I am trying to configure Log4j2 to pick up a configuration file location from a property file outside the deployed .war. I need to set the location of the external property file at runtime so I am passing the following -D argument to my Tomcat:
-Dapplication.home=file:///C:/dev/env1-home
I have an application.properties file in the env1-home directory listed above and the various properties are picked up by the spring container and used throughout the application. One of the properties is:
application.log4j.config=C:/dev/env1-home/log4j2.xml
I would like to feed that application file property into a context parameter in the web.xml, but the following value in the web.xml:
<context-param>
<param-name>log4jConfiguration</param-name>
<param-value>$${sys:application.log4j.config}</param-value>
</context-param>
Gives me the following error:
ERROR StatusLogger Unable to access file:/C:/dev/Tomcat/apache-tomcat-8.5.31-8090/bin/$%7Bsys:application.log4j.config%7D
java.io.FileNotFoundException: C:\dev\Tomcat\apache-tomcat-8.5.31-8090\bin\${sys:application.log4j.config} (The system cannot find the file specified)
...
I have created a standalone boot.jar that I need to start integrating into our higher environments.
Each environment has a property file that contains database specific connection information. Since this does not live in my boot jar, I would like to somehow add the path to this database.properties file and then read the entries based on key. Used to create a bean like so:
<bean id="propertyLoader" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="locations">
<value>classpath:application.properties</value>
</property>
but with boot, I am not sure how to do this: But I want to point to this below property example and pull out my values and somehow populate the application.properties values that I hardcoded in dev.
server:/config/database.properties
jdbc.username=TEST
jdbc.password=CHANGEME
Updating my application.properites:
spring.datasource.username='jdbc.username'
spring.datasource.password='jdbc.password'
Something like that do I can parameterize my application.properties file.
SpringBoot offers profiles, which basically allows you to have separate application.properties file for each environment.
You can have something like this:
public interface DataSourceConfig {}
#Component
#Profile("dev")
public DevDataSourceConfig implements DataSourceConfig{}
#Component
#Profile("prod")
public ProdDataSourceConfig implements DataSourceConfig{}
If you have the spring profile "dev" set as active, only the DevDataSourceConfig bean will be instantiated and in Spring Environment the properties that will be injected, will be read from the application-dev.properties file.
Similarly when you have the "prod" profile activated, only the ProdDataSourceConfig will be instantiated and the properties will be loaded from application-prod.properties file.
This allows you to have:
---
application-dev.properties
spring.datasource.username='jdbc.username'
spring.datasource.password='jdbc.password'
---
application-prod.properties
spring.datasource.username='PROD_jdbc.username'
spring.datasource.password='PROD_jdbc.password'
If you want to load the configuration from a custom location on the file system - you can check how to pass the location with command line arguments (docs)
Example:
java -jar boot.jar --spring.config.location=classpath:/database.properties
you already told you can not have property files inside your jar, still there are multiple options.
1> passing a property file for respective env.
java -jar myproject.jar --spring.config.location=classpath:/database.properties
2> pass properties while calling the jar
java -jar app.jar --spring.datasource.username="jdbc.username" --spring.datasource.password="jdbc.password"
Read a lot of other options here `
I would go with option 1, because passing credentials is never advisable in arguements.
I can't find an answer to this question on stackoverflow hence im asking here so I could get some ideas.
I have a Spring Boot application that I have deployed as a war package on Tomcat 8. I followed this guide Create a deployable war file which seems to work just fine.
However the issue I am currently having is being able to externalize the configuration so I can manage the configuration as puppet templates.
In the project what I have is,
src/main/resources
-- config/application.yml
-- config/application.dev.yml
-- config/application.prod.yml
-- logback-spring.yml
So how can I possibly load config/application.dev.yml and config/application.prod.yml externally and still keep config/application.yml ? (contains default properties including spring.application.name)
I have read that the configuration is load in this order,
A /config subdirectory of the current directory.
The current directory
A classpath /config package
The classpath root
Hence I tried to load the configuration files from /opt/apache-tomcat/lib to no avail.
What worked so far
Loading via export CATALINA_OPTS="-Dspring.config.location=/opt/apache-tomcat/lib/application.dev.yml"
however what I would like to know is,
Find out why loading via /opt/apache-tomcat/lib classpath doesn't work.
And is there a better method to achieve this ?
You are correct about load order. According to Spring boot documentation
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).
[Note]
You can also use YAML ('.yml') files as an alternative to '.properties'.
This means that if you place your application.yml file to /opt/apache-tomcat/lib or /opt/apache-tomcat/lib/config it will get loaded.
Find out why loading via /opt/apache-tomcat/lib classpath doesn't work.
