Question

这是代码，我想要做的就是测试是否在数据库中输入了用户凭据，然后通过 toast 消息显示成功，如果它们是，并且他们不返回错误消息

以下是我从其中一个网站上发现的所有说明，最终得到了以下代码，但吐司仍然没有显示出来。

using Android.App;
using Android.Widget;
using Android.OS;
using Android.Views.InputMethods;
using Android.Content;
using Android.Views;
using System;
using System.Net;
using System.Collections.Specialized;
using Org.Json;
using System.Text;
namespace App
{
    [Activity(Label = "App", MainLauncher = true, Icon = "@drawable/icon")]
    public class MainActivity : Activity, Android.Views.View.IOnClickListener
    {
        EditText username, password;
        Button signIn;
        signInAsync sn;
        protected override void OnCreate(Bundle bundle)
        {

            base.OnCreate(bundle);
            SetContentView(Resource.Layout.Main);
            initialize();
        }
        public void initialize()
        {
            username = (EditText)FindViewById(Resource.Id.editText1);
            password = (EditText)FindViewById(Resource.Id.editText2);
            signIn = (Button)FindViewById(Resource.Id.button1);
            signIn.SetOnClickListener(this);
        }
        public override bool OnTouchEvent(MotionEvent e)
        {
            InputMethodManager imm = (InputMethodManager)GetSystemService(Context.InputMethodService);
            EditText username = (EditText)FindViewById(Resource.Id.editText1);
            EditText password = (EditText)FindViewById(Resource.Id.editText2);
            //Cisti fokus
            username.ClearFocus();
            password.ClearFocus();
            //imm.HideSoftInputFromWindow(username.WindowToken, 0);
            return base.OnTouchEvent(e);
            //Sklanja tastaturu s ekrana na klik na pozadinu.
        }
        public void OnClick(View v)
        {
            switch (v.Id)
            {
                case Resource.Id.button1:
                    sn = new signInAsync(this);
                    sn.Execute();
                    break;
            }
        }
        public class signInAsync : AsyncTask
        {
            MainActivity mainActivity;

            public signInAsync(MainActivity mainActivity)
            {
                this.mainActivity = mainActivity;
            }
            string username, password;
            protected override void OnPreExecute()
            {
                base.OnPreExecute();

                username = mainActivity.username.Text;
                password = mainActivity.password.Text;
            }
            protected override Java.Lang.Object DoInBackground(params Java.Lang.Object[] @params)
            {

                WebClient client = new WebClient();
                client.UseDefaultCredentials = true;
                client.Proxy.Credentials = CredentialCache.DefaultCredentials;
                Uri uri = new Uri("http://192.168.1.198/android/login.php");
                NameValueCollection parameters = new NameValueCollection();
                parameters.Add("username", username);
                parameters.Add("password", password);
                var response = client.UploadValues(uri, parameters);
                var responseString = Encoding.UTF8.GetString(response);
                JSONObject ob = new JSONObject(responseString);
                if (ob.OptString("success").Equals("1"))
                {
                    mainActivity.RunOnUiThread(() =>
                    {
                        Toast.MakeText(mainActivity, "Uspješno ste se ulogovali", ToastLength.Short).Show();
                    });
                };
                if (ob.OptString("error").Equals("2"))
                    Toast.MakeText(mainActivity, "Pogresno", ToastLength.Short).Show();
                if (ob.OptString("error").Equals("3"))
                    Toast.MakeText(mainActivity, "Error", ToastLength.Short).Show();
                return null;
            }

        }
    }
}

这是我的PHP文件：

<?php
$db_name = "korisnici";
$mysql_username = "Mpro";
$mysql_password = "prolinet";
$server_name = "192.168.1.198";
$conn = mysqli_connect($server_name,$mysql_username,$mysql_password,$db_name);

if($conn) {
    echo "Connection success";
} else {
    echo "Faliure to connect";
}

if(isset($_POST['username']) && isset($_POST['password'])) {
    $user_name = $_POST['username'];
    $user_pass = $_POST['password'];
    $mysql_qry = "SELECT * FROM korisnici WHERE username = '$user_name' AND password = '$user_pass'";
    if(mysql_fetch_row($mysql_qry)){
        $response["success"] = 1;
        echo json_encode($response);

    } else{
        $response["error"]=2;
        echo json_encode($response);
    }

} else {
    $response["error"] = 3;
    echo json_encode($response);
}
?>

很抱歉打扰你，但我是Xamarin初学者。

Answer 1

更改此行

Toast.MakeText(mainActivity, "Uspješno ste se ulogovali", ToastLength.Short).Show();

使用：

Toast.MakeText(mainActivity.this, "Uspješno ste se ulogovali", ToastLength.Short).Show();

希望有所帮助

Answer 2

如果你想在asntectask的onPostExcute（）方法中使用asyntask成功或失败时显示toast成功或失败。

Answer 3

首先确保执行显示吐司的行。您可以使用Log类打印一些消息，以确保应用程序中发生的事情。

Answer 4

使用getContext或mainActivity.this代替mainActivity

Answer 5

您无法确定您的活动是否在后台线程中处于有效状态。因此，而不是在应用程序上下文中传递活动传递，如下所示：

public signInAsync(Context appContext)

然后做：

Toast.MakeText(appContext,...

因此，当您从某个活动中调用signInAsync时，您会将其称为：

signInAsync(this.ApplicationContext)

单击按钮时不会弹出Toast消息

5 个答案: