Develop Reference

Micronaut + Maven - Environment variables

I am working on a Micronaut + Maven project.
I need to parametrize some values of my application.yml such as passwords and connection strings, to avoid committing them.
I know values can be parametrized this way:
secret-value: '${SECRET_VALUE}'
But i cant find any other way to set SECRET_VALUE except setting bash value in .bashrc or .profile or .envoirment script files.
I would like to use a .env file somehow, in order to commit a .env.example file in git repo.
Any thoughts?

According to maven-resource-plugin documentation :
https://maven.apache.org/plugins/maven-resources-plugin/examples/filter.html
You could add a filter file in the filters tag of your pom file.
See we can separate "your.name" from the POM by specifying a filter file my-filter-values.properties containing: in the documentation above.

I may have lost the focus of the question.
the solution was simply create a .env file with required values, then run application using something like
dotenv run ./mvnw mn:run.
regarding #yunandtidus solution:
That is good only for "fixed" variables, such as application name, as pom values are shared between environments (assuming that we have an application-dev.yml and an application-prod.yml).
Keep in mind that, unlike .env, pom.xml must be committed, as any application.yml you have.

Create ZNodes without cmd in Zookeeper

I am trying to implement Configuration Management through Zookeeper. I have created few ZNodes from command line as:
create /config ""
create /config/mypocapp ""
create /config/mypocapp/name "John Doe"
Here, name is one of the properties that I want to access in my app called mypocapp.
Since we will be having a lot of properties in our application, we just can't use command line to create each and every property like this.
Is there a way we can create the properties in zookeeper through some UI or directly in a file (and import it to zookeeper).
I am completely new to zookeeper and not getting any help in this direction. Please help.

Exhibitor is one of the options that one can try to insert, modify or delete properties in ZNodes.
One can follow the steps given below:
Download the pom file of Exhibitor UI from the Github.
Built the pom file using maven that will generate a jar file.
Run the jar file as: java -jar <jar-file-name>.jar -c file
Go to your browser and type in: localhost:8080 to access Exhibitor UI.
Here, you can configure your Zookeeper ensemble and can edit the properties.
Please note that each instance of Zookeeper will have corresponding Exhibitor UI.
In order to run exhibitor on a different port, you can run:
java -jar <jar-file-name>.jar -c file --port <port-of-your-choice>

There are now also VS Code extensions that allow viewing and editing the Zookeeper node hierarchy and data, like this one:
https://marketplace.visualstudio.com/items?itemName=gaoliang.visual-zookeeper

How can I dynamically use externalized properties file with Spring jar?

With this setup (from Eclipse using Windows10)
I was able to correctly start my SpringBoot application. This one worked too (same directory pattern):
Now I'm packaging my project as JAR and I want to use an external properties file. I had an teste32.yml file beside my JAR at the same directory (also tried to use it inside /config directory, as show here, but it didn't work either)
I want to dynamically use a properties file beside my JAR file everytime. Doesn't matter at which directory they are, I wanted to dynamically point to a properties file always at the same directory as the JAR is. I want to say to my client: "take this JAR and this file, put them wherever you want and run this command X and everything will be alright". I'm trying to discover command X but before I add some dynamic path, I'm trying with absolutes paths. I'm using this:
java -jar myJar.jar -Dspring.config.name=teste32 -Dspring.config.location=C:\workspace\myProject\target\
I manually copied teste32 inside target\ to test this. But this didn't work. This didn't work either (only spring.config.location variants):
-Dspring.config.location=file:C:\workspace\myProject\target\
-Dspring.config.location=classpath:/
-Dspring.config.location=file:C:/workspace/myProject/target/
I also tried with no spring.config.location, only name
So my questions are:
What does classpath: and file: mean? Until now I got the 2 correct setups by pure luck and I would like to understand when to use them.
When I have my project package as a JAR, what classpath becomes?
Finally, which combination is necessary to dynamically use a properties always at the same directory as the JAR?
UPDATE
Using --debug at the correct example got me this line at the very begging (Spring banner was still visible):
2018-09-25 15:45:14.480 DEBUG 11360 --- [ main] o.s.b.c.c.ConfigFileApplicationListener : Loaded config file 'file:src/main/resources/xirulei/teste32.yml' (file:src/main/resources/xirulei/teste32.yml)
But after moving myJar.jar and teste32.yml to a specific directory and running java -jar myJar.jar -Dspring.config.name=teste32 --debug (without spring.config.location, since teste32 is at the same directory as JAR), I simply didn't get any ConfigFileApplicationListener debug line.