However, if you place application.dev.yml to that path, it will not be loaded because application.dev.yml is not filename Spring is looking for. If you want Spring to read that file as well, you need to give it as option
--spring.config.name=application.dev or -Dspring.config.name=application.dev.
But I do not suggest this method.
And is there a better method to achieve this ?
Yes. Use Spring profile-specific properties. You can rename your files from application.dev.yml to application-dev.yml, and give -Dspring.profiles.active=dev option. Spring will read both application-dev.yml and application.yml files, and profile specific configuration will overwrite default configuration.
I would suggest adding -Dspring.profiles.active=dev (or prod) to CATALINA_OPTS on each corresponding server/tomcat instance.
I have finally simplified solution for reading custom properties from external location i.e outside of the spring boot project. Please refer to below steps.
Note: This Solution created and executed windows.Few commands and folders naming convention may vary if you are deploying application on other operating system like Linux..etc.
1. Create a folder in suitable drive.
eg: D:/boot-ext-config
2. Create a .properties file in above created folder with relevant property key/values and name it as you wish.I created dev.properties for testing purpose.
eg :D:/boot-ext-config/dev.properties
sample values:
dev.hostname=www.example.com
3. Create a java class in your application as below
------------------------------------------------------
import org.springframework.boot.context.properties.ConfigurationProperties;
import org.springframework.context.annotation.PropertySource;
#PropertySource("classpath:dev.properties")
#ConfigurationProperties("dev")
public class ConfigProperties {
private String hostname;
//setters and getters
}
--------------------------------------------
4. Add #EnableConfigurationProperties(ConfigProperties.class) to SpringBootApplication as below
--------------------------------------------
#SpringBootApplication
#EnableConfigurationProperties(ConfigProperties.class)
public class RestClientApplication {
public static void main(String[] args) {
SpringApplication.run(RestClientApplication.class, args);
}
}
---------------------------------------------------------
5. In Controller classes we can inject the instance using #Autowired and fetch properties
#Autowired
private ConfigProperties configProperties;
and access properties using getter method
System.out.println("**********hostName******+configProperties.getHostName());
Build your spring boot maven project and run the below command to start application.
-> set SPRING_CONFIG_LOCATION=<path to your properties file>
->java -jar app-name.jar
I am trying to make my first bean in Spring but got a problem with loading a context.
I have a configuration XML file of the bean in src/main/resources.
I receive the following IOException:
Exception in thread "main" org.springframework.beans.factory.BeanDefinitionStoreException: IOException parsing XML document from class path resource [src/main/resources/beans.xml]; nested exception is
java.io.FileNotFoundException: class path resource [src/main/resources/beans.xml] cannot
be opened because it does not exist
but I don't get it, since I do the following code test:
File f = new File("src/main/resources/beans.xml");
System.out.println("Exist test: " + f.exists());
which gives me true! resources is in the classpath. What's wrong?
Thanks, but that was not the solution. I found it out why it wasn't working for me.
Since I'd done a declaration:
ApplicationContext context = new ClassPathXmlApplicationContext("beans.xml");
I thought I would refer to root directory of the project when beans.xml file was there.
Then I put the configuration file to src/main/resources and changed initialization to:
ApplicationContext context = new ClassPathXmlApplicationContext("src/main/resources/beans.xml");
it still was an IO Exception.
Then the file was left in src/main/resources/ but I changed declaration to:
ApplicationContext context = new ClassPathXmlApplicationContext("beans.xml");
and it solved the problem - maybe it will be helpful for someone.
Edit:
Since I get many people thumbs up for the solution and had had first experience with Spring as student few years ago, I feel desire to explain shortly why it works.
When the project is being compiled and packaged, all the files and subdirs from 'src/main/java' in the project goes to the root directory of the packaged jar (the artifact we want to create). The same rule applies to 'src/main/resources'.
This is a convention respected by many tools like maven or sbt in process of building project (note: as a default configuration!). When code (from the post) was in running mode, it couldn't find nothing like "src/main/resources/beans.xml" due to the fact, that beans.xml was in the root of jar (copied to /beans.xml in created jar/ear/war).
When using ClassPathXmlApplicationContext, the proper location declaration for beans xml definitions, in this case, was "/beans.xml", since this is path where it belongs in jar and later on in classpath.
It can be verified by unpacking a jar with an archiver (i.e. rar) and see its content with the directories structure.
I would recommend reading articles about classpath as supplementary.
Try this:
new ClassPathXmlApplicationContext("file:src/main/resources/beans.xml");
file: preffix point to file system resources, not classpath.
file path can be relative or system (/home/user/Work/src...)
I also had a similar problem but because of a bit different cause so sharing here in case it can help anybody.
My file location
How I was using
ClassPathXmlApplicationContext("beans.xml");
There are two solutions
Take the beans.xml out of package and put in default package.
Specify package name while using it viz.
