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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T00:56:18+00:00 2026-05-20T00:56:18+00:00

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

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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!

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  1. Editorial Team
    Editorial Team
    2026-05-20T00:56:19+00:00Added an answer on May 20, 2026 at 12:56 am

    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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