Processing is related to Android mode 'RTSP (RTP, SDP)'. Help me

GWAK · May 14, 2019, 4:07am

Processing is related to Android mode ‘RTSP (RTP, SDP)’.

The ‘processing.video’ library of the JAVA program makes it easy to resolve ‘RTSP’.

By the way, as far as I know
The ‘processing.video’ library does not support Android mode.
Question 1) Does the ‘processing.video’ library work in Android mode?

import processing.video.*;
 
Movie movie;
 
void setup() {
  size(640, 360);
  Movie.supportedProtocols[0] = "rtsp";
 
  background(0);
  try
    {
    movie = new Movie(this, "rtsp://127.0.0.1:8554/");
    }
  catch (Exception e)
    {
    }
  }
 
void movieEvent(Movie m) {
  m.read();
}
 
void draw() 
  {
  image(movie, 0, 0, width, height);
  }

Question 2) Is there a good way to do ‘RTSP’ in Android mode?

I need your help.

akenaton · May 14, 2019, 8:18am

@GWAK ===

question 1= no
question 2= yes, you can using android videoView or android native mediaPlayer - Be sure that the file referred by your rtsp is ok with android supported media codecs.
more details here: https://developer.android.com/guide/topics/media/media-formats

GWAK · May 15, 2019, 12:45am

@akenaton

Thank you for your kind answer.

GWAK · May 16, 2019, 1:03pm

Can I run that library for processing?
In the above manner,
I want to perform rtsp function.

akenaton · April 30, 2020, 6:22pm

@GWAK ===
difficult to help you without knowing exactly what you are doing:

what kind of codec is used by your cam
how the streaming is made (vlc?)
what is the stream output codec and url

GWAK · April 30, 2020, 9:28pm

@akenaton

Please check the image.

I would like to check the video and sound of CCTV video on mobile.
You cannot hear the sound with ‘MJPEG’.
So, I think it is necessary to implement ‘RTSP’ on mobile.

[JAVA MODE]

import VLCJVideo.*;

VLCJVideo video;

void setup() {
size(640, 360);
video = new VLCJVideo(this);
video.openAndPlay(“rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/”);
}

void draw() {
background(0);
image(video, 0, 0);
}

[INFORMATION]

ID : admin
PW : aa852456!
ip address : myksj1105.iptime.org:8081
(port : 8081)

Please review. I want to solve it in ‘Android Mode’.

akenaton · May 1, 2020, 1:18pm

@GWAK ===
ok, i understand what you are wanting;
one more question: are you sure that you have to use RTP ? Why? - Using http should be more simple with a webView.
Anyway it is not easy to code that using a videoView without any library; there are a lot of problems with codecs and as for now i think that the most sure way is to add the genuine VLC library to your app. It exists for android but is it possible to add it to Processing, i dont know. I have to see.

GWAK · May 1, 2020, 11:11pm

@akenaton,

I can`t be implemented as a webView.
(You can also implement it as ‘webView’?)

I tried, but it doesn’t.
Is there a workaround(solution)?

act = this.getActivity();
wbc = new WebViewClient();

webview = new WebView(act);
webview.setLayoutParams(new RelativeLayout.LayoutParams( 1000, 1500 ));
webview.setX(0);
webview.setY(0);
webview.setWebViewClient(wbc);
webview.getSettings().setJavaScriptEnabled(true);
webview.loadUrl(“rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/”);

fl = (FrameLayout)act.findViewById(0x1000);
fl.setVisibility(View.VISIBLE);
fl.addView(webview);

akenaton · May 2, 2020, 10:30am

@GWAK === what error message, if any, with the webView?

GWAK · May 2, 2020, 2:57pm

@akenaton

link : Can't play this video error in android WebView using RTSP stream - Stack Overflow

act = this.getActivity();
wbc = new WebViewClient();

webview = new WebView(act);
webview.setLayoutParams(new RelativeLayout.LayoutParams( 1000, 1500 ));
webview.setX(0);
webview.setY(0);
webview.setWebViewClient(wbc);
webview.getSettings().setJavaScriptEnabled(true);
webview.loadUrl(“rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/”);

fl = (FrameLayout)act.findViewById(0x1000);
fl.setVisibility(View.VISIBLE);
fl.addView(webview);

There is no error in the code above.

However, in order to execute ‘rtsp’, ‘ACTION_VIEW’ must be set.
Please refer to the link above.

Can you make the above source code for ‘Processing’?

Please review. Help. And thank you for the answer.

akenaton · May 2, 2020, 3:00pm

@GWAK === yes, i know that there is no error in this code and that ACTION_VIEW must be set; but when you set it to your URL what happens? What is the error code or what is the message inside the app?

GWAK · May 2, 2020, 10:24pm

@akenaton

import android.app.Activity;
import android.content.Context;
import android.widget.FrameLayout;
import android.view.ViewParent;
import android.widget.RelativeLayout;
import android.text.Editable;
import android.graphics.Color;
import android.widget.Toast;
import android.os.Looper;
import android.view.WindowManager;
import android.view.inputmethod.InputMethodManager;
import android.os.Bundle;
import android.os.Environment;
import android.view.View;
import android.view.ViewGroup;
import android.webkit.WebView;
import android.webkit.WebViewClient;
import java.net.URI;
import java.net.URISyntaxException;
import android.content.Intent;

Activity act;
FrameLayout fl;

WebView webview;
WebViewClient wbc;

