步

1) 访问http://www.jsonschema2pojo.org/

2) 将您的回复粘贴到那里并输入包和类名

3) 选择目标语言为 Java 或 Kotlin（如果您正在使用）

4) 源类型为 Json

5) 注释样式为 Gson

6) 点击预览

7) 将这些类复制并粘贴到您的应用程序包中