我想从xamarin android c#中发送数据到一个外部PHP脚本。

Question

这是我在login.cs下面的代码

var user = User.Text;
var pass = Pass.Text;
 try
    {

        var postData = new List<KeyValuePair<string, string>>();
        postData.Add(new KeyValuePair<string, string>("username", user));
        postData.Add(new KeyValuePair<string, string>("password", pass));

        var content = new FormUrlEncodedContent(postData);

        HttpClient client = new HttpClient();

        client.BaseAddress = new Uri("Http://10.0.2.2:3307");

        var response = await client.PostAsync("Http://10.0.2.2:3307/login.php", content);
        result = response.Content.ReadAsStringAsync().Result;

    }
    catch (Exception ex)
    {
        await DisplayAlert("Error", ex.ToString(), "Ok");
        return;
    }

但我在下面这行得到一个错误，说 Java.net.protocol: 意外的状态行 "Y."

    var response = await 
    client.PostAsync("Http://10.0.2.2:3307/login.php", content);

Answer 1

我设法解决了这个问题。

问题是我试图连接到一个错误的端口，我把它从3307改成了80，而且android模拟器和xampp使用相同的ip地址，所以我不得不检查如何连接到外部本地服务器的文档。你可以在这里查看。

https:/developer.android.comstudiorunemulator-networking。

而且我用json来解析用户模型类，而不是使用KeyValuePair，因为它没有那么好用。

var user = User.Text;
var pass = Pass.Text;
try{

    User us = new User();
    us.username = user;
    us.password = pass;

    string json = JsonConvert.SerializeObject(us);

    var content = new StringContent(json, Encoding.UTF8, "application/json");

    HttpClient client = new HttpClient();

    Uri uri = new Uri("Http://10.0.2.2:80/api/login.php");       

    client.BaseAddress = new Uri("Http://10.0.2.2:80");

    HttpResponseMessage response = await client.PostAsync(uri, cont);
    string  result =  await response.Content.ReadAsStringAsync();
    result = result.Trim('[', ']');

    dynamic output = JsonConvert.DeserializeObject(result);

}
catch (Exception ex)
{
    await DisplayAlert("Error", ex.ToString(), "Ok");
    return;
}