Android - How to open the Hangout app from a Hangout URL displayed in the webview? -


i made android app displaying website in webview.
webpage has many links google hangouts url (https://talkgadget.google.com/hangouts/_/g23uuqx.......) in there people in call.

enter image description here

when link clicked on desktop's browser, url gets opened in new tab , user able join google hangout call.

when link clicked on mobile's browser, hangout app get's opened , user able join call. how can have same effect when link clicked in webview?

when open hangout url in mobile app(webview), opens google play webpage in app(webview) , asks me install hangout have installed.

enter image description here

instead of going google play store within webview of app,

(when google hangouts app installed on android phone)
1) how open hangout app app has hangout urls of form(https://talkgadget.google.com/hangouts/_/g23uuqx.......) in webview of app?

(when google hangouts app not installed on android phone)
2) how open google playstore app , inform user install google hangout app in order join call?

kindly help. thanks.

you can use webviewclient.shouldoverrideurlloading catch these links , start appropriate activity. example of how launch google play app way described here: allow 'market://' links inside webview open google play store


Comments

Post a Comment

Popular posts from this blog

c++ - Difference between pre and post decrement in recursive function argument -

i have following sample code used pre-decrement void function(int c) { if(c == 0) { return; } else { cout << "dp" << c << endl; function(--c); cout << c << endl; return; } } this function give output input 4 : dp3 dp2 dp1 dp0 0 1 2 3 but when use post decrement void function(int c) { if(c == 0) { return; } else { cout << "dp" << c << endl; function(c--); cout << c << endl; return; } } output goes in infinite loop dp4 can explain me in detail why happens ? this happens because function(c--) called same value of c , when finishes c decremented. recursively called, hence called same value no chance of return , hit stack overflow error. assume this, int x = 1, y = 0; y = x++ result y == 1 , x == 2. if y = ++x , bot x , y 2.
Read more

php - Nothing but 'run(); ' when browsing to my local project, how do I fix this? -

i added new project. made mistake in projectname.conf file in sites-available apache2, led me reinstalling apache2. after reinstalling , re-enabling projects on every page of every project .. <body> run(); </body> .. i've run composer update, restarted service, restarted pc, checked config files, tried disabling new project. there nothing in error log, , access.log: 127.0.0.1 - - [15/may/2015:01:40:47 +0200] "get / http/1.1" 304 181 "-" "mozilla/5.0 (x11; ubuntu; linux x86_64; rv:38.0) gecko/20100101 firefox/38.0" maybe i'm forgetting apache2 reinstall? don't think did different last time installed it. looks haven't setup php correctly. take @ this great guide , maybe you're forgetting link php config apache2.
Read more

php - How can I echo out this array? -

i don't know how go grabbing data in variable , put foreach loop. class tree{ private $info = array( array('name'=>'jim','state'=>'nu','company'=>'nu','phone'=>array('cell'=>'5615111111','office'=>'5611111111'),'email'=>array('primary'=>'exs@example.com','ex@there.com')), array('name'=>'joe smith','city'=>'phoenix','phone'=>'4805551111','email'=>'jsmith@some_email.com'), array('name'=>'john doe','city'=>'chandler','company'=>'doe co.','email'=>array('jdoe@gmail.com','personal'=>'email@email.com'),'phone'=>'6025550002') ); } if you're inside of class, can access private variable using $this->variablename . example, i...
Read more