How to really get the name of a class without the package?

Question

My question is intentionally titled nearly exactly like this one, but what I want is to strip the package name and nothing else. My classes look like

class AProperlyAndDescriptivelyNamedClass {
    class First {...}
    class Next {...}
    class Last {...}
}

class AnotherTopLevelLongNamedClass {
    class First {...}
    class Next {...}
    class Last {...}
}

As you can see, I'm using some long and descriptive class names for my top-level classes and repeating all the stuff in the nested classes would be pretty nonsensical. Using some strange shortcut would be no better. I always refer to the classes using their enclosing class (and no package) and that's how I want to print them too.

I know I could get the package name and strip a corresponding number of characters from the full class name, but I guess somebody did it already?

score 3 · Answer 1 · answered Dec 14 '13 at 12:25

I guess you need to implement a short method like this:

private static String getSimpleName(Class<?> clazz) {
    String simpleName = clazz.getSimpleName();
    if (clazz.isMemberClass()) {
        Class<?> enclosingClass = clazz.getEnclosingClass();
        simpleName = enclosingClass.getSimpleName() + "." + simpleName;
    }
    return simpleName;
}

Then

String simpleName = getSimpleName(AnotherTopLevelLongNamedClass.First.class);
System.out.println(simpleName);

will Output

AnotherTopLevelLongNamedClass.First

The above solution will only work with one level of enclosure, whereas `cls.getName().replace(cls.getPackageName() + ".", "")` is simpler and works with multiple levels. — Easty77, Dec 08 '22 at 10:54

score 2 · Answer 2 · answered Jul 09 '22 at 16:13

2

There is a method getSimpleName() in java.lang.Class<T>, which does indeed return only the name of the class.

Source

answered Jul 09 '22 at 16:13

Krumuvecis

121
3

dsafcdsge4rfdse · Answer 3 · 2013-12-14T12:53:15.977

Have a look at this:

    public static String findName (Object o){
        List<String> list = new ArrayList<String>();
        Class<? extends Object> clazz = o.getClass();

        list.add(clazz.getSimpleName());
        while (clazz.getEnclosingClass() != null) {
            clazz = clazz.getEnclosingClass();
            list.add(clazz.getSimpleName());
        }

        StringBuilder builder = new StringBuilder();
        for (int i=list.size()-1; i>=0; i--){
            builder.append(list.get(i));
            builder.append(".");
        }
        builder.delete(builder.length()-1, builder.length());
        return builder.toString();

    }

score 0 · Answer 4 · answered Mar 14 '21 at 03:43

Already you describe the best method, I think.

I know I could get the package name and strip a corresponding number of characters from the full class name, but I guess somebody did it already?

The earlier answers don’t take this approach, but I think they suffer for it. Roughly it would be:

cl.getName().substring( cl.getPackageName().length() )

Taking account of edge cases, here’s an implementation:

/** Returns `cl.getName` stripped of any leading package qualifier.
  */
String nameWithoutPackage( Class cl ) {
    String name = cl.getName();
    if( !cl.isArray() ) { // Then a package qualifier might lead the name.
        int strip = cl.getPackageName().length();
        if( strip > 0 ) name = name.substring( strip + /*dot separator*/1 ); }
    return name; }

How to really get the name of a class without the package?

4 Answers4

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