1

Well i tried some samples in my code to open app on clicking on an link in the browser but still i can't able to do what i want in my app. I want trough these links xamarin and this. Can anyone suggest me what to do, the right way to get the soultion for my question, advance thanks.

My Code:

[IntentFilter (new[]{Intent.ActionView}, 
 Categories = new [] {Intent.CategoryDefault}, 
 DataScheme = "superduperapp",
 DataHost = "something")]

link is for example: "superduperapp://my_code_is_here"

Umair M
  • 10,298
  • 6
  • 42
  • 74
Vicky
  • 921
  • 1
  • 11
  • 33

1 Answers1

4

You might be missing CategoryBrowsable. Try this:

[IntentFilter ( 
    new [] { Intent.ActionView }, 
    Categories = new [] { Intent.CategoryDefault, 
    Intent.CategoryBrowsable }, 
    DataScheme = "superduperapp", 
    DataHost = "my_code_is_here")]

Also, remember that you can easily test your intents via adb:

adb shell am start -a android.intent.action.VIEW -d superduperapp://my_code_is_here
Umair M
  • 10,298
  • 6
  • 42
  • 74
Eugen Timm
  • 744
  • 6
  • 13
  • well i have my link as "superduperapp://email="; + email + "&activationCode=" + activationCode + "&type=" + type; what to set in host and scheme? i have set scheme = superduperapp and host = ? – Vicky Jul 01 '15 at 09:39
  • If I understand this other question (http://stackoverflow.com/questions/22921637/android-intent-data-uri-query-parameter) correctly, you should be fine with just providing "email" as your DataHost and "superduperapp" as your DataScheme. When you actually receive the data inside your app you should be able to get the full query string. – Eugen Timm Jul 01 '15 at 21:22