Android Kotlin open Link with button

Question

i try to open an link with a button in Kotlin, but if i use this code

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)
}

And in my button i use

openNewTabWindows("https://Google.com/")

It say it need context After url?

What does that mean?

Answer 1

openNewTabWindow(urls: String, context: Context) function needs 2 paramters, a String and a Context.

And in my button i use openNewTabWindows("https://Google.com/")

You just called this function with 1 parameter, then of course

It say it need context After url.

You need to pass a Context as the second parameter. Since you say you are implementing the action of clicking a button (which is, inside @Override public void onClick(View v) {} in Java, or a Lambda with type (View) -> Unit in Kotlin), which is probably inside an Activity, and the reference of this may be changed, you can pass getContext() or for example MainActivity.this as the context needed for program, or

openNewTabWindows("https://Google.com/", context) // Kotlin version of getContext()
openNewTabWindows("https://Google.com/", this@MainActivity) // Kotlin version of MainActivity.this

May both OK.