如何从多部分/表单数据响应中提取数据

Question

我正在使用POST方法将图像上传到storageAPI，内容类型为多部分/表单数据。api返回一个对象响应，如下所示：

{
"id": "6d50c066-cf65-4748-8b9a-183c3526f49b",
"name": "hotel_6.jpg",
"fileKey": "lv/im/5d9feb8e-2ea8-439d-a550-1e937081e085-hotel_6.jpg",
"fileExtension": ".jpg",
"mimeType": "image/jpeg",
"catalogueUrl": {
"mainUrl": "https://xy.abc.com/lv/im/5d9feb8e-2ea8-439d-a550-1e937081e085-hotel_6.jpg",
"thumbnailUrls": []
},
"createdAt": "2021-11-25T06:40:40.0869466+00:00"

}

如何提取变量"；mainUrl"；以便我可以将其值分配给_pictureController？以下是我所做的：

uploadFile() async {
var accessToken = await sharedPref.read(key);
var postUrl = '$baseUrl/catalogue?thumbnail=${param.thumbnailTrueFalse}';
Map < String, String > headers = {
"Authorization": "Bearer $accessToken",
};
// multipart request object
var request = http.MultipartRequest("POST", Uri.parse(postUrl));
request.headers.addAll(headers);
// add selected file with request
request.files.add(http.MultipartFile("file", imageStream, imageSize,
filename: imageName));
// Send request
var response = await request.send();
// Read response
var result = await response.stream.bytesToString();
print('readResponse: $result');
if (response.statusCode == 200) {
var data = StorageResponse.fromJson(jsonDecode(result));
print('data: $data');
setState(() {
_pictureController.text = data.catalogueUrl!.mainUrl!;
});
return data;
} else {
throw Exception('Failed to upload photo.');
}
}

"；StorageResponse"；类别如下：

@JsonSerializable()
class StorageResponse {
var id;
var name;
var fileKey;
var fileExtension;
var mimeType;
Catalogue ? catalogueUrl;
var createdAt;
StorageResponse({
this.id,
this.name,
this.fileKey,
this.fileExtension,
this.mimeType,
this.catalogueUrl,
this.createdAt,
});
factory StorageResponse.fromJson(Map < String, dynamic > json) =>
_$StorageResponseFromJson(json);
Map < String, dynamic > toJson() => _$StorageResponseToJson(this);
@override
toString() {
String output =
'{id:${this.id},name:${this.name},fileKey: ${this.fileKey},fileExtension:${this.fileExtension},mimeType: ${this.mimeType}mimeType},catalogueUrl: ${this.catalogueUrl},,createdAt: ${this.createdAt}}';
return output;
}
}

Answer 1

您可以使用以下结构将Json文件转换为类，反之亦然。

以下结构工作正常。

import 'dart:convert';
class StorageResponse {
final String id;
final String name;
final String fileKey;
final String fileExtension;
final String mimeType;
Catalogue catalogueUrl;
final DateTime createdAt;
StorageResponse(
this.id,
this.name,
this.fileKey,
this.fileExtension,
this.mimeType,
this.catalogueUrl,
this.createdAt,
);
factory StorageResponse.fromMap(Map<String, dynamic> json) {
return StorageResponse(
json['id'],
json['name'],
json['fileKey'],
json['fileExtension'],
json['mimeType'],
Catalogue.fromMap(json['Catalogue']),
DateTime.parse(json['createdAt']));
}
Map<String, dynamic> toJson() => {
'id': id,
'name': name,
'fileKey': fileKey,
'fileExtension': fileExtension,
'mimeType': mimeType,
'Catalogue': catalogueUrl.toJson(),
'createdAt': createdAt
};
@override
toString() {
String output =
'{id:${this.id},name:${this.name},fileKey: ${this.fileKey},fileExtension:${this.fileExtension},mimeType: ${this.mimeType}mimeType},catalogueUrl: ${this.catalogueUrl},,createdAt: ${this.createdAt}}';
return output;
}
}
class Catalogue {
final String mainUrl;
final List<String> thumbnailUrls;
Catalogue(this.mainUrl, this.thumbnailUrls);
factory Catalogue.fromMap(Map<String, dynamic> json) {
return Catalogue(json['mainUrl'], jsonDecode(json['thumbnailUrls']));
}
Map<String, dynamic> toJson() =>
{'mainUrl': mainUrl, 'thumbnailUrls': jsonEncode(thumbnailUrls)};
}

用于

if (response.statusCode == 200) {
var data = StorageResponse.fromMap(jsonDecode(result));
print('data: $data');
setState(() {
_pictureController.text = data.catalogueUrl!.mainUrl!;
});
return data;
} else {
throw Exception('Failed to upload photo.');
}