从REST获取响应时使用Retrofit的NullPointer
从REST获取响应时使用Retrofit的NullPointer
提问于 2018-03-10 01:45:26
我正在使用Retrofit与Android上的REST合并，但是我得到了一个错误NullPointerException，如下所示。我尝试使用postman，API运行良好，并得到响应。
ukmutilizer.project.com.ukm_utilizer.model.CheckEmail.getData()‘：尝试在空对象引用上调用虚拟方法的java.util.List java.util.List
这是我的活动课
private void sendRequest(String checkEmail){
ApiInterface apiService = ApiClient.getClient().create(ApiInterface.class);
Call<CheckEmail> call = apiService.getEmailStatus(checkEmail);
call.enqueue(new Callback<CheckEmail>() {
@Override
public void onResponse(Call<CheckEmail> call, Response<CheckEmail> response) {
CheckEmailData emailDataList = response.body().getData();
Log.d("Numer of Data : ", String.valueOf(response.body().getData()));
}
@Override
public void onFailure(Call<CheckEmail> call, Throwable t) {
Toast.makeText(CheckEmailPage.this, "Something went wrong!", Toast.LENGTH_SHORT).show();
Log.e("Error Retrofit : ", String.valueOf(t));
}
});这是ApiInterface
public interface ApiInterface {
@POST("users/check_status")
Call<CheckEmail> getEmailStatus(@Body String email);
}这是改造实例
`public class ApiClient {
public static final String BASE_URL = "https://f49d9d29-8471-4126-95b0-1ec3d18eda94.mock.pstmn.io/v1/";
private static Retrofit retrofit = null;
public static Retrofit getClient(){
HttpLoggingInterceptor logging = new HttpLoggingInterceptor();
logging.setLevel(HttpLoggingInterceptor.Level.BODY);
OkHttpClient httpClient = new OkHttpClient.Builder().addInterceptor(logging).build();
if(retrofit == null){
retrofit = new Retrofit.Builder()
.baseUrl(BASE_URL)
.addConverterFactory(GsonConverterFactory.create())
.build();
}
return retrofit;
}
}`这是json的回应
{
"code": 1000,
"message": "OK",
"data": {
"id": "1",
"email": "test@gmail.com",
"status": "1",
"name": "test",
"category": "2"
}
}这是POJO
`public class CheckEmail {
@SerializedName("code")
@Expose
private Integer code;
@SerializedName("message")
@Expose
private String message;
@SerializedName("data")
@Expose
private CheckEmailData data;
public Integer getCode() {
return code;
}
public void setCode(Integer code) {
this.code = code;
}
public String getMessage() {
return message;
}
public void setMessage(String message) {
this.message = message;
}
public CheckEmailData getData() {
return data;
}
public void setData(CheckEmailData data) {
this.data = data;
}
}`CheckEmailData POJO
`public class CheckEmailData {
@SerializedName("id")
@Expose
private String id;
@SerializedName("email")
@Expose
private String email;
@SerializedName("status")
@Expose
private String status;
@SerializedName("name")
@Expose
private String name;
@SerializedName("category")
@Expose
private String category;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
public String getStatus() {
return status;
}
public void setStatus(String status) {
this.status = status;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getCategory() {
return category;
}
public void setCategory(String category) {
this.category = category;
}
}`回答 4
Stack Overflow用户
回答已采纳
发布于 2018-03-10 14:31:23
我已经解决了这个问题
我在界面上添加了标题，如下所示
public interface ApiInterface {
@Headers({
"Content-Type:application/x-www-form-urlencoded"
})
@POST("v1/users/check_status")
Call<CheckEmail> getEmailStatus(@Body String email);
}Stack Overflow用户
发布于 2018-03-10 02:14:10
从json响应的api中只能得到一个数据，json对象
{
"code": 1000,
"message": "OK",
"data": {
"id": "1",
"email": "test@gmail.com",
"status": "1",
"name": "test",
"category": "2"
}
}然而，您可以将数据声明为一个List对象，在该对象中，它期望上面的data是一个json数组格式。
您应该将列表更改为
@SerializedName("data")
@Expose
private CheckEmailData data;我相信会没事的。
Stack Overflow用户
发布于 2018-03-10 02:14:26
在那个JSON中，"data“是一个对象，而不是一个数组。在CheckEmail类中，将private List<CheckEmailData> data;更改为private CheckEmailData data;
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/49204531
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