Android 改造 POST ArrayList

Question

正在尝试发送 List<String> 到服务器并出现错误请求错误。

public interface PostReviewApi {
        @FormUrlEncoded
        @POST("/")
        void addReview(@Field("review[place_id]") int placeId, @Field("review[content]") String review,
                       @Field("review[rating]") float rating, @Field("review[tag_list]") List<String> tagsList, Callback<ReviewEntity> callback);
    }

...
    postApi.addReview(mRestaurantId, reviewText, (float) mRating, tagsList, new Callback<ReviewEntity>() {
                @Override
                public void success(ReviewEntity reviewEntity, Response response) {
                }

                @Override
                public void failure(RetrofitError error) {
                }
            });
        }

使@Field("review[tag_list[]]")也无济于事

Answer 1

尝试使用 @Body 注释而不是 @Field 并传递单个 ReviewBody 对象。

class ReviewBody {

    public Review review;

    public ReviewBody(int placeId, float rating, String content, List<String> tagList) {
        review = new Review(placeId, rating, content, tagList);
    }

    public class Review {
        @SerializedName("place_id")
        public int placeId;
        public float rating; 
        public String content;
        @SerializedName("tag_list")
        public List<String> tagList;

        public Review(int placeId, float rating, String content, List<String> tagList) {
            this.placeId = placeId;
            this.rating = rating;
            this.content = content;
            this.tagList = tagList;
        }
    }
}

@POST("/")
void addReview(@Body ReviewBody body, Callback<ReviewEntity> callback);

（没有@FormUrlEncoded）

Answer 2

@字段注解用于表单编码请求。例如，

 @FormUrlEncoded
 @POST("/")
 Call<ResponseBody> example(
     @Field("name") String name,
     @Field("occupation") String occupation);


Calling with foo.example("Bob Smith", "President") yields a request body of name=Bob+Smith&occupation=President.

但是您将无法使用@Field 注释发送ArrayList tagList。

而是创建一个配置 class 并将 class 变量注释为 @SerializedName ("XYZ") 后跟变量名称以启用 Retrofit 中数组列表的序列化。请找到以下代码以供参考，

  public class Hotels {

  private @SerializedName("place_id") int mRestaurantId;
  private @SerializedName("content") String review;
  private @SerializedName("rating") float rating;
  private @SerializedName("tag_list") List<String> tagsList;

  public int getmRestaurantId() {
    return mRestaurantId;
  }

  public void setmRestaurantId(int mRestaurantId) {
    this.mRestaurantId = mRestaurantId;
  }

  public String getReview() {
    return review;
  }

  public void setReview(String review) {
    this.review = review;
  }

  public float getRating() {
    return rating;
  }

  public void setRating(float rating) {
    this.rating = rating;
  }

  public List<String> getTagsList() {
    return tagsList;
  }

  public void setTagsList(List<String> tagsList) {
    this.tagsList = tagsList;
  }

您还可以定义您的界面，如下所示，

@POST("/") Call<ReviewEntity> addReview(@Body Hotels hotels);

希望这对未来的前景有所帮助

Answer 3

在Multipart的情况下：

@Part("item_ids[]") items : List<Int>

完整示例：

假设您要注册（发送用户数据 - 上传任何图片和列表）

@Multipart
@POST(REGISTER_LINK)
suspend fun register(
    @PartMap partMap: MutableMap<String, RequestBody>, // (name, email..etc)
    @Part file: MultipartBody.Part?, // image file
    @Part("item_ids[]") listOf: List<Int> // list of
): NetworkResponse<>

Android 改造 POST ArrayList

Android Retrofit POST ArrayList

android

android-networking

retrofit