Does there exist such a thing, where you send it a string as data, and it knows to open the application with the entry for that string?
Thanks.
if you want to allow user to search something new. Than you can not depend on other applications. This will launch the browser(which is present in all the android devices) with your text as the url so that user can search it using your app.
try this:
String url = "the text you want to search";
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i);
Or
the dictionary must have Broadcast Reciever
Related
I am using an intent to open up a specific string literal(the website address) in an Android device's browser.
If I go with
String myIntentString = "url:www.google.com"
Uri myInputUri = Uri.parse(myIntentString);
Intent intent = new Intent(Intent.ACTION_VIEW, myInputUri);
startActivity(intent);
Then the app crashes.
However, if I change my string to "http://www.google.com" (the correct format), then it obviously works.
Yes, this is a school assignment and it is specified that the string MUST be "url:www.google.com"
If it helps, this is the first assignment where we need to turn in the Android Manifest file. Does this mean it has to do with that?
What am I missing?
is it possible to ask website visitors to save your website on their home-screen?
android
Check out this answer here, on how to create a shortcut:
How to add android bookmark on homescreen from web page?
If you're writing an Android app, you can use this approach. However, if this is a website, this approach will not work, as it looks like the Android Intent system ignores website generated actions.
you can create a web page shortcut on home screen using this code. Provide the necessary info like url, title etc..
final Intent in = new Intent();
final Intent shortcutIntent = new Intent(Intent.ACTION_VIEW, Uri.parse(url));
long urlHash = url.hashCode();
long uniqueId = (urlHash << 32) | shortcutIntent.hashCode();
shortcutIntent.putExtra(Browser.EXTRA_APPLICATION_ID, Long.toString(uniqueId));
in.putExtra(Intent.EXTRA_SHORTCUT_INTENT, shortcutIntent);
in.putExtra(Intent.EXTRA_SHORTCUT_NAME, title);
in.putExtra(Intent.EXTRA_SHORTCUT_ICON_RESOURCE,
Intent.ShortcutIconResource.fromContext(
BrowserBookmarksPage.this,
R.drawable.ic_launcher_shortcut_browser_bookmark));
in.setAction("com.android.launcher.action.INSTALL_SHORTCUT");
//or in.setAction(Intent.ACTION_CREATE_SHORTCUT);
sendBroadcast(in);
I was wondering if there is a way to start a Browser with a yahoo search query. For example user can select a certain word or phrase and click a button and the activity will start the browser with the yahoo search query.
The quickest way is with intents, like so:
String query = "my search terms";
String url = "http://search.yahoo.com/search?p=" + query;
Intent i = new Intent(Intent.ACTION_VIEW);
i.setData(Uri.parse(url));
startActivity(i);
Android will automatically attempt to open this in the application associated with URLs which, by default, should be their default web browser.
Try appending the search keyword to Yahoo search api url and accessing it via uri onClick new button in your activity.
I'm trying to launch the Amazon Kindle app from my android application but based on a specific book that the user clicks.
I'm able to determine the books available and which book the user has selected but I have only been able to launch the kindle application (using the package name com.amazon.kindle) to launch the kindle app.
Does anyone know of any additional commands I can send to specify a book to open? I know this is possible as there is a widget on the google play store where the user selects a boo and it creates a button on the homescreen that launches the kindle app and opens the book.
Thanks in advance!
First we need to set the intent to an ACTION_VIEW intent.
Then we need to define an Uri for the data which is actually a link that looks something like: kindle://book/?action=open&book_id=AMZNID0/B000FC1GHO/0/, where in this case the section B000FC1GHO corresponds to the ID of the book.
Finally we can then start the activity. In my case I had to set some flags on the intent to launch a new activity.
The code I'm using is as follows:
if(intent.getAction().contains("BOOK_ACTION_"))
{
Log.w("LOG", "We have a book selected");
bookID = intent.getAction().substring(12);
Log.w("LOG", bookID);
Intent readbook = new Intent(Intent.ACTION_VIEW);
readbook.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
Uri file = Uri.parse("kindle://book/?action=open&book_id=AMZNID0/" + bookID + "/0/");
readbook.setData(file);
context.startActivity(readbook);
}
I'm overriding the onReceive method in this case so that I can perform some additional steps on each book. Presumably because I'm just setting an ACTION_VIEW intent this could have been handles in the other class that does the onClickListener for the imageview that holds the book I want.
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.