a) java -jar myJar.jar -Dspring.config.name=teste32 -Dspring.config.location=C:\workspace\myProject\target
Did you check content of target dir? I'm pretty sure your cfg file is placed to target\classes\xirulei and it is why Spring cannot find it in target
b) When you place teste32.yml in the same directory as jar file then Spring must be able to find it (given this directory is working directory) without -Dspring.config.location (but you still need to provide -Dspring.config.name=teste32)
c) When you use -jar and do not provide additional class paths then classpath: points to the root of packages inside jar. Spring cannot find your file at classpath:/ because your file is at classpath:/xirulei/

Well, after all it was a simple mistake. As documentation says and as already pointed here, it should be
java -jar myproject.jar --spring.config.name=myproject
and not
java - jar myproject.jar -Dspring.config.name=myproject
As stated on question, only when using Eclipse -D(JVM argument) is necessary. When using bash/cmd, just --(program argument) is the correct option:

Property File Outisde of .WAR file

I need to be able to read in a property file, that lives outside of my war. My problem is that a need a solution that will allow me to tell my war file where my property file is located. Can this be done through bashrc variables and windows env variables?
I need to do this because I need to be able to drop the property file in different locations that could be away from the war file.
I am struggling to come up with a solution.

This will very much depends what the property file is for. Some libraries will have have the possibility of setting it on the command-line and others will allow you to explicitly load them in code.
In case of the latter, System.getProperties() can prove helpful, since it allows you to read properties passed to the JVM using the '-D' flag. For example
java -jar -Dfilelocation="yourfilelocation" yourapp.jar
would populate the system property 'filelocation' with the string 'yourfilelocation'. This could then be used in your code to load the property file (or whatever you want to do with it).
Since you are running inside some sort of application server, there are different ways you can accomblish this. For jetty you can put them in start.ini (or simply pass them on the command-line when you start jetty)
where tomcat uses an enviroment variable called JAVA_OPTS, so
JAVA_OPTS='-Dfilelocation=yourfilelocation' start.sh
would set the system property when you start tomcat.

one way is to provide specific location by passing java argument or setting up environment varialble and read it from app to determine the location
and as a fallback (default) app should be announcing to look at
${user.home}/appname/some.peroperties

You could defines a list of possible directories ... and try to read each one ...
I have a solution that read a property file in the file system using a default directory structure.
Eg: c:\properties\code-suit\prd\application.properties
Where:
code-suit is a variable defined by application name;
prd is a environment that I will use
Good luck

Android - use ant to create build configurations that change configuration values

What I want is a way to have settings that are dependent on build configuration. To give a specific example, my android application connects to a web service. In development, I want the service url to be pulled in from a configurable value. In Test, I want a different value pulled in. In production, yet another value.
So, in code I have something like this:
public class HttpRequestHelper
{
private static String GetServiceUrl(ServiceAction action)
{
return serviceUrl + action.toString();
}
}
By default (when debugging/running through eclipse), I want that url to be http://localhost:1234
In Test I want https://test.mydomain.com
In Production I want https://mydomain.com
I am new to eclipse and ant and it has been a long time since I used java. How do I go about setting this up? What should the build.xml look like? I understand that when I want to build the test/prod versions I will need to use the command line. That's okay. But I don't know how to get this serviceUrl auto-set dependent on the build. I'm not even sure the best place to put this information (a resource, a properties file?). I really want to avoid setting it, building, setting it, building, etc.

As answers mentioned above says, you have to place the URLs in a property file like dev.properties, test.properties, prod.properties etc..
Now only thing that you need to do is making your build intelligent enough to choose a property file depending upon environment.
That can be done by passing a parameter to ANT, something like:
$ ant -file MyBuild.xml -DcurrentEnv=dev (For Development environment)
$ ant -file MyBuild.xml -DcurrentEnv=test (For Test)
$ ant -file MyBuild.xml -DcurrentEnv=prod (For Production)
Inside your build script, this is how you can include your property file:
<target name="jarMe">
<jar destfile="sample.jar" basedir="src" includes="${currentEnv}.properties"/>
</target>
With this in place, whatever name you supply at the time of build, property file with that name will be picked up.

You could try to have a following property file in your build.properties file:
service.url=*
And you could have http://localhost:1234 or https://test.mydomain.com in local.properties for your development and integration testing, and it could be set to https://mydomain.com in default.properties.
By do ing this, you have will get different value for service.url in different build environment. You could use that value to generate a config file, and parse it into your code, or set it to env variable, or just put it into a resource file, and Android will read it for you:
<?xml version="1.0" encoding="utf-8"?>
<resources>
<string name="service-url">##toben_to_be_replaced_during_build_time##</string>
</resources>

I would start by placing the urls into a properties file that you can then place onto the classpath. Make a test and a production properties file. Then depending on the build place the correct file onto the classpath and pull the properties at runtime.

Found a tutorial which goes through all the details of using ant to automate a build system, to create and use build configurations, as well as to build the release project with one command. Here it is: http://www.androidengineer.com/2010/06/using-ant-to-automate-building-android.html
Seems a little long, but it goes through all the steps and details involved.