Develop Reference

loading of property file outside of war file in spring boot

I need to load the application.properties file from outside the spring boot war which going to be deployed in tomcat.
I tried various solution missing something
Tried setting environmental variable as below in windows
name : SPRING_CONFIG_NAME
value:D:/test/application.properties
i tried multiple values for above value like file:/// in prefix and only file: as perfix .Nothing worked
Tried having context parameter is tomcat like mentioned in below SO answer
https://stackoverflow.com/a/44697239/2751962
Tried loading like this in main file which extends SpringBootServletIntializer
protected SpringApplicationBuilder configure(SpringApplicationBuilder application) {
return application.sources(Application.class)
.properties(getProperties());
}
public static void main(String[] args) throws Exception {
SpringApplication.run(Application.class, args);
SpringApplicationBuilder springApplicationBuilder = (SpringApplicationBuilder) (new SpringApplicationBuilder(Application.class))
.sources(Application.class)
.properties(getProperties())
.run(args);
}
static Properties getProperties() {
Properties props = new Properties();
props.put("spring.config.location", "file:///D:/test/application.properties​");
return props;
}
I not sure what i missed , Kindly help.

External Configuration in Spring Boot
When using Spring Boot, there are documented naming conventions and directory structure. A Spring Boot app searches for properties to load from a prioritized list, so there are to suggestions for you to consider:
Use command-line flag spring.config.location to target specific file or directory from which to load properties sources. You can use this to specify directories to search or individual files to load. Be cautious loading individual files though, if you intend to use profile-based properties. (add flag in command like this: java -jar MyJar.jar --spring.config.location=D:\test\)
By default, Spring Boot will look for a ./config/ directory where the WAR is and the directory of the WAR itself, so you may place "application.properties" in either position and it will be loaded.
Pivotal provides a super great reference for Spring Boot. Section 24 covers properties more extensively than I can in a post.
https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html (*links to most recent release's reference)
Note: I am not a Windows user, so be careful pasting in that filepath above. Edit Me.
Extending Configuration to Deployable Packages
Normally Spring Boot packages into an executable WAR or JAR that has an embedded servlet container engine that is used for the runtime. In your case, however, you are packaging a conventional WAR and deploying that to an external instance of Tomcat, so the configuration parameters must be propagated through Tomcat, using the JAVA_OPTS variable.
For a Apache Tomcat, the convention is to place your properties in ${catalina_base}/conf where catalina.base points to the location of the Tomcat instance. I created a working demo just now following these steps:
First, follow section 88.1 of the reference to setup a base WAR app
mvn package
Place application.properties in conf directory within Tomcat
set JAVA_OPTS=-Dspring.config.location=${catalina.base}/conf/
"%CATALINA_HOME%"\bin\startup
Deploy
It's not the cleanest deployment pipeline, but if you must use an external Tomcat instance, then this will work. However, to run multiple apps with separate property files on the same Tomcat instance would complicate things. In that case, using Spring Framework (not Boot) would be easier to configure.

You can try setting properties via XML and or Java configuration and #PropertySource.
#Configuration
#PropertySource("classpath:foo.properties")
public class PropertiesWithJavaConfig {
//...
}
source :- https://www.baeldung.com/properties-with-spring

Spring boot deployed as WAR on tomcat, How to Externalize application.properties for different profiles(Dev,Test,Staging,Prod)

I am in a scenario where I am deploying by spring boot as a WAR on tomcat. Here in this application I have application.properties which has database username/password , some URL which help in consuming rest services(urls vary depending on the environment). Now I need to get the DB credentials and URL's depending on the environment it is deployed to. how to achieve it.

On external container ex tomcat
You can pass configuration using Jndi variable in context.xml
This var override local property defined in application.properties
Or pass -Dspring.profiles.active=env
in tomcat startup script ,for select environment specific application.properties
if you prefer to have configuration inside the war

As per 24. Externalized Configuration it should be enough to place a profile specific properties e.g. application-dev.properties on the classpath. The property precedence is:
...
Profile-specific application properties outside of your packaged jar (application-{profile}.properties and YAML variants).
Profile-specific application properties packaged inside your jar (application-{profile}.properties and YAML variants).
Application properties outside of your packaged jar (application.properties and YAML variants).
Application properties packaged inside your jar (application.properties and YAML variants).
...
This can however work slight differently if you are packaging as JAR as per 24.3 Application Property Files. The property precedence is:
A /config subdirectory of the current directory
The current directory
A classpath /config package
The classpath root

according to the discussion with Karol, I guess using properties sources referencing file system path should be ok:
#Configuration
#PropertySource("${mywebapp.config.dir}/application.properties")
public class SpringConfig {
}
And just path JVM args at the startup of tomcat like
-Dmywebapp.config.dir=file:/etc/mywebapp
This way you can deploy one application.properties different for each environment.
As for your comment:
#PropertySource(value={"file:C:/Users/foo/apache-tomcat-8.5.28/webapps/application.properties}"})
as it's a windows system path may you have to double backslash your path: C:\\Users\\foo ...

