This is my code below login.cs
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;
}
But I get a error at this line below saying Java.net.protocol: Unexpected status line "Y. "
var response = await
client.PostAsync("Http://10.0.2.2:3307/login.php", content);
I managed to fix the problem :
The problem was that i tried to connect to a wrong port and i changed it from 3307 to 80, and also the android emulator uses the same ip address as xampp so i had to check the documentation on how to connect to an external local server. You can check here:
https://developer.android.com/studio/run/emulator-networking
And i used json to parse the user model class instead of using KeyValuePair because it didn't work as well.
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;
}