Dart json.encode返回带有没有引号的键值的json字符串

Question

我正在尝试将字典转换为json字符串。但是我没有得到任何字符串的引号。我正在使用飞镖2。这就是我所拥有的

  var resBody = {};
  resBody["email"] = "employerA@gmail.com";
  resBody["password"] = "admin123";
  var user = {};
  user["user"] = resBody;
  String str = json.encode(user);

输出是：

{user: {email: employerA@gmail.com, password: admin123}}

我希望这就像一个真正的json对象

{"user": {"email": "employerA@gmail.com", "password: admin123"}}

我怎么能告诉飞镖在它周围加上引号？ 我查看了this线程，并且正在完成对用户有用的功能 我做错了吗？

Answer 1

这是按预期工作的

Invoke-WebRequest -Uri 'http://localhost:3000/api/add/user' -Method POST  -ContentType 'application/json' -Body (ConvertTo-Json $postParams -Compress)

打印

import 'dart:convert';

void main() {
  var resBody = {};
  resBody["email"] = "employerA@gmail.com";
  resBody["password"] = "admin123";
  var user = {};
  user["user"] = resBody;
  String str = json.encode(user);
  print(str);
}

DartPad example

Answer 2

如果响应内容太长，那么So below是一个功能，也可以帮助将长而精确的JSON数据打印到您的Dart终端中。可以像在Dart中使用print('some message')一样使用此函数，但是参数必须是JSON字符串。

import 'dart:convert'; //Don't forget to import this

void printJson(String input) {
  const JsonDecoder decoder = JsonDecoder();
  const JsonEncoder encoder = JsonEncoder.withIndent('  ');
  final dynamic object = decoder.convert(input);
  final dynamic prettyString = encoder.convert(object);
  prettyString.split('\n').forEach((dynamic element) => print(element));
}

您会使用/称呼它

  Future<dynamic> getYourApiData({double lon, double lat}) async {
    final String url = yourApiUrl;
    final http.Response response = await http.get(url);
    response = {
      "user": {"email": "employerA@gmail.com", "password": "admin123"}
    }; //Suppose your response is like this
    /*
  Below we are calling the printJson() with the response.body
  But you need to make sure it's in String format.
  */
    printJson(response.body.toString());
  }

它的输出将是这样