I have a spring boot application jar
#SpringBootApplication
#EnableScheduling
#PropertySource({"classpath:1.properties","classpath:2.properties"})
public class MyClass {
//My class
}
1.properties is in src/main/resources
2.properties is in server location project location
| MyClass.jar
| config
| 2.properties
But I am getting file not found for 2.properties when starting the application.Please let me know what I could be missing here.
Like mentioned in the comment, your 2.properties file is not under your classpath. You can only use classpath if your file really exists in your jar or war.
To get the 2.properties you should use the command file: instead of classpath:.
#PropertySource("classpath:1.properties","file:${application_home}2.properties")
I am not quite sure if its still necessary to have an OS environment variable or a system property to set the path to your property file. In this case I named it application_home.
Related
Been searching for others that have run into this issue, and not finding much out there, so it can't be that common.
I have a spring-boot project that I want to convert into a jar project, running with embedded tomcat. It's using yml files (application.yml and then the profile versions - eg appplication-dev.yml.) It ran fine as war with the yml files, however, when I convert it to a jar, and kick off the jar, the embedded tomcat never starts UNLESSS I add an empty application.properties file as well. (No errors just no Tomcat startup unless the empty application.properties file is added.)
I believe it's somehow related to one of our internal jar dependencies (also spring), since if I remove that dependency from the pom (and any of the code referencing it) I can get the jar to startup the embedded tomcat just fine (without providing the empty application.properties file.)
I could also, of course, forgo using yml files and just use .properties files, but I'd like to use yml files if possible. Why adding an empty applcation.properties file causes things to work has me stumped.
If it helps, the config in the dependency project that causes the issue we're seeing is set up as:
#Configuration
#EnableConfigurationProperties(OracleDataSourceProperties.class)
#EnableTransactionManagement
#ComponentScan(basePackages = {"com.foo.data.services","com.foo.data.domain", "com.foo.utility", "com.foo.cipher.utility"})
#MapperScan(value = {"com.foo.data.services.mapper","com.foo.data.services.batchmapper"})
public class DataServicesPersistenceConfig { ... }
and the OracleDataSourceProperties class:
#ConfigurationProperties(prefix="oradb", ignoreUnknownFields = true)
public class OracleDataSourceProperties extends BaseVO implements InitializingBean{
Created a jar file for a spring boot multimodule application and ran the jar file using java -jar command. While starting the application, it gives ResourceFinderException. When I analyzed it, the issue is happening because in my ResourceConfig file, i have used the package for my api end points. If I use register(service.class), the application starts fine. Any suggestion how can I provide the package instead of using register? The reason I want to use package is because I have lots of services inside multiple packages and the code looks very ugly if i use register for all the services. The ResourceConfig file looks like below.
public class AppResourceConfig extends ResourceConfig {
public AppResourceConfig {}{
super();
property("jersery.config.beanValidation.enableOutputValidationErrorEntity.server");
**packages("com.api");**
register(GsonProvider.class);
register(RequestContextFilter.class);
register(NotFoundExceptionMapper.class);
register(DefaultExceptionMapper.class);
}
}
Here the issue is with highlighted line: packages("com.api")
If I comment out this code application will be up. Otherwise it is giving org.glassfish.jersey.server.internal.scanning.ResourceFinderException: java.io.FileNotFoundException: api-01.03.00.04-snapshot.jar (No such file or direcotry)
Note: api-01.03.00.04-snapshot.jar is the jar file for one of the module in a project
In spring boot application how do I give an external windows path using #Value Spring annotation and Resource
The below example works fine that look into resources folder but I want to give the path outside of application like c:\data\sample2.csv
#Value("classPath:/sample2.csv")
private Resource inputResource;
...
#Bean
public FlatFileItemReader<Employee> reader() {
FlatFileItemReader<Employee> itemReader = new FlatFileItemReader<Employee>();
itemReader.setLineMapper(lineMapper());
itemReader.setLinesToSkip(1);
itemReader.setResource(inputResource);
and if I want to get the value from properties file in annotaion, whats the format to put the path in windows?
i tried these, none of them worked:
in code
#Value("${inputfile}")
in properties file:
inputfile="C:\Users\termine\dev\sample2.csv"
inputfile="\\C:\\Users\\termine\\dev\\sample2.csv"
inputfile="C:/Users/termine/dev/sample2.csv"
inputfile="file:\\C:\Users\termine\dev\sample2.csv"
inputfile="file://C://Users//termine///dev//sample2.csv"
When you use classpath spring will try to search with the classpath even if you provide the outside file path.
so instead of using classpath: you can use file:
Ex.
#Value("file:/sample2.csv") //provide full file path if any
Use the key spring.config.location in properties to set the config location. Spring-boot will by default load properties from the locations, with precedence like below :
A /config subdir of the current directory.
The current directory
A classpath /config package
The classpath root
and apart from this when you start the jar or in application.properties you can provide the location of the config file like :
$ java -jar myproject.jar --spring.config.location=classpath:/default.properties,classpath:/override.properties
You can serve static files from the local disk, by making the resource(s) "sample2.csv" as a static resource. An easy way to do this is by adding spring.resources.static-locations configuration to your applicaiton.properties file. Example:
spring.resources.static-locations=file:///C:/Temp/whatever/path/sample2.csv",classpath:/static-files, classpath:/more-static-resource
When I did this in one of the projects, I was able to access the file form the browser using localhost:8080/sample2.csv.
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 a Spring Boot application. Everything is fine once I deploy to the fat jar file with everything contained in it. But, what I actually want is the configuration files to be located externally. for example I have the following directory structure:
bin - contains startup and shutdown scripts
conf - all configurations. i.e. application.properties, logback.xml i18n.properties
logs - log files
libs - app.jar
If I use this directory structure and execute the jar using
java -cp ./conf -jar ../libs/app.jar
then the properties in the conf directory are not loaded or recognized. Is there a better way to do this maintaining the directory structure above? Or, what is the alternative/best practice?
Boot external config is what you are looking for.
Especially it mentions:
SpringApplication will load properties from application.properties
files in the following locations and add them to the Spring
Environment:
A /config subdir of the current directory.
The current directory
A classpath /config package
The classpath root
So I would say adding the config folder on classpath is good step. Them it should find application.properties and load it automatically.
For different config files I use:
#Configuration
#EnableAutoConfiguration
#PropertySource({
"classpath:path/some.properties",
"classpath:another/path/xmlProperties.xml"
})
public class MyConfiguration {
// ...
}
Edit:
As Dave pointed out (Thank Dave!) there is either -cp or -jar, so you can't add it to classpath like that. But there are options. This should help you to solve the problem: Call "java -jar MyFile.jar" with additional classpath option.
Additionally #PropertySource doesn't require the resources to be classpath resources if I'm not mistaken.
It should also be mentioned that there is a spring.config.location parameter that allows one to specify a file system / classpath location for externalized configuration files. This is documented in the following section of the Spring Boot reference guide:
http://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html#boot-features-external-config-application-property-files