I have a Spring MVC application and in it I am running a periodic job using a class with method annotated as #Scheduled
In this method, I want to get the base application path i.e. http://localhost:8080/ or http://www.mywebsite.com/ based on whether this is my local system or production system.
How can I do this? I do not have access to HttpServletRequest because this is not a Controller class.
Any hints would be appreciated
In my opinion it is a good idea to use profiles and store properties like base application path in properties file - where each environment has its own property file: config_dev.properties, config_production.properties
Once they are there you can load them in job-like classes using Environment (described on SpringSource blog).
How to configure Tomcat and Spring to use profiles: Spring 3.1 profiles and Tomcat configuration
Put a myconfiguration.properties out of your application, to let the application know that whether its running locally or in production. And then in your method annotated as #Scheduled just read the Property file.
String configPath = System.getProperty("config.file.path");
File file = new File(configPath);
FileInputStream fileInput = new FileInputStream(file);
Properties properties = new Properties();
properties.load(fileInput);
And provide the agrument,
-Dconfig.file.path=/path/to/myconfiguration.properties
when running your application server (or container). This can be done by putting,
JAVA_OPTS="$JAVA_OPTS -Dconfig.file.path=/path/to/myconfiguration.properties"
at the beginning (roughly) of the script, which is used while running your application server.
For tomcat its catalina.sh
For Jboss AS its run.sh
For weblogic its setDomainEnv.sh
And After doing that start your server and deploy your application. Finally, your #Scheduled method should know the information it needs. As the property file is outside of the application, you can change the value of the property when you want without rebuilding the application or without even disturbing it!
just add this code in your web.xml
<context-param>
<param-name>webAppRootKey</param-name>
<param-value>my.root.path</param-value>
</context-param>
and use it your code as a system properties
Related
I use WAS Liberty Server container which provides server.xml and server.env file for configuring many things in addition to configuring DataSource properties such as url, username, password etc.
For security reasons, these properties are not known to developers for production Liberty servers. However, developers use same server.xml/server.evn files but with different DataSource properties so they can do their work.
So, I might have two server.env files like:
PRODUCTION server.env: dataSourceUrl="server-A-URL" (this URL is not known to developers)
DEVELOPMENT server.env: dataSourceUrl="server-B-URL" (this URL is known to developers)
, then the dataSourceUrl is used in server.xml files in production and development to set the url accordingly.
So, the structure of server.xml/server.env file is same for developers and production, only the DataSource url, username, password are different. This way developers can work using their own DataSource properties and once ready to deploy they app, it is handed to other team which then just changes the DataSource properties to the production values and deploys the application to production server.
With Springboot JPA, I know we can use application.properties file to set these DataSource values. But, we would like to be able to set these to the values located in server.env file. Basically to have something like this in application.properties file:
spring.datasource.url=dataSourceUrl //dataSourceUrl is set in server.env
, then be able to use this value in Java code using #Value injection like:
public class MyClass {
#Value("${spring.datasource.url}")
String dsUrl;
...
}
I have been reading about externalizing properties but I am not able to figure out how to do this
You can use Liberty's jndiEntry elements to make configured values available in JNDI. You will need the jndi-1.0 feature, after which you can configure,
<jndiEntry jndiName="spring/datasource/url" value="${dataSourceUrl}"/>
and access it in your application as:
String dsUrl = InitialContext.doLookup("spring/datasource/url");
Or, from a web or ejb component as:
#Resource(lookup = "spring/datasource/url")
String dsUrl;
I am running hibernate search with spring boot. I have written a working configuration for my application. How ever, i want to externalize my configuration and use ./config/hibernate.properties instead of src/main/resources/hibernate.properties. After copying my properties file to the desired location, i am getting and exception:
nested exception is java.io.FileNotFoundException: class path resource [hibernate.properties] cannot be opened because it does not exist
Anyone with any idea on how i should tell spring to read my configuration file?
Move your configuration to an src/main/resources/application.properties file and prepend spring.jpa.properties. everywhere, so hibernate.dialect will become spring.jpa.properties.hibernate.dialect, for example.
