When to use which constructor for ComponentName in Android?

android constructor components
Zordid picture Zordid · Feb 13, 2011 · Viewed 25.1k times · Source

I am a little bit confused about the ComponentName class in Android.

There are different ways to get to a component name object, but I don't know when to use which... and why!

Example:

  • Application package is de.zordid.sampleapp
  • but widget provider class is de.zordid.sampleapp.widget.WidgetProvider

Using 

ComponentName cn = new ComponentName("de.zordid.sampleapp.widget",
    "WidgetProvider");

I got this component info: ComponentInfo{de.zordid.sampleapp.widget/WidgetProvider}, but I could not use this - the component is unknown! But the JavaDoc says I should give the package and the class within that package - and that is what I did, didn't I??

Using

ComponentName cn = new ComponentName(context, WidgetProvider.class);

yields ComponentInfo{de.zordid.sampleapp/de.zordid.sampleapp.widget.WidgetProvider} - and that works fine!!

There is even another way to get a ComponentName - by context and a string. Which one should be used where and when??

Thanks!

Answer

Robert Tupelo-Schneck picture Robert Tupelo-Schneck · Feb 13, 2011

The ComponentName constructor taking two Strings can be used to refer to a component in another application. But, the first argument is not the package name of the class; it is the package name of the application---the package attribute of the manifest element in that application's AndroidManifest.xml. So your first example should be

ComponentName cn = new ComponentName("de.zordid.sampleapp",
    "de.zordid.sampleapp.widget.WidgetProvider");

That constructor could certainly be used to refer to components in your own application, but since you already have hold of a Context from your own application you might as well use it and use one of the other constructors. In my opinion, the one taking a Class should be preferred whenever usable. You could use the one taking a String if you only know the class dynamically for some reason; in that case, it should take the fully-qualified class name as above.

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