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Accepted
Aritzbn
Asked: 2020-03-29 01:26:09 +0800 CST 2020-03-29 01:26:09 +0800 CST

View stream in Xamarin

  • 772

I'm trying to decode a stream to bitmap so I can pass it through an imageview to play a stream.

Alternative 1:

I have based it on this StackOverflow question: How to load an ImageView by URL in Android? [closed] taking what I thought I needed, but when I run the App by calling I ca Funcion()n't show the image.

What am I doing wrong? I have taken the equivalent functions of javapara xamarin, the objects and others coincide so, I don't understand if my code or the stream is wrong or...

In my case I have tried to use several retransmissions taken from different places. Youtube, twitch, some television that broadcasts live or a local IP where I see stream through http.

private void Funcion()
{
   ImageView image = FindViewById<ImageView>(Resource.Id.vView);
   Bitmap stream = Task.Run(() => doInBackground()).Result;
   image.SetImageBitmap(stream);

}

protected Bitmap doInBackground()
{
    Bitmap stream;
    stream = BitmapFactory.DecodeStream(new Java.Net.URL("URL").OpenStream());

    return stream;
}

This is the view I use

<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:background="#000000"
    tools:showIn="@layout/activity_main">

    <ImageView
        android:layout_width="500dp"
        android:layout_height="500dp"
        android:background="#ffffff"
        android:id="@+id/vView" />
</RelativeLayout>

He OnCreate()only has the basics:

protected override void OnCreate(Bundle savedInstanceState)
{
    base.OnCreate(savedInstanceState);

    SetContentView(Resource.Layout.activity_main);

    Funcion();

}

The code has no more.

Alternative 2

Use VideoView:

VideoView video = FindViewById<VideoView>(Resource.Id.vView);


video.SetVideoURI(Android.Net.Uri.Parse("http://ip:puerto/"));
video.RequestFocus();
video.Prepared += delegate
{
   video.Start();
};

This option seems the simplest to me, but when I start I get a dialog in which it tells me: ** The video cannot be played.** And that's the end of it.

http

If I enter from the browser with http://ip:port/ and inspect the element, what I see is this:

<Body>
    <img style="-webkit-user-select: none;" src="http://ip:puerto/">
</Body>
c#

1 Answers

  1. Best Answer

    I have found a solution for this, as the stream can be seen from any browser I have chosen to use it WebViewto be able to view the "stream", which actually seems to be a sequence of images and deny the zoom and movements of the webviewso that it looks like a video .

    C# code:

    var wPlayer = FindViewById<WebView>(Resource.Id.webView1);
//Zoom Out para ajustar contenido a pantalla por width
wPlayer.Settings.LoadWithOverviewMode = true; 
wPlayer.Settings.UseWideViewPort = true;

wPlayer.SetPadding(0, 0, 0, 0);

wPlayer.LoadUrl("http://ip:port/?params");

    Axml:

    <?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
xmlns:app="http://schemas.android.com/apk/res-auto"
xmlns:tools="http://schemas.android.com/tools"
android:layout_width="match_parent"
android:layout_height="match_parent"
app:layout_behavior="@string/appbar_scrolling_view_behavior"
tools:showIn="@layout/activity_main">

<android.webkit.WebView
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:id="@+id/webView1"
/>


</RelativeLayout>
    • 3