ClassPathXmlApplicationContext("com/mypackage/beans.xml");
src/main/resources is a source directory, you should not be referencing it directly. When you build/package the project the contents will be copied into the correct place for your classpath. You should then load it like this
new ClassPathXmlApplicationContext("beans.xml")
Or like this
new GenericXmlApplicationContext("classpath:beans.xml");
This is because applicationContect.xml or any_filename.XML is not placed under proper path.
Trouble shooting Steps
1: Add the XML file under the resource folder.
2: If you don't have a resource folder. Create one by navigating new by Right click on the project new > Source Folder, name it as resource and place your XML file under it.
use it
ApplicationContext context = new FileSystemXmlApplicationContext("Beans.xml");
You have looked at src directory. The xml file indeed exist there. But look at class or bin/build directory where all your output classes are set. I suspect you will need only resources/beans.xml path to use.
I suspect you're building a .war/.jar and consequently it's no longer a file, but a resource within that package. Try ClassLoader.getResourceAsStream(String path) instead.
Note that the first applicationContext is loaded as part of web.xml; which is mentioned with the below.
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>META-INF/spring/applicationContext.xml</param-value>
</context-param>
<servlet>
<servlet-name>myOwn-controller</servlet-name>
<servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>META-INF/spring/applicationContext.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
Where as below code will also tries to create one more applicationContext.
private static final ApplicationContext context =
new ClassPathXmlApplicationContext("beans.xml");
See the difference between beans.xml and applicationContext.xml
And if appliationContext.xml under <META-INF/spring/> has declared with <import resource="beans.xml"/> then this appliationContext.xml is loading the beans.xml under the same location META-INF/spring of appliationContext.xml.
Where as; in the code; if it is declared like below
ApplicationContext context = new ClassPathXmlApplicationContext("beans.xml");
This is looking the beans.xml at WEB-INF/classes OR in eclipse src/main/resources.
[If you have added beans.xml at src/main/resources then it might be placed at WEB-INF/classes while creating the WAR.]
So totally TWO files are looked up.
I have resolved this issue by adding classpath lookup while importing at applicationContext.xml like below
<import resource="classpath*:beans.xml" />
and removed the the line ClassPathXmlApplicationContext("beans.xml") in java code, so that there will be only one ApplicationContext loaded.
In Spring all source files are inside src/main/java. Similarly, the resources are generally kept inside src/main/resources. So keep your spring configuration file inside resources folder.
Make sure you have the ClassPath entry for your files inside src/main/resources as well.
In .classpath check for the following 2 lines. If they are missing add them.
<classpathentry path="src/main/java" kind="src"/>
<classpathentry path="src/main/resources" kind="src" />
So, if you have everything in place the below code should work.
ApplicationContext ctx = new ClassPathXmlApplicationContext("Spring-Module.xml");
Gradle : v4.10.3
IDE : IntelliJ
I was facing this issue when using gradle to run my build and test. Copying the applicationContext.xml all over the place did not help. Even specifying the complete path as below did not help !
context = new ClassPathXmlApplicationContext("C:\\...\\applicationContext.xml");
The solution (for gradle at least) lies in the way gradle processes resources. For my gradle project I had laid out the workspace as defined at https://docs.gradle.org/current/userguide/java_plugin.html#sec:java_project_layout
When running a test using default gradle set of tasks includes a "processTestResources" step, which looks for test resources at C:\.....\src\test\resources (Gradle helpfully provides the complete path).
Your .properties file and applicationContext.xml need to be in this directory. If the resources directory is not present (as it was in my case), you need to create it copy the file(s) there. After this, simply specifying the file name worked just fine.
context = new ClassPathXmlApplicationContext("applicationContext.xml");
Beans.xml or file.XML is not placed under proper path. You should add the XML file under the resource folder, if you have a Maven project.
src -> main -> java -> resources
I did the opposite of most. I am using Force IDE Luna Java EE and I placed my Beans.xml file within the package; however, I preceded the Beans.xml string - for the ClassPathXMLApplicationContext argument - with the relative path. So in my main application - the one which accesses the Beans.xml file - I have:
ApplicationContext context =
new ClassPathXmlApplicationContext("com/tutorialspoin/Beans.xml");
I also noticed that as soon as I moved the Beans.xml file into the package from the src folder, there was a Bean image at the lower left side of the XML file icon which was not there when this xml file was outside the package. That is a good indicator in letting me know that now the beans xml file is accessible by ClassPathXMLAppllicationsContext.
This is what worked for me:
new ClassPathXmlApplicationContext("classpath:beans.xml");
If this problem is still flummoxing you and you are developing using Eclipse, have a look at this Eclipse bug: Resources files from "src/main/resources" are not correctly included in classpath
Solution seems to be look at properties of project, Java build path, source folders. Delete the /src/main/resources dir and add it again. This causes Eclipse to be reminded it needs to copy these files to the classpath.