I am able to read the properties file from tomcat/webapp location
for every environment I can ask system engineers to drop the file at the location,
I dont know if its a good solution. Please suggest
#PropertySource(value={"file:C:/Users/foo/Downloads/apache-tomcat-8.5.28/webapps/application.properties"})

Spring boot external configuration of property file

I have a spring boot application that I can package in a war that I want to deploy to different environments. To automate this deployment it'd be easier to have the configuration file externalized.
Currently everything works fine with a application.properties file in src/main/resources. Then I use ´mvn install´ to build a war deployable to tomcat.
But I would like to use a .yml file that does not need to be present on mvn install but that would be read from during deployment of the war and is in the same or a directory relative to my war.
24. externalized configuration shows where spring boot will look for files and 72.3 Change the location of external properties of an application gives more detail on how to configure this but I just do not understand how to translate this to my code.
My application class looks like this:
package be.ugent.lca;
Updated below
Do I need to add a #PropertySource to this file? How would I refer to a certain relative path?
I feel like it's probably documented in there as most spring boot documentation but I just don't understand how they mean me to do this.
EDIT
Not sure if this should be a separate issue but I think it's still related.
Upon setting the os variable the error of yaml file not found went away. Yet I still get the same error again as when I had no application .properties or .yml file.
Application now looks like this:
#Configuration
**#PropertySource("file:${application_home}/application.yml")**
#ComponentScan({"be.ugent.lca","be.ugent.sherpa.configuration"})
#EnableAutoConfiguration
#EnableSpringDataWebSupport
public class Application extends SpringBootServletInitializer{
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
The application_home OS variable
$ echo $application_home
C:\Masterproef\clones\la15-lca-web\rest-service\target
My application.yml file(part it complains about):
sherpa:
package:
base: be.ugent.lca
Error upon java -jar *.war
All variations upon:
Caused by: java.lang.IllegalArgumentException: Could not resolve placeholder 'sherpa.package.base' in string value "${sherpa.package.base}"
at org.springframework.util.PropertyPlaceholderHelper.parseStringValue(PropertyPlaceholderHelper.java:174)
at org.springframework.util.PropertyPlaceholderHelper.replacePlaceholders(PropertyPlaceholderHelper.java:126)
at org.springframework.core.env.AbstractPropertyResolver.doResolvePlaceholders(AbstractPropertyResolver.java:204)
at org.springframework.core.env.AbstractPropertyResolver.resolveRequiredPlaceholders(AbstractPropertyResolver.java:178)
at org.springframework.context.support.PropertySourcesPlaceholderConfigurer$2.resolveStringValue(PropertySourcesPlaceholderConfigurer.java:172)
at org.springframework.beans.factory.support.AbstractBeanFactory.resolveEmbeddedValue(AbstractBeanFactory.java:808)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.doResolveDependency(DefaultListableBeanFactory.java:1027)
at org.springframework.beans.factory.support.DefaultListableBeanFactory.resolveDependency(DefaultListableBeanFactory.java:1014)
at org.springframework.beans.factory.annotation.AutowiredAnnotationBeanPostProcessor$AutowiredFieldElement.inject(AutowiredAnnotationBeanPostProcessor.java:545)
... 142 more

