我有此URL
https://zozor54-whois-lookup-v1.p.rapidapi.com/?rapidapi-key=MYAPIKEYb&domain=DOMAINTOCHECK&format=FORMATTYPE
快速API提供两个标头和其他内容
我尝试了此代码通过探索HTTP包但不工作:
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);
}
1条答案
按热度按时间jpfvwuh41#
您需要使用参数对url进行编码
参见How do you add query parameters to a Dart http request?