This bug affected me when using the "Neon" release of Eclipse. (And was very frustrating until I realized the simple fix just described)
I was experiencing this issue and it was driving me nuts; I ultimately found the following lying in my POM.xml, which was the cause of the problem:
<resources>
<resource>
<directory>src/main/resources</directory>
<filtering>true</filtering>
<includes>
<include>**/*.properties</include>
</includes>
</resource>
</resources>
I was not sure to write it but maybe someone save a few hours:
mvn clean
may do the job if your whole configuration is already perfect!
I have stuck in this issue for a while and I have came to the following solution
Create an ApplicationContextAware class (which is a class that implements the ApplicationContextAware)
In ApplicationContextAware we have to implement the one method only
public void setApplicationContext(ApplicationContext context) throws BeansException
Tell the spring context about this new bean (I call it SpringContext)
bean id="springContext" class="packe.of.SpringContext" />
Here is the code snippet
import org.springframework.beans.BeansException;
import org.springframework.context.ApplicationContext;
import org.springframework.context.ApplicationContextAware;
public class SpringContext implements ApplicationContextAware {
private static ApplicationContext context;
#Override
public void setApplicationContext(ApplicationContext context) throws BeansException {
this.context = context;
}
public static ApplicationContext getApplicationContext() {
return context;
}
}
Then you can call any method of application context outside the spring context for example
SomeServiceClassOrComponent utilityService SpringContext.getApplicationContext().getBean(SomeServiceClassOrComponent .class);
I hope this will solve the problem for many users
I am on IntelliJ and faced the same issue. Below is how i resolved it:
1. Added the resource import as following in Spring application class along with other imports: #ImportResource("applicationContext.xml")
2. Saw IDE showing : Cannot resolve file 'applicationContext.xml' and also suggesting paths where its expecting the file (It was not the resources where the file applicationContext.xml was originally kept)
3. Copied the file at the expected location and the Exception got resolved.
Screen shot below for easy ref:
But if you would like to keep it at resources then follow this great answer link below and add the resources path so that it gets searched. With this setting exception resolves without #ImportResource described in above steps:
https://stackoverflow.com/a/24843914/5916535
Sharing my case and how I debugged it, maybe helps someone:
this will only be relevant if you have first checked you actually have the resources folder in correct place and correctly named
create some temporary folder somewhere, preferably out of any git projects (e.g. mkdir playground) and move there (cd playground)
copy the java archive there (e.g. cp /path/to/java.war .) that is missing that beans.xml
unpack it (e.g. unzip java.war on ubuntu)
find if there's any .xml files in there (for example in WEB-INF/classes) (the unpacking process should show a list of files being unpacked, most of them will probably be other dependencies as archives, these are not relevant)
if you don't see a beans.xml, just read the other .xml files (e.g. cat root-config.xml), you might find something like root-config.xml there or similar, in there you might either have some other <import resource="somethingelse.xml"> records or nothing.
if this is the case, this means you do have that file (root-config.xml here) present in the project or if not, continue going up parent projects to where the archive is getting packaged from. Find that file, add <import resource="beans.xml"> and run mvn package.
Now verifying the fix by doing the steps in 1.-5. should result in that file (root-config.xml here) in the newly packaged archive having the beans.xml defined and once you deploy it, it should work.
Make sure that beans.xml is located in the resources folder.
I have a web application with mutiple Spring configuration files. These files are loaded using "contextConfigLocation" in web.xml. Like this:
<context-param>
<param-name>contextConfigLocation</param-name>
<param-value>
classpath*:META-INF/*beans.xml
</param-value>
</context-param>
Everything works as desired.
Now I am having to write a command line application that must load the same files as the web application. Currently I am using ClassPathXmlApplicationContext and manually specifying each configuration file name. But sooner or later somebody is going to add another file and expect it to be read by the CLI, just like the web app. Currently that will not happen because each file is explicitly specified in my CLI. So I need my CLI to load configuration files just like the web app i.e. load all configuration files that match a pattern. Is there a way to do this using ClassPathXmlApplicationContext or any other way?
I think you can do this using ClassPathXmlApplicationContext, This will load any context file that is in class path ending with name Beans
public class LoadContext {
/**
* #param args
*/
public static void main(String[] args)
{
ApplicationContext ctx = new ClassPathXmlApplicationContext("classpath:*Beans.xml");
}
}
Why don't you create a new configuration file spring-all.xml and load only this. In this file, use the import element to import all other xml configuration files that you need.
<beans>
<import resource="spring-services.xml"/>
<import resource="spring-daos.xml"/>
<import resource="spring-controllers.xml"/>
...
</beans>