我如何将此快速 API 的 HTTP Url 转换为 Dart Http 请求

Question

我有这个 URL

https://zozor54-whois-lookup-v1.p.rapidapi.com/?rapidapi-key=MYAPIKEYb&domain=DOMAINTOCHECK&format=FORMATTYPE

快速 API 给出了两个标题和其他东西

我通过探索 HTTP package 尝试了这段代码，但没有工作：

import 'package:http/http.dart' as http;

void main() async {
  var url = 'https://zozor54-whois-lookup-v1.p.rapidapi.com/?domain=sendrank.com&format=json';
  var headers = {
    'X-Rapidapi-Key': APIKEyY
    'X-Rapidapi-Host': 'zozor54-whois-lookup-v1.p.rapidapi.com',
    'Host': 'zozor54-whois-lookup-v1.p.rapidapi.com'
  };

  var response = await http.get(url, headers: headers);
  print(response.body);
}

Answer 1

您需要使用参数对 url 进行编码

final queryParameters = {
  'domain': 'sendrank.com',
  'format': 'json',
};

final uri = Uri.https('zozor54-whois-lookup-v1.p.rapidapi.com', '/', queryParameters);

final response = await http.get(uri, headers: {
 'X-Rapidapi-Key': APIKEyY
    'X-Rapidapi-Host': 'zozor54-whois-lookup-v1.p.rapidapi.com',
    'Host': 'zozor54-whois-lookup-v1.p.rapidapi.com'
});

请参阅如何将查询参数添加到 Dart http 请求？