Using external properties files
The answer lies in the Spring Boot Docs, I'll try to break it down for you.
First of all, no you should not use #PropertySource when working with Yaml configuration, as mentioned here under the Yaml shortcomings :
YAML files can’t be loaded via the #PropertySource annotation. So in the case that you need to load values that way, you need to use a properties file.
So, how to load propery files? That is explained here Application Property Files
One is loaded for you: application.yml , place it in one of the directories as mentioned in the link above. This is great for your general configuration.
Now for your environment specific configuration (and stuff like passwords) you want to use external property files, how to do that is also explained in that section :
If you don’t like application.properties as the configuration file name you can switch to another by specifying a spring.config.name environment property. You can also refer to an explicit location using the spring.config.location environment property (comma-separated list of directory locations, or file paths).
So you use the spring.config.location environment property.
Imagine you have an external config file: application-external.yml in the conf/ dir under your home directory, just add it like this:
-Dspring.config.location=file:${home}/conf/application-external.yml as a startup parameter of your JVM.
If you have multiple files, just seperate them with a comma. Note that you can easily use external properties like this to overwrite properties, not just add them.
I would advice to test this by getting your application to work with just your internal application.yml file , and then overwrite a (test) property in your external properties file and log the value of it somewhere.
Bind Yaml properties to objects
When working with Yaml properties I usually load them with #ConfigurationProperties, which is great when working with for example lists or a more complex property structure. (Which is why you should use Yaml properties, for straightforward properties you are maybe better of using regular property files). Read this for more information: Type-Safe Configuration properties
Extra: loading these properties in IntelliJ, Maven and JUnit tests
Sometimes you want to load these properties in your maven builds or when performing tests. Or just for local development with your IDE
If you use IntelliJ for development you can easily add this by adding it to your Tomcat Run Configuration : "Run" -> "Edit Configurations" , select your run configuration under "Tomcat Server" , check the Server tab and add it under "VM Options".
To use external configuration files in your Maven build : configure the maven surefire plugin like this in your pom.xml:
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-surefire-plugin</artifactId>
<configuration>
<argLine>-Dspring.config.location=file:${home}/conf/application-external.yml</argLine>
</configuration>
</plugin>
When running JUnit tests in IntelliJ:
Run → Edit Configurations
Defaults → JUnit
add VM Options -> -ea -Dspring.config.location=file:${home}/conf/application-external.yml

Yes, you need to use #PropertySource as shown below.
The important point here is that you need to provide the application_home property (or choose any other name) as OS environment variable or System property or you can pass as a command line argument while launching Spring boot. This property tells where the configuration file (.properties or .yaml) is exactly located (example: /usr/local/my_project/ etc..)
#Configuration
#PropertySource("file:${application_home}config.properties")//or specify yaml file
#ComponentScan({"be.ugent.lca","be.ugent.sherpa.configuration"})
#EnableAutoConfiguration
#EnableSpringDataWebSupport
public class Application extends SpringBootServletInitializer{
public static void main(String[] args) {
SpringApplication.run(Application.class, args);
}
}

There is a very simple way to achieve this.
Inside your original application.properties file you can just specify the following line:
spring.config.import=file:Directory_To_The_File/Property_Name.properties
It will automatically sync all the properties from the external property file.
Now lets say that you have a situation where you need to get properties from multiple property files. In that case, you can mention the same line in the external property file which in turn will take the remaining properties from the second property file and so on.
Consider the following example.
application.properties:
spring.config.import=file:Resources/Custom1.properties
Custom1.properties:
server.port=8090
.
.
.
spring.config.import=file:Resources/Custom2.properties

One of the easiest way to use externalized property file using system environment variable is, in application.properties file you can use following syntax:
spring.datasource.url = ${OPENSHIFT_MYSQL_DB_HOST}:${OPENSHIFT_MYSQL_DB_PORT}/"nameofDB"
spring.datasource.username = ${OPENSHIFT_MYSQL_DB_USERNAME}
spring.datasource.password = ${OPENSHIFT_MYSQL_DB_PORT}
Now, declare above used environment variables,
export OPENSHIFT_MYSQL_DB_HOST="jdbc:mysql://localhost"
export OPENSHIFT_MYSQL_DB_PORT="3306"
export OPENSHIFT_MYSQL_DB_USERNAME="root"
export OPENSHIFT_MYSQL_DB_PASSWORD="123asd"
This way you can use different value for same variable in different environments.

Use below code in your boot class:
#PropertySource({"classpath:omnicell-health.properties"})
use below code in your controller:
#Autowired
private Environment env;

spring boot: add new yml files to application config

i want developers to be able to locally override some configuration properties. (let's say we work on google drive and everyone should test it on its own account).
i don't want to override properties using command line (because it has to be set inside every IDE configuration and on every CLI run).
what i want is: application should use all the standard spring boot config files (application.yml etc) and also look for e.g. local.yml (on the classpath) or some file inside user.home. and those additional files should override other settings.
how to add new yml resources and order them correctly?
edit: i know spring's default orders and locations. question is about adding new ones

If you look in the Spring Boot documentation about the locations for configuration files (http://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#boot-features-external-config), you can see, that they are loaded from the following places (amongst others):
Profile-specific application properties outside of your packaged jar (application-{profile}.properties and YAML variants)
Application properties outside of your packaged jar (application.properties and YAML variants).
There are two default locations where they are loaded from ( see http://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#boot-features-external-config-application-property-files):
A /config subdirectory of the current directory.
The current directory
Current directory in this case means the working directory for the Java process (Usually the directory where the JAR is located, or in case of running with in the IDE, usually the project root folder). So the developers just can place their own configuration files in that places and they are automatically loaded (and will override properties within the JARs). Add that files to .gitignore (or .svnignore or ...) and they won't accidentally committed into your repository.

