Create Logger on Package Level Without Class

Question

I have a kotlin file with a couple of package-level functions and without any class. I would like to add logging to this class but struggle to find an elegant way to give the logger an identifier.

This is an example

package com.example.myproject.my_package

import org.slf4j.LoggerFactory


private val log = LoggerFactory.getLogger("com.example.myproject.my_package")

fun bla(term: String) {
   log.info("invoked with $term")
}

There are very good best practices to use classes to find a good identifier: link 1 link 2. What's the approach if there are no classes?

I would like to avoid writing the identifier by hand and adjust it when the package name changes. Is there a way to get the package name in kotlin?

Top-level functions are good when they are utility functions. But logic should be located in a class (or object). — Miha_x64
– Miha_x64, Commented Jul 27, 2017 at 10:00
why that? object oriented programming is only one way to develop your code, there are other paradigm that kotlin perfectly is suited for. — linqu
– linqu, Commented Jul 27, 2017 at 10:28
That's not because of OOP paradigm. (By the way, Singleton, which object stands for, is a procedural pattern.) That's because your functions are gathered in a file, which is not a fully-functional scope — it has no representation in a programming language. — Miha_x64
– Miha_x64, Commented Jul 27, 2017 at 11:55

GaetanZ · Accepted Answer · 2017-07-30 17:37:36Z

13

This line should do the job:

private val log = LoggerFactory.getLogger(object{}::class.java.`package`.name)

The object is doing nothing but nevertheless kotlin will create a class to hold the code. You can then access the class, the package, the package name.

Note the usage of backtick because package is a reserved keyword. In java it is not a problem because the real method is getPackage(). The shorter syntax of kotlin transform this method call by a direct access to the property which name now collides with reserved keywords.

edited Jul 30, 2017 at 17:37

answered Jul 29, 2017 at 12:57

GaetanZ

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2 Comments

linqu Over a year ago

Great answer, thanks. Finally I went with the solution given by Frank's comment. If you are fine with this extra dependency his solution is even more elegant.

GaetanZ Over a year ago

Of course, I was more focused on your precise question "Is there a way to get the package name in kotlin?" than what I would do myself.

hunterwb · Accepted Answer · 2017-11-12 13:00:26Z

5

You can create a top-level logger for the generated class like this:

private val logger = LoggerFactory.getLogger(MethodHandles.lookup().lookupClass())

Or for the package like this:

private val logger = LoggerFactory.getLogger(MethodHandles.lookup().lookupClass().`package`.name)

This has the benefit of not creating an additional class file but it requires atleast Java 7. I have created a library which assists with this and more at https://github.com/kxtra/kxtra-slf4j.

answered Nov 12, 2017 at 13:00

hunterwb

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1 Comment

dQw4w9WyXcQ Over a year ago

better answer. good info, should be set as the accepted.

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