📜  字符串不是在颤振中输入 Uri - 任何代码示例

📅  最后修改于: 2022-03-11 14:58:27.945000             🧑  作者: Mango

代码示例1
// Simply do this if your link has no parameters or just a string
var url = Uri.parse(url_string)

// According to an answer @stackoverflow
final _authority = "example.com";
final _path = "/api";
final _params = { "q" : "dart" };
final _uri =  Uri.https(_authority, _path, _params);