使用 Retrofit 从 Room 持久性库发送数据

Question

我最近接手了一个使用 Android 的 Room 持久性库以及 Retrofit 的项目。不幸的是，我对这两个库都不是很了解。

我目前正在尝试将使用 Room 保存的 JSON 记录数组发送到使用改造的 API。

我的实体如下所示：

@Entity(tableName = "location_table")
public class LocationEntity implements Serializable {

    @NonNull
    @PrimaryKey(autoGenerate = true)
    private int id;

    @SerializedName("job_id")
    @ColumnInfo(name = "job_id")
    @Expose
    private String job_id;

    @SerializedName("coordinates")
    @ColumnInfo(name = "coordinates")
    @Expose
    private String coordinates;

    @SerializedName("created_at")
    @ColumnInfo(name = "created_at")
    @Expose
    private String created_at;

    public int getId() { return id; }
    public void setId(int id) { this.id = id; }

    public String getJob_id() { return job_id; }
    public void setJob_id(String job_id) { this.job_id = job_id; }

    public String getCoordinates() { return coordinates; }
    public void setCoordinates(String coordinates) { this.coordinates = coordinates; }

    public String getCreated_at() { return created_at; }
    public void setCreated_at(String created_at) { this.created_at = created_at; }

    @Ignore
    public LocationEntity() {
        // DATA TO IGNORE
    }

    public LocationEntity(String job_id, String coordinates, String created_at) {
        this.job_id = job_id;
        this.coordinates = coordinates;
        this.created_at = created_at;
    }
}

在我的 DOA 中，我有：

@Dao
public interface LocationDAO {
    @Query("SELECT * from location_table ORDER BY id ASC")
    List<LocationEntity> getAll();
}

并且存储库有：

public class LocationRepository {

    private static BudtrackDatabase budtrackDatabase;
    public LocationRepository(Context context) {
        budtrackDatabase = Room.databaseBuilder(context, BudtrackDatabase.class, AppConstants.DATABASE_NAME).build();
    }

    public static List<LocationEntity> getLocations() {
        return budtrackDatabase.locationDAO().getAll();
    }
}

我的要求：

public interface LocationRequest {

    /**
     * Send through multiple locations
     * @param token
     * @param coordinates
     */
    @FormUrlEncoded
    @POST("api/coordinates")
    Call<String> sendLocations(@Field("token") String token,
                               @Field("coordinates[]") List<LocationEntity> coordinates);
}

然后我尝试使用改造发送数据，如下所示：

String token = "12345";
List<LocationEntity> locations = locationRepository.getLocations();

Call<String> callLocation = sendLocationService.sendLocations(token, locations);
callLocation.enqueue(new Callback<String>() {
    @Override
    public void onResponse(Call<String> call, Response<String> response) {
        if (response.isSuccessful()) {
            // Logic for success
        } else {
            try {
                JSONObject jObjError = new JSONObject(response.errorBody().string());
                Toast.makeText(activity, jObjError.getString("message"), Toast.LENGTH_LONG).show();
            } catch (Exception e) {
                Toast.makeText(activity, e.getMessage(), Toast.LENGTH_LONG).show();
            }
        }
    }

    @Override
    public void onFailure(Call<String> call, Throwable t) {
        Log.e("Failed To Send Location", t.getMessage());
    }
});

然后我像这样在 API 端获取数据：

"coordinates" : [ 
        "com.yard8.budtrack.data.location.LocationEntity@4d87c17", 
        "com.yard8.budtrack.data.location.LocationEntity@6c60804", 
        "com.yard8.budtrack.data.location.LocationEntity@fdad1ed", 
        "com.yard8.budtrack.data.location.LocationEntity@66fc822", 
        "com.yard8.budtrack.data.location.LocationEntity@8dec2b3", 
        "com.yard8.budtrack.data.location.LocationEntity@f9da070", 
];

发送数据以使其作为 JSON 数组通过的正确方法是什么？

Answer 1

我可能一直试图让事情变得过于复杂，而我需要做的就是改变

@Field("coordinates[]") List<LocationEntity> coordinates

到

@Field("coordinates") String coordinates

然后在调用之前使用...

Gson gson = new Gson();
String locations = gson.toJson(locationRepository.getLocations());

...并将它作为我的第二个参数传递。

Call<String> callLocation = sendLocationService.sendLocations(token, locations);

我真的认为这仅适用于单个对象而不是它们的列表，因为我一直在接收数据服务器端。

Answer 2

你有没有尝试过这样的事情

1.创建新类

class ParentEntity{

        private List<LocationEntity> locationEntities;

        public List<LocationEntity> getLocationEntities() {
            return locationEntities;
        }

        public void setLocationEntities(List<LocationEntity> locationEntities) {
            this.locationEntities = locationEntities;
        }
    }

2。然后在您的响应改造中

// ....... //
@Override
public void onResponse(Call<String> call, Response<String> response) {
    if (response.isSuccessful()) {
        // Logic for success
        ParentEntity readObj = response.body();
        List<LocationEntity> locations = readObj.getLocationEntities();

        // DO SOMETHING YOU WANT

    } else {
        try {
            JSONObject jObjError = new JSONObject(response.errorBody().string());
            Toast.makeText(activity, jObjError.getString("message"), Toast.LENGTH_LONG).show();
        } catch (Exception e) {
            Toast.makeText(activity, e.getMessage(), Toast.LENGTH_LONG).show();
        }
    }
}
// ....... //

更新 尝试使用@Body，可能结构如下 1.新建类

class ParentEntity{

        private LocationEntity location;
        private String token;

        public LocationEntity getLocation() {
            return location;
        }

        public void setLocation(LocationEntity location) {
            this.location = location;
        }

        public String getToken() {
            return token;
        }

        public void setToken(String token) {
            this.token = token;
        }
    }

1. 使用正文而不是字段，也没有@FormUrlEncoded

   @POST("/") 调用 sendLocations(@Body ParentEntity parentEntity);