Question

这是我的API响应，我希望所有建筑物名称都包含在字符串列表中

{
    "entity_id": "86",
    "building_name": "Burj Khalifa",
    "location": "Al  Ttay",
    "image_field": "1595916594oad.jpeg"
},
{
    "entity_id": "87",
    "building_name": "Azmair",
    "location": " Eyal Nasser ",
    "image_field": "1596541099s.jpeg"
},

]

我尝试了这个但没用

List<String> _buildlist = (jsonDecode(response['building_name']) as List<dynamic>).cast<String>();
print(_buildlist);

Answer 1

您需要解码JSON，然后将每个元素map转换为相关的字符串，然后将它们重新组成一个列表：

  var decoded = json.decode(j) as List;
  var names = decoded.map<String>((e) => e['building_name']).toList();
  print(names); // prints [Burj Khalifa, Azmair]

Answer 2

使用“ dart：convert”包尝试以下操作：

class examClass:
    def __init__(self, logger_tag):
        self.name = 'logger'
        self.logger = logging.getLogger(logger_tag + "." + self.name)
        self.logger.info("Created an instance of " + self.name) 

    def __del__(self):
        class_name = self.__class__.__name__
        self.logger.info("%s destroyed" % class_name)

您可以在https://dartpad.dev/中运行代码并对其进行测试

Answer 3

首先，您需要在数组列表中获取所有字段的值。为此，您首先需要解析所有响应。为使用app.quicktype.io的任何响应创建适当的模型类。这是来自app.quicktype.io的模型类，

import 'dart:convert';

List<ParseResponse> parseResponseFromJson(String str) => List<ParseResponse>.from(json.decode(str).map((x) => ParseResponse.fromJson(x)));

String parseResponseToJson(List<ParseResponse> data) => json.encode(List<dynamic>.from(data.map((x) => x.toJson())));

class ParseResponse {
    ParseResponse({
        this.entityId,
        this.buildingName,
        this.location,
        this.imageField,
    });

    String entityId;
    String buildingName;
    String location;
    String imageField;

    factory ParseResponse.fromJson(Map<String, dynamic> json) => ParseResponse(
        entityId: json["entity_id"],
        buildingName: json["building_name"],
        location: json["location"],
        imageField: json["image_field"],
    );

    Map<String, dynamic> toJson() => {
        "entity_id": entityId,
        "building_name": buildingName,
        "location": location,
        "image_field": imageField,
    };
}

现在，如果您调用parseResponseFromJson（resonseString）;那么您将获得列表，然后可以从这里开始。

Answer 4

您首先需要使用json.decode（）将JSON转换为Maps 列表。然后迭代地图列表以创建新的建筑物名称列表。切记要明确声明尽可能多的变量，并减少使用'var'关键字。它可以帮助编译器更快地编译程序，并提高代码的可读性。

以下代码显示了我的方法：

import 'dart:convert';
void main() {
  
 var response =[
    {
    "entity_id": "86",
    "building_name": "Burj Khalifa",
    "location": "Al  Ttay",
    "image_field": "1595916594oad.jpeg"
    },
    {
    "entity_id": "87",
    "building_name": "Azmair",
    "location": " Eyal Nasser ",
    "image_field": "1596541099s.jpeg"
    },
 ];
 List<String> _buildlist = List();
 List<Map<String,String>> data = json.decode(response);
 for(Map<String,String> i in data)
    _buildlist.add(i["building_name"]);

 print(_buildlist);
}

如何在颤振中将Json值转换为List <String>

4 个答案: