Kotlin kotlinClass.class.getName() cannot return package name but only simple class name

Question

AClass.class.getName();

if AClass is a java class, this method will return package name and class name. but when i convert AClass java file to Kotlin file ,it will only return a class name. so system cannot find this class path

the code above

I'm not a professional in Kotlin, but you could try AClass.class.getPackage().getName(); and append that to your AClass.class.getName(); Like so: String location = AClass.class.getPackage().getName() + "." + AClass.class.getName(); — CJ Burkey
– CJ Burkey, Commented Jun 14, 2017 at 4:01
thanks, Class.forName(fragmentName) if fragmentName is a Kotlin class ,it will throws ClassNotFoundException. not matter fragmentName is complete name. so i was confused — 叶钦富
– 叶钦富, Commented Jun 14, 2017 at 6:34
val c = Class.forName(fragmentName)?.newInstance() fragment=c as Fragment fragmentName is a KClass complete name, it solves my problem — 叶钦富
– 叶钦富, Commented Jun 15, 2017 at 8:12

holi-java · Accepted Answer · 2017-06-14 10:09:35Z

55

there are many ways to get the full qualified name of a java Class in kotlin:

get name via the property KClass.qualifiedName:

val name = AClass::class.qualifiedName;

OR get name via the property Class.name:

val name = AClass::class.java.name;

OR get name via the method Class#getName:

val name = AClass::class.java.getName();

the table of the qualified name of a class as below:

|-----------------------|-----------------------|-----------------------|
|                       |          Class        |     Anonymous Class   |
|-----------------------|-----------------------|-----------------------|
| KClass.qualifiedName  |    foo.bar.AClass     |         null          |
|-----------------------|-----------------------|-----------------------|
| Class.name            |    foo.bar.AClass     |    foo.bar.AClass$1   |
|-----------------------|-----------------------|-----------------------|
| Class.getName()       |    foo.bar.AClass     |    foo.bar.AClass$1   |
|-----------------------|-----------------------|-----------------------|

edited Jun 14, 2017 at 10:09

answered Jun 14, 2017 at 9:20

holi-java

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4 Comments

Eugen Pechanec Over a year ago

If you added an example input sample and output of each described option, that'd be fantastic.

holi-java Over a year ago

@EugenPechanec thanks for your feedback, sir. how about it now?

叶钦富 Over a year ago

thanks, actually this is want i want

Will Chen Over a year ago

I believe the second and third techniques are exactly the same, as Kotlin exposes Java methods beginning with .get as properties: kotlinlang.org/docs/java-interop.html#getters-and-setters

CoolMind · Accepted Answer · 2019-10-09 08:39:37Z

10

If it is a java fragment

var fragmentSimpleName = FragmentName::class.java.simpleName as String

edited Oct 9, 2019 at 8:39

CoolMind

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answered Apr 16, 2018 at 20:55

LinuxFelipe-COL

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1 Comment

CoolMind Over a year ago

Thanks. As I understood, the same is for Kotlin fragment.

Nitin Patel · Accepted Answer · 2017-06-14 04:46:30Z

9

Try below solution::-

var name = MainActivity::class.java.canonicalName as String

answered Jun 14, 2017 at 4:46

Nitin Patel

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2 Comments

叶钦富 Over a year ago

Actually，It just a few code be covered into Kotlin, AClass.class.getName();this code above was invoked in java class , so i cannot use Kotlin syntax like your answer.

叶钦富 Over a year ago

Actually，It just a few code be covered into Kotlin, AClass.class.getName();this code above was invoked in java class , so i cannot use Kotlin syntax like your answer. i want a complete file name,because Class.forName(fragmentName) will be invoked in next. fragmentName should be replace by AClass complete name. but AClass is a Kotlin class now, it throws ClassNotFoundException

Kashif Anwaar · Accepted Answer · 2019-09-30 15:40:39Z

6

This is what I use to get class-name.

    val TAG = javaClass.simpleName

For Android developer's, it's very useful to declare as a field, and call to print logs.

answered Sep 30, 2019 at 15:40

Kashif Anwaar

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4 Comments

CoolMind Over a year ago

And what is javaClass?

Kashif Anwaar Over a year ago

javaClass is the class in which you type. just type javaClass in any class it'll work.

CoolMind Over a year ago

Thanks! It shouldn't be used as a constant in companion object, rather like a variable. In compaion object we can use val TAG = YourClass::class.java.simpleName.

Kashif Anwaar Over a year ago

@CoolMind That is long solution, you've to type your class name everywhere you use it. Whereas my line of code can be copied to any class and can be used without any modification.

Josep Bigorra · Accepted Answer · 2018-09-05 12:19:00Z

1

Maybe I am a little bit late for the party, but I do it using hash code of the new instance of the fragment. It is an Int so allows all kinds of tests.

 private val areaFragment by lazy { Area_Fragment.newInstance() }

    var fragmentHashCode = fragment.hashCode()
            when (fragmentHashCode) {
                areaFragment.hashCode() -> {
                    myNavigationView.setCheckedItem(R.id.nav_area)
            }

answered Sep 5, 2018 at 12:19

Josep Bigorra

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