Then you can use Spring features to move your configuration wherever you want. To move it to ./config/application.properties I suppose you will have to add #PropertySource("./config/application.properties") to one of your #Configuration classes, or something similar.
I'm sure you can also keep the hibernate configuration in a separate file (separate from the rest of your application configuration).
See https://docs.spring.io/spring-boot/docs/current/reference/html/boot-features-external-config.html for more details about externalizing configuration in Spring Boot.
For some reason, it seems hibernate-search will prevent application from starting as long as a hibernate.properties configuration file does not exist. After trying for a while without success, i found a work around for my problem.
First, i created an empty hibernate.properties file and place it under src/main/resources.
Secondly, i moved all hibernate-search configurations to application.properties as follows:
spring.jpa.properties.hibernate.search.default.indexmanager = elasticsearch
spring.jpa.properties.hibernate.search.default.elasticsearch.host = http://my-server.com
spring.jpa.properties.hibernate.search.default.elasticsearch.index_schema_management_strategy = CREATE
spring.jpa.properties.hibernate.search.default.elasticsearch.required_index_status = yellow
This way, the application will start and spring will get all configuration from the externalized configuration as documented here.
It is hard to understand but for my application a required format. I have some custom libraries which are included at runtime and so they are not in the spring application context. To get apis from spring boot application I catched required apis and overhand this to my external classes.
To show an example:
HashValueService hashValueService
= (HashValueService) appContext.getBean("hashValueServiceImpl");
ServiceList srvList = new ServiceList();
srvList.setHashValueService(hashValueService);
In this way I'm able to get access to my database, which is in my application context.
I have a lot of properties distributed in the whole application. So I want to use the default application.properties to centralized often used properties in my application, like the keystore.
For that I edited application.properties with this line:
application.keystore=server.jks
But of course the usage of the Spring's #Value does show me a null for that attribute, because this class is not in my application context:
#Value("${application.keystore}")
private String keystore;
Do you have an idea to overhand this properties to this customer libraries? Maybe the creation of a new property file whould help? Thank u a lot.
Majority of Spring magic is done by BeanPostProcessors. Take a good look at them - link.
#Value wiring (and much more) is performed by AutowiredAnnotationBeanPostProcessor, you can use it for your purpose:
AutowiredAnnotationBeanPostProcessor beanPostProcessor =
appContext.getBean(AutowiredAnnotationBeanPostProcessor.class);
ServiceList srvList = new ServiceList();
beanPostProcessor.processInjection(srvList);
After that, your ServiceList should have String keystore field initialized.
I have a spring-boot application with annotations instead of context.xml.
In my src/main/resources folder I have: application-dev.properties and application-test.properties.
which work perfectly for different profiles (while running with VM option like -Dspring.profiles.active=dev)
Now I need to externalize this properties with file in /opt/software/Tomcat8/conf/app.properties
Some props override each other, some don't.
in Tomcat config context.xml I say:
<Environment name="app.properties"
value="file:///opt/software/Tomcat8/conf/app.properties"
type="java.lang.String" override="false"/>
How to use it via JNDI in my application configuring app with no XML but annotations in Spring-bot application class?
I need it to have priority to inner jar properties according to
Link to Spring-boot.doc
One solution I found was to have the vm argument -Dloader.path with the external path when executing the application. Please keep in mind if you're using a fat jar you may need to create the package in Zip model, otherwise it will not work.
I am trying to override the tomcat session manager with an embedded tomcat.
These are the steps preformed in-order to load the context.xml that defines the manager entity.
..
Context context = tomcat.addWebapp(contextPath, appBase);
File configFile = new File ("D:\\context.xml");
context.setConfigFile(configFile.toURI().toURL());
tomcat.start();
..
The session manager seems to be recognized as it's constructor is being invoked but the startInternal() method is never invoked and the session manager being used is the old tomcat session manager.
The weird thing is that when defining the same configuration in a non embedded tomcat, the session manager is being overridden without problems.
Would appriciate any help on the subject.
Non embedded tomcat uses server.xml and embedded tomcat does not uses server.xml file rather we need to pass the arguments in the method.