There's a new way to do this, after Spring Boot v2.4, by using spring.config.import: https://spring.io/blog/2020/08/14/config-file-processing-in-spring-boot-2-4#importing-additional-configuration
By adding this part to your application.yml file, you should be able to import the additional configuration:
spring:
config:
import: local.yml
The article also has this section:
Imports can be considered as additional documents inserted just below the document that declares them. They follow the same top-down ordering as regular multi-document files: An import will only be imported once, no matter how many times it is declared.
So the contents of local.yml should be handled as if they were appended to the end of application.yml, thereby allowing you to override any property in application.yml.

From Spring Boot Documentation : Application property files:
SpringApplication will load properties from application.properties files in the following locations and add them to the Spring Environment:
A /config subdirectory of the current directory.
The current directory
A classpath /config package
The classpath root
The list is ordered by precedence (properties defined in locations higher in the list override those defined in lower locations).
This also goes for yaml, so you everyone can add application.yml under config directory, under the directory you run the spring boot jar from.
You can also customize the extra configuration file to be local.yml if you'd like by using spring.config.location:
--spring.config.location=classpath:/application.yml,classpath:/local.yml
Note however:
spring.config.name and spring.config.location are used very early to determine which files have to be loaded so they have to be defined as an environment property (typically OS env, system property or command line argument).

To provide the configuration from external config file in spring-boot application -
-Dspring.config.location=file:/home/vfroot/Workspace/project/MODULE_HOME/application.yaml
this command can be run with terminal:
mvn clean install -Dspring.config.location
= file:/home/vfroot/Workspace/MODULE_HOME/application.yaml
or need to set in Eclipse VM argument.
Also to set the active profiles :
-Dspring.profiles.active=dev

Well, since i am new in Spring Boot & Restfull Web Services. However, i managed to add a new .yml file to mange database and server port.
Instructions that i followed:
Project File.
Other Sources
src/main/resources
default package
right click on "default package"
add new YAML FILE
Or of YAML File option not available
5. right click on "default package"
6. then in categories: other --> File Types: YAML File

Override the classpath in the MANIFEST.MF of a jar [duplicate]

I have a JAR file that contains an application as well as configuration files for that application. The application loads configuration files from the classpath (using ClassLoader.getResource()), and has its dependencies completely satisfied using the configuration files baked into the JAR file.
On occasion I want the application to be run with a slightly different configuration (specifically I want to override the JDBC URL to point to a different database) so I create a new configuration file, store it in the correct directory structure (which means in a directory /config of a classpath entry), and I want to do something like this:
java -cp new-config:. -jar application.jar
But I can't get the classpath to have the new-config path entry before the application JAR's contents. Is it hard-coded that the JAR's content is always the first thing on the classpath?

Why not just invoke the application without specifying -jar and instead name the application main class explicitly? This will allow you to put both your new-config and the application.jar on the classpath in the required order:
e.g. (assuming "new-config" is a directory containing the overridden properties file)
java -cp new-config:application.jar Application.Main.Class
I believe the name of main class can be found in the MANIFEST.MF file inside the jar ....

When you use the -jar option to launch your application:
... the JAR file is the source of all
user classes, and other user class
path settings are ignored.
as described here. A workaround would be to specify the classpath in the jar file's manifest to include the additional path (described here).
However, given that you are only talking about amending configuration you may want to take a different approach that is not reliant on the classpath. For example, I typically configure my applications through Spring using property files to determine the location of databases, etc. My Spring configuration is consistent across test, QA and live environments but I pass a different property file as a command line argument when launching the app.
Spring Configuration Snippet
<bean id="MyDataSource" class="org.springframework.jdbc.datasource.SingleConnectionDataSource">
<property name="url" value="jdbc:microsoft:sqlserver://${dbServer}:${dbPort};DatabaseName=${dbName}"/>
<property name="username" value="${dbUserName}"/>
<property name="password" value="${dbPassword}"/>
<property name="suppressClose" value="false"/>
</bean>
Property File Snippet
dbServer=MyServer
dbPort=1433
dbName=MyDb
dbUserName=Me
dbPassword=foobar

The JAR archive specified by the -jar option, overrides all other values.
You would have to generally do it with an outside config file or build your own solution withod ClassLoader.getResource().
We use a custom solution to solve this - we load the internal properties like so:
final Properties p = new Properties();
p.load(DefaultConfiguration.class.getResourceAsStream("config.properties"));
We then load the external file in the same way and overwrite the internal values with the external ones.
For info on how class loading works see:
http://java.sun.com/javase/6/docs/technotes/tools/findingclasses.html

It may not be possible using just the CLASSPATH. There are ways to get make the call to ClassLoader.getResource() use a static path to find the resource. If it is doing that, it is bypassing the CLASSPATH.