邮递员中的 Google book api 工作正常但在颤振中给出状态码 404答案

【问题标题】：Google book api in postman working fine But in flutter giving status code 404邮递员中的 Google book api 工作正常但在颤振中给出状态码 404
【发布时间】：2021-06-18 12:18:29
【问题描述】：

当我在邮递员中运行 google book API 时，它工作正常并给出状态码 200。但在颤振中，它给出了以下错误。并给出 404 状态码。我已经生成了 API 密钥，还启用了 Google Book API。所以如果有人知道我做错了什么，请帮助我。

TestLayout.dart

   class TestLayout extends StatefulWidget {
      static String id = "Test_Layout";
    
      @override
      _TestLayoutState createState() => _TestLayoutState();
    }
    
    class _TestLayoutState extends State<TestLayout> {
      Future<void> getData() async {
        String ApiKey = "..........";
    
        final response = await http.get(
            Uri.https("books.googleapis.com",
                "books/v1/volumes?q=flowers+inauthor:keyes&key=$ApiKey"),
            headers: {"Content-Type": "application/json"});
    
        var jsonBody = jsonDecode(response.body);
    
        //book b1=book.fromJson(jsonBody);
    
        print(response.statusCode);
        print(b1.id);
      }
    
      @override
      Widget build(BuildContext context) {
        return Scaffold(
          appBar: AppBar(
            title: Text("Flutter"),
          ),
          body: Column(
            children: [
              TextButton(
                  onPressed: () async {
                    await getData();
                  },
                  child: Text("Submit")),
            ],
          ),
        );
      }
    }

Book.dart：

class book {

  final int id;

   book({this.id});

   factory book.fromJson(Map<String,dynamic> json)
   {
     return book(
       id: json["totalItems"]
     );
   }
}

错误：

    E/flutter (27854): [ERROR:flutter/lib/ui/ui_dart_state.cc(199)] Unhandled Exception: 
    FormatException: Unexpected character (at character 1)
    E/flutter (27854): <!DOCTYPE html>
    E/flutter (27854): ^
    E/flutter (27854): 
    E/flutter (27854): #0      _ChunkedJsonParser.fail (dart:convert- 
    patch/convert_patch.dart:1404:5)
    E/flutter (27854): #1      _ChunkedJsonParser.parseNumber (dart:convert- 
     patch/convert_patch.dart:1271:9)
    E/flutter (27854): #2      _ChunkedJsonParser.parse (dart:convert- 
    patch/convert_patch.dart:936:22)
    E/flutter (27854): #3      _parseJson (dart:convert-patch/convert_patch.dart:40:10)
    E/flutter (27854): #4      JsonDecoder.convert (dart:convert/json.dart:506:36)
    E/flutter (27854): #5      JsonCodec.decode (dart:convert/json.dart:157:41)
    E/flutter (27854): #6      jsonDecode (dart:convert/json.dart:96:10)
    E/flutter (27854): #7      _TestLayoutState.getData 
    (package:freebookshare/TestFile/TestLayout.dart:22:20)
    E/flutter (27854): <asynchronous suspension>
    E/flutter (27854): #8      _TestLayoutState.build.<anonymous closure> 
    (package:freebookshare/TestFile/TestLayout.dart:40:17)
    E/flutter (27854): <asynchronous suspension>
    E/flutter (27854):

【问题讨论】：

标签： flutter dart

【解决方案1】：

我使用了 Uri.parse() 方法，它成功了。

String url="https://books.googleapis.com/books/v1/volumes?q=flowers+inauthor:keyes&key=$ApiKey";

final response = await http.get(Uri.parse(url),
    headers: {"Content-Type": "application/json"});

【讨论】：

【解决方案2】：

这是一个老问题，我以前也遇到过。在进行了一些研究之后，我通过删除 API 调用中的 Uri 构造函数获得了必要的输出。尝试删除 Uri Constructor 并添加最终响应的代码如下

    final response = await http.get("https://www.googleapis.com/books/v1/volumes?q=flowers+inauthor:keyes&key=$apiKey",
        headers: {"Content-Type": "application/json"});

之后TestLayout的代码如下

    class TestLayout extends StatefulWidget {
  static String id = "Test_Layout";

  @override
  _TestLayoutState createState() => _TestLayoutState();
}

class _TestLayoutState extends State<TestLayout> {
  Future<void> getData() async {
    String apiKey = "..........";

    final response = await http.get("https://www.googleapis.com/books/v1/volumes?q=flowers+inauthor:keyes&key=$apiKey",
        headers: {"Content-Type": "application/json"});

    var jsonBody = jsonDecode(response.body);
    
    book b1=book.fromJson(jsonBody);
    
    print(response.statusCode);
    print(b1.id);
  }

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(
        title: Text("Flutter"),
      ),
      body: Column(
        children: [
          TextButton(
              onPressed: () async {
                await getData();
              },
              child: Text("Submit")),
        ],
      ),
    );
  }
}

希望这对你有用

【讨论】：

我删除了 Uri 构造函数，但它给出了一个错误，例如“参数类型 'String' 不能分配给参数类型 Uri”
您是否正确更改了最终回复行，如我输入的行中所述？因为它对我来说完全没问题。（我刚刚创建了一个虚拟项目，结果是一样的。状态码是 200，甚至我能够打印 JSON 对象）
我认为新版本的dart get方法不接受String作为参数。