//@Override
public void onStart() {
super.onStart();

act = this.getActivity();

wbc = new WebViewClient();

webview = new WebView(act);
webview.setLayoutParams(new RelativeLayout.LayoutParams( 1200, 800 ));
webview.setX(100);
webview.setY(50);
webview.setWebViewClient(wbc);
webview.getSettings().setJavaScriptEnabled(true);

webview.loadUrl(“rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/”);

webview.setWebViewClient(new WebViewClient() {
public boolean shouldOverrideUrlLoading(WebView view, String url) {
    if (url.startsWith("rtsp")) {
        URI uri = URI.parse(url);
        Intent intent = new Intent(Intent.ACTION_VIEW, uri);
        act.startActivity(intent);
        return true;
    }
    return super.shouldOverrideUrlLoading(view, url);
}
});

fl = (FrameLayout)act.findViewById(0x1000);
fl.addView(webview);
}

void settings() {
fullScreen();
}

void setup() {
orientation(PORTRAIT);
background(255, 250, 0);
}

void draw() {
}

ERROR in C:\Users\user\AppData\Local\Temp\android6658805089143585074sketch\src\processing\test\webview_test\webVIEW_TEST.java (at line 90)
URI uri = URI.parse(url);
^^^^^
The method parse(String) is undefined for the type URI

ERROR in C:\Users\user\AppData\Local\Temp\android6658805089143585074sketch\src\processing\test\webview_test\webVIEW_TEST.java (at line 91)
Intent intent = new Intent(Intent.ACTION_VIEW, uri);
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
The constructor Intent(String, URI) is undefined

5 problems (2 errors, 3 warnings)

I don’t know how to add the source code.

akenaton · May 3, 2020, 2:12pm

@GWAK ===
the code you have put cannot run; there are at least 3 errors:

fl (the frame layout) cannot be acceeded as you do (it was possible before, but now it does not work) ; you have to code:
fl = (FrameLayout) act.getWindow().getDecorView().getRootView();

your shouldOverrideUrlLoading(WebView view, String url){ method is absolutely useless in your case and it is deprecated; this method is useful when the user wants to change the url and that is not your case; more details here: https://developer.android.com/reference/android/webkit/WebViewClient#shouldOverrideUrlLoading(android.webkit.WebView,%20java.lang.String)
so you comment all these lines since webView.set…But if you want to use ACTION_VIEW instead of loadUrl then comment loadURL() and write:

String url = "rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/";
   Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
    act.startActivity(intent);

Last problem is that for parsing your import is not the good one; you have to import
import android.net.Uri;

Now i think that the code will run… Of course i cannot test as i dont know if your cam is running… but with another url it works.
dont forget to ask for internet permission!

GWAK · May 3, 2020, 3:15pm

@akenaton

I solved it thanks to you.

It wasn’t solved within the application screen, but it was connected to ‘video’ and played.

thank you. Thank you very much.

akenaton · May 3, 2020, 3:16pm

@GWAK === ok, put answered for others

noel · May 3, 2020, 3:17pm

Yes please. Show that part of the code!

GWAK · May 3, 2020, 3:19pm

import android.app.Activity;
import android.content.Context;
import android.widget.FrameLayout;
import android.view.ViewParent;
import android.widget.RelativeLayout;
import android.text.Editable;
import android.graphics.Color;
import android.widget.Toast;
import android.os.Looper;
import android.view.WindowManager;
import android.view.inputmethod.InputMethodManager;
import android.os.Bundle;
import android.os.Environment;
import android.view.View;
import android.view.ViewGroup;
import android.webkit.WebView;
import android.webkit.WebViewClient;
//import java.net.URI;
//import java.net.URISyntaxException;
import android.content.Intent;
import android.webkit.WebResourceError;
import android.webkit.WebResourceRequest;
import android.webkit.WebView;
import android.webkit.WebViewClient;

import android.net.Uri;

Activity act;
FrameLayout fl;

WebView webview;
WebViewClient wbc;

//@Override
public void onStart() {
super.onStart();

act = this.getActivity();

wbc = new WebViewClient();

webview = new WebView(act);

webview.setLayoutParams(new RelativeLayout.LayoutParams( 1000, 1500 ));
webview.setX(100);
webview.setY(50);
// webview.setWebViewClient(wbc);
webview.getSettings().setJavaScriptEnabled(true);
// webview.loadUrl(“rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/”);
String url = “rtsp://admin:aa852456!@myksj1105.iptime.org:8081/11/”;
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
act.startActivity(intent);

// fl = (FrameLayout)act.findViewById(0x1000);
fl = (FrameLayout) act.getWindow().getDecorView().getRootView();
fl.addView(webview);

}

void settings() {
fullScreen();
}

void setup() {
orientation(PORTRAIT);
background(255, 250, 0);
}

void draw() {
}

GWAK · May 3, 2020, 3:32pm

@akenaton
@noel

What we want to implement through this ‘solution’ is as follows.
Development of real-time verification system through CCTV

thank you.

noel · May 3, 2020, 3:53pm

I still don’t understand where the server resides. Because it is is on a local LAN isn’t it.
Is it the camara, that creates a virtual server?

GWAK · May 3, 2020, 3:59pm

@noel

Normally, a floating ip is given.

However, cctv requires a static ip address.
So the workaround can be implemented via ‘DDNS’.

exam1)
site1 ip address: 116.148.4.8
url: myksj1105.iptime.org

exam2)
site1 ip address: 116.148.4.20
url: myksj1105.iptime.org

You can also check it from outside through ‘DDNS’.

CCTV inside,
‘HTTP’
‘RTSP’
‘HTTPS’
Etc. The server is built-in.

We just need to match the ip address.

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