Develop Reference

Can't open link like "intent://" in android app

I want to open link like "intent://qr.nspk.ru/url-something/#Intent;scheme=someScheme;package=com.example.android;end#Intent;scheme=someScheme;end" from an android application.
If I write something like this:
val i = Intent(Intent.ACTION_VIEW, Uri.parse(androidUrl))
startActivity(i)
I get this error:
No Activity found to handle Intent { act=android.intent.action.VIEW
dat=intent://qr.nspk.ru/url-something/#Intent;scheme=someScheme;package=com.example.android;end#Intent;scheme=someScheme;end
But If I go to this link from browser, it opens and it open an app that I need. Can you help me to understand why I can't open link from app and how to fix it?

The intent: scheme is used for the browser to encode an Android Intent. If you want to use this URL from an Android app, you need to create an Intent from the URL like this:
val i = Intent.parseUri(androidUrl, Intent.URI_INTENT_SCHEME)
startActivity(i)

In Kotlin how do I open a link in new Window

I am a tyro in Kotlin but I have a good Knowledge of Android and Core java. I am stuck on one condition while developing an android app via the Kotlin assistance.
I want that when user clicks a link present on the pdf document; the link should be opened on a browser (and if browser is opened then the link should open on the new window not new tab of the same window ).
I have achieved much of my objective but I didn't found out how to open a link in the new Window if the browser is open already?
I have tried the code below(when the link on the pdf is clicked then it redirects to the below function call):
fun web_page_open(urls: String) { // for more than one url
val uris = Uri.parse(urls)
val intents = Intent(Intent.ACTION_VIEW, uris)
startActivity(intents)
}
I have tried my level-best to explain my problem and also searched a lot(on github as well) but all my efforts went in vein.
Any help is warmly welcomed.
EDIT: Let's consider an instance if the user has already opened the default browser (say ABZfox) then when the link inside the pdf(or a doc) is clicked then the new Window of ABZfox opens instead of the same window in which the user was previously working. I'm sure the question makes some sense now!!!

You can try this one, might be helpful, open new tab of web browser like
fun openNewTabWindow(urls: String, context : Context) {
val uris = Uri.parse(urls)
val intents = Intent(Intent.ACTION_VIEW, uris)
val b = Bundle()
b.putBoolean("new_window", true)
intents.putExtras(b)
context.startActivity(intents)
}

You may use chrome custom tabs instead
To use it you need to add below dependency to your gradle
compile 'com.android.support:customtabs:23.1.1'
Now use below code to open the url
CustomTabsIntent.Builder builder = new CustomTabsIntent.Builder();
builder.enableUrlBarHiding();
builder.setShowTitle(true);
builder.setToolbarColor(Color.TRANSPARENT);
CustomTabsIntent customTabsIntent = builder.build();
customTabsIntent.launchUrl(getActivity(), Uri.parse(url));

Open URL on phone from watch face

I have a watch face that I've created and I was looking into adding an "About" screen in the watch settings (on the watch). This screen would show the current version and show a button to open a URL leading to a changelog. I want the user to be able to click the button on their watch, and have it open the specified URL on the phone.
Is this possible without me creating a mobile companion application?

You can do this if the watch is running Android Wear 2 using a RemoteIntent, as such:
Intent intent = new Intent(Intent.ACTION_VIEW)
.addCategory(Intent.CATEGORY_BROWSABLE)
.setData(Uri.parse("http://www.google.com"));
RemoteIntent.startRemoteActivity(context, intent, null);
...replacing http://www.google.com with your desired URL, of course.
AFAIK, there's no way to accomplish this on Wear 1.x without sending a message to your handheld app and programming it to open the URL itself.

