java get class name with package
getPackage() method is used to return the package of this class, we find the package of the class by using class loader. An example of this is java.util.ArrayList. Java Naming conventions.
In my particular case, I needed to find out every Class and Sub-Classes inside a package, thus reparteed within several Jars. If the class was loaded by the bootstrap class loader the set of packages loaded from CLASSPATH is searched to find the package of the class. As a Java developer you probably used to know about reflexion. Here's the explanation, taken from the exercise descriptionsourcedir can be either absolute (starting with “/”) or relative to where we run the program fromsourcedir is a given argument for the main method. In C# we have Type.FullName and Type.Name for getting the name of a type (class in this case) with or without the namespace (package in java-world).. What is the java equivalent to Type.Name?. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under By using our site, you acknowledge that you have read and understand our By clicking “Post Your Answer”, you agree to our To subscribe to this RSS feed, copy and paste this URL into your RSS reader.
This entity can be a class, an array, an interface, etc. Frame java.awt.Frame java.awt ---- ClassUtils ClassUtils ---- GregorianCalendar java.util.GregorianCalendar java.util Thanks to L. Janovszki for the idea Alternate way to get only the package name (null if there is no package) : This method returns the fully-qualified name of this package as defined in the Java Language Specification, Third Edition 6.5.3, for example, java.lang Package Utils: 12. Following is the declaration for java.lang.Package.getName() method. As a Java developer you probably used to know about reflexion.
To get the package name of a class we can use the getClass().getPackage() method. Java program to get class name without package In this example we are making a simple example which describes you that how you can get the class name without package.
Stack Overflow works best with JavaScript enabled I need to check the argument starts with a '/' (mention in the description). If there is no simple name it should throw an exception. Where developers & technologists share private knowledge with coworkersProgramming & related technical career opportunitiesI did a quick search on this and was not to impressed by the first results that I found on Google so consider this as a freebie.It also maps the internal name of an anonymous class to an empty String and does stuff with array class names.I didn't know it could return an empty string, and IMHO that's a design flaw. in my product, their are two classes in two different PROJECT with the Same Class name & package name. In fact, the previous example does just that: It defines a Rectangle class when there is already a Rectangle class in the java.awt package. if it does, just use it as absolute.
package org.kodejava.example.lang; import java.util.Date; public class … Note that a package / class name ("a.b.C") is different from the path of the .class files (a/b/C.class), and that using the package name / class name to derive a path is typically bad practice. By using our site, you acknowledge that you have read and understand our The fully-qualified class name can be obtained using the getName() method.
The fully-qualified class name can be obtained using the getName() method.
By clicking “Post Your Answer”, you agree to our To subscribe to this RSS feed, copy and paste this URL into your RSS reader. The class loader of this class is used to find the package. That is the class name including all package names. Get all information about a package: 5.
In particular the simple name of an array whose component type is anonymous is "[]".It is actually stripping the package information from the name, but this is hidden from you.The following function will work in JDK version 1.5 and above.Thanks for contributing an answer to Stack Overflow!
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