I want to read properties file keys and I found that there is the classpath attribute to set in the #PropertySource annotation :
#Configuration
#PropertySources({
#PropertySource("classpath:config.properties"),
#PropertySource("classpath:db.properties")
})
public class AppConfig {
//...
}
Where should the properties file be placed and how to know the classpath in the annotation ?
What is class path ?
The class path is the path that the Java runtime environment searches for classes and other resource files.
How to identify classpath ?
Check your IDE, what all in build path. Folders in Build Path are in classpath.
You can add any number of folders and files in classpath explicitly.
You can access anything within folders/packages of classpath can be accessed relatively :)
Classpath
Refer
You have two options
Define shared.loader in your /conf/catalina.properties. Example:
shared.loader = D:\data\properties
If you are using maven, put your properties in /src/main/resources
I found the location in this tuto.
Related
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've created my application.properties file:
spring.config.additional-location=C:\Users\user\
spring.datasource.url=jdbc:postgresql://<db>:<port>/<db>
I need to feed Spring with an additional file located on C:\Users\user\application.properties:
spring.datasource.username=user
spring.datasource.password=password
As you can see I've tried to use spring.config.additional-location property into my application.properties file.
However, bootstrap tells me that no authentication has been provided.
you can use another file name by specifying a spring.config.name environment property. You can also refer to an explicit location by using the spring.config.location environment property (which is a comma-separated list of directory locations or file paths). The following example shows how to specify a different file name:
$ java -jar myproject.jar --spring.config.name=myproject
Reference URL:
https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html
You can specify your alternative properties using #PropertySources like this:
#PropertySources({
#PropertySource({"classpath:application.properties"}),
#PropertySource(value = {"file:${conf-dir}}/application-override.properties" },ignoreResourceNotFound = true)
})
public class AppConfig {
...
The properties in the bottom PropertySource will override properties from first one, if the file exists.
The documentation says to use file:C:/Users/user/
the "file:" or "classpath:" part is important.
There are numerous other ways to do this, too.
- profiles (and per profile application-<profile>.properties)
- #Configuration + #PropertySource
- ...
I have a spring boot application jar
#SpringBootApplication
#EnableScheduling
#PropertySource({"classpath:1.properties","classpath:2.properties"})
public class MyClass {
//My class
}
1.properties is in src/main/resources
2.properties is in server location project location
| MyClass.jar
| config
| 2.properties
But I am getting file not found for 2.properties when starting the application.Please let me know what I could be missing here.
Like mentioned in the comment, your 2.properties file is not under your classpath. You can only use classpath if your file really exists in your jar or war.
To get the 2.properties you should use the command file: instead of classpath:.
#PropertySource("classpath:1.properties","file:${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.
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 getting the following error:
Caused by: java.io.FileNotFoundException: class path resource [request-
ws/src/main/resources/application.yml] cannot be opened because it does
not exist
Class with the issue:
#Configuration
#PropertySource("classpath:request-ws/src/main/resources/application.yml")
public class RequestDataSource {
Now I am trying to access the yml file in from a different module. The module name is request-ws. The goal to to create two data sources. Any advice would be greatly appreciated.
The classpath for application.yml should be under: src/main/resources
according to: classpath:request-ws/src/main/resources/application.yml
you obviously don't have that yml file under:
src/main/resources/request-ws/src/main/resources/application.yml
Try to create your custom folder under: src/main/resources
Since my project is broken down into modules when the war file is made my application.yml file ends up in WEB-INF Put this on on your class:
#Configuration
#EnableConfigurationProperties
#PropertySource("classpath:WEB-INF/classes/application.yml")
public class DataSourceConfig {