Note the new way to do this is with RemoteActivityHelper:
val remoteActivityHelper = RemoteActivityHelper(context, Executors.newSingleThreadExecutor())
val result = remoteActivityHelper.startRemoteActivity(
Intent(Intent.ACTION_VIEW)
.addCategory(Intent.CATEGORY_BROWSABLE)
.setData(
Uri.parse("http://www.google.com/")
),
null
)
and if you want to show the nice "Open on phone" animation, use ConfirmationActivity:
startActivity(
Intent(this, ConfirmationActivity::class.java)
.putExtra(ConfirmationActivity.EXTRA_ANIMATION_TYPE, ConfirmationActivity.OPEN_ON_PHONE_ANIMATION))

Android: starting a default application or one which may be different on different devices

I would like to start a default application: browser, contact-book, phone, email, music app, etc. I have found many q/a, like browser opening a specific URL or blank, and here the answer is even "No not possible". But I would like to just open/launch it without telling it to go to a specific URL or sending a mail to someone, etc.
However, I also saw some Home applications where this seems to be working (at least for some apps). On my colleague's device there is for example a different contact-book (no google) which is detected and opened correctly.
I have seen in the Android documentation some intent categories that point to these problems, but these are only >= API.11. So I can't use/test them on my device.
Question: Is it not somehow possible to launch a default application (having the app chooser is of course ok) without providing extra data? If no, what do you think are these Home apps doing (perhaps workarounds are somehow possible).
PS: for the phone app I think, I have a workaround using Intent.ACTION_DIAL without any other information which will open simply the dialer.
UPDATE: I modified the title. Some applications like the address book may not be the same on different devices. So in this case I would like to start the address-book app, whichever this is.

This answer is not a 100% answer, but some workarounds on some typical applications.
Still open are: music player, address book
Browser: I get a list of applications that handle "http"-data intents, and then I look if one is available in the list of preferred applications.
Intent appFilter = new Intent(Intent.ACTION_VIEW);
appFilter.setData(Uri.parse("http://www.google.com"));
List<ResolveInfo> browserInfoList = pm.queryIntentActivities(appFilter, 0);
List<IntentFilter> outFilters = new ArrayList<IntentFilter>();
List<ComponentName> outActivities = new ArrayList<ComponentName>();
pm.getPreferredActivities(outFilters, outActivities, null);
if(outActivities.size() > 0) {
for(ComponentName cn : outActivities) {
String cnClass = cn.getClassName();
String cnPkg = cn.getPackageName();
for (ResolveInfo info : browserInfoList) {
if(info.activityInfo.name.equals(cnClass) &&
info.activityInfo.packageName.equals(cnPkg)) {
return cn;
}
}
}
}
In case no default is found, I open a browser chooser dialog, see here.
Phone: as described in the question:
Intent intent = new Intent(Intent.ACTION_DIAL);
startActivity(intent);

You can start apps by the function "startActivity" if you know about the canonical app name
like "android.com.browser". Do this simple by searching for AndroidManifest.xml in the app
source code (look at Codeaurora.com or at github/Cyanogenmod) and grab the app name you want.
After you know about the App name ("Activity") implement the code as follows:
Intent intent = new Intent();
intent.setClassName(this, "com.android.browser");
intent.setCategory(Intent.ACTION_MAIN);
startActivity(intent);
THIS is only a example, sometimes you have to put intent extras or data values, this information can be found in the app's AndroidManifest.xml too.

Android - Call default browser with and redirect to a designated url

Hi
I want to write an app to call default browser and redirect to a designated url.
Any suggestion to 1)call the default browser, 2)redirect to a designated url.
Thanks

you just want to launch an ACTION_VIEW intent with the Uri of the webpage as your data element :
Intent httpIntent = new Intent(Intent.ACTION_VIEW);
httpIntent.setData(Uri.parse("http://www.bbc.co.uk"));
startActivity(httpIntent);

To open the default browser use an Intent with the action VIEW. To tell the browser which page to load us the data-part of the Intent.
Example:
Intent browse = new Intent(Intent.ACTION_VIEW, Uri.parse("http://stackoverflow.com"));
startActivity(browse);
Since this is a basic task in Android you might want to read some basics about Intents in Android.