You can use the loader.path parameter to define a location for an external lib folder. All jars under this folder will be added to the classpath. For example, if you'd like to define C:\extLib as your external lib folder, you can do the following:

java -Dloader.path=/C:/extLib/ -jar aapName.jar

For this to work, you need to use the PropertiesLauncher. There are two ways to do that:

Option 1

Update the project pom.xml and add the following tag:

<configuration>  <!-- added -->
  <layout>ZIP</layout> <!-- to use PropertiesLauncher -->
</configuration>

Effective build tag, the post-update looks like below:

<build> 
    <plugins>
        <plugin>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-maven-plugin</artifactId>
            <configuration>  <!-- added -->
                <layout>ZIP</layout> <!-- to use PropertiesLauncher -->
            </configuration>
        </plugin>
    </plugins>
</build>

Option 2

Use the PropertiesLauncher when launching the application from the commandline:

java -cp aapName.jar -Dloader.path=/C:/extLib/ org.springframework.boot.loader.PropertiesLauncher

References:
How to add jars to SpringBoot classpath with jarlauncher

/src/main/resources is a Maven project structure convention. It's a path inside your project where you place resources. During the build step, Maven will take files in there and place them in the appropriate place for you to use them in your runtime classpath, eg in an executable .jar, some physical file system location used in the classpath (with java's -cp option), etc.

I could choose to build my application myself or with a different build tool. In such a case, /src/main/resources would not exist. However, the intention is for the classpath to be the same, ie. to contain the same resources and .class files.

The Spring boot documentation talks about the classpath because it shouldn't make assumptions about how your project is set up.

It seems like you are running your project from command line. Run following command :

java -classpath spring.jar;spring-sec.jar,......so on com.example.UrMainClass

while specifying jar in command, make sure you provide complete path of jar. Also check this link https://stackoverflow.com/a/10122038/1065180

On Linux:

java -cp MyApp.jar:/home/sleeper/thirdparty/lib -Dloader.main=myMainApplicationClass org.springframework.boot.loader.PropertiesLauncher

On Windows:

java -cp MyApp.jar;/home/sleeper/thirdparty/lib -Dloader.main=myMainApplicationClass org.springframework.boot.loader.PropertiesLauncher

This will avoid messing with the manifest or the Spring Boot Maven plugin configuration as in the other answers. It will launch your app with the PropertiesLauncher, which allows you to specify the main class in loader.main. As mentioned earlier, for some reason if you use PropertiesLauncher with loader.path, it will not add resource files to the classpath. This works around the issue by using -cp instead of -jar.

EDIT As mentioned by Pianosaurus in the comment, use ":" instead of ";" as separator in the classpath on Linux

I just check that no matter if you use .properties or .yaml file you will have always classpath:app\mock-templates as an injected property value. And that's ok. Spring does not complete classpath: prefix to the real path in a time of reading from properties. That will be done next by ClassPathResource.

P.S. I think quotest also don't make any sense here.

Updated answer:

You can follow the practice in my original answer, but we recently dropped this for a simpler and cleaner option that is more standard (the "Java" way). We made the change because we needed to dynamically load dependent libraries at runtime that were not available at compile time (in their exact version). In our case we wanted to load dependent jars only from separate folder(s) and not from an executable jar. We ended up having duplicate dependencies in the executable jars and in separate folder(s), so we decided to drop the executable jar Properties Launcher and instead only load dependencies from separate folders. This is often NOT the best option and should be evaluated for your use case. I prefer reading the standard Java classpath.

To run a Spring Boot app without an executable jar, we used Maven Assembly to put the dependent jars in a /libs directory and dropped the spring-boot-maven-plugin. The steps and some code for this are below:

1. Remove the spring-boot-maven-plugin that creates the executable jar in ZIP format

2. Add the following to your assembly XML

   <dependencySets> <dependencySet> <outputDirectory>where you want the libs to go</outputDirectory> <useProjectArtifact>whether you want to include the project artifact here</useProjectArtifact> </dependencySet> </dependencySets>

3. Run your code from the main class and include the dependent jar folder(s) on the classpath. Use the standard classpath notation on your OS and not the custom, awkward PropertiesLauncher loader path syntax

   java -cp <standard-classpath> <main-class>

An example of an actual call: java -cp $CLASSPATH:./lib/*:./cfg/*:my-app.jar Application.class

In this way you execute the Spring Boot app via standard java execution call, no custom Spring loading syntax. You just need to ensure that all of your dependencies are available on the classpath at runtime. We found this much easier to maintain and made this the standard for all of our apps.

Original answer:

After some researching, and thanks to @TuyenNguyen's helpful answer I was able to get the following working:

I added the following to my spring-boot-maven-plugin so that when I run from the command line it uses the PropertiesLauncher instead of the JarLauncher:

<configuration>
     <mainClass>${mainClass}</mainClass>
     <layout>ZIP</layout> //THIS IS THE IMPORTANT PART
</configuration>

See here and here for more about the PropertiesLauncher options. It allows you to set the classpath, among other things. See here, here, and here for where I found the answer to this problem. Using format ZIP makes the PropertiesLauncher be used.

From there, I was able to use this command to launch the application as I intended:

java -Dloader.path=../config,../ -Dloader.config.location=classpath:application.properties -jar ../app-exec.jar

Another important note: when specifying the -Dloader.path make sure to use comma-separated values and only directories and files, as described here. Also, be sure to put the -D args before you specify -jar jar or they will not be set.

If anyone has any suggestions or edits to further improve this answer or the original question in order to help additional users, please let me know or make the edits yourself!

You can define your own classpath by the command-line. Lets suppose your jar is myapp.jar and you wand add one extra directory /var/application/config/, so you can execute with the following command line:

java -cp myapp.jar:/var/application/config/ -Dloader.main=myapp.Application org.springframework.boot.loader.PropertiesLauncher

ps: if you are using Windows use ; instead of : to separate your classpath items.