I need to pass a folder (java.io.File) as a function parameter.
I tried to just declare the location of the folder, but it looks in SERVER_HOME (/home/user/tomcat).
So my next try is to inject a File (directory) which is located in WEB-INF/myFolder.
my first try failed:
<bean name="path" class="java.io.File">
<constructor-arg value="classpath:WEB-INF/myFolder" />
</bean>
But it looks for /home/user/tomcat/classpath:WEB-INF/myFolder
I've been messing around but I can't figure out how to do it.
Any help or advice would be great.
Thank you all!
You are not supposed to access the file system from a webapp, especially not to make assumptions about the webapps own file representation. WEB-INF/myfolder may or may not be a directory, depending if you are dealing with a WAR (no) or an exploded WAR (yes). If you absolutely need a file system resource, try to acquire it using System.getPreference("java.io.tmpdir").
OK, here are a few hints:
Use a Factory bean to retrieve the File. Let it
Create a temporary directory
Extract the contents of the jar to that temporary directory
return the temporary directory in it's getObject() method
Then inject the factory bean into your bean:
<!-- This is a Factorybean that creates a file -->
<bean class="com.yourcompany.ConfigFolderCreator" id="configDir">
<property name="packageToUnpack" value="com.yourcompany.yourpackage" />
</bean>
<bean class="com.your.legacy.Api">
<!-- inject the factory bean instead of a file -->
<property name="configFolder" ref="configDir" />
</bean>
You could use spring's FileSystemResource class
<bean id="bean" class="org.springframework.core.io.FileSystemResource">
<constructor-arg><value>your.file</value></constructor-arg>
</bean>
Related
I have developed a webapplication which is having jsp and java code. Right now I have placed all the key-value into a env/lifecycle specific properties file (like conf-dev.properties,conf-stg.properties,conf-prod.properties).
I want to externalize these properties file so that it can be placed outside of war(without effecting the war).
right now war file is tightly coupled with properties file. if i have to modify any thing i have to build and make war and deploy.
I have very limited access on deployment server machine (only have access for one folder where i can put my configuration files) & deployment process is handled by CI(jenkin & automated script).
I explored on internet and came to know that we can achieve this using spring, would like to know what is the best way to achieve this?
As you are using Spring I suppose you already use PropertyPlaceholderConfigurer. If not you should ;)
The location of a property file can be anything that can be resolved as spring Resource. This includes classpath, servletcontext and also file references as URIs (file:///... For absolute paths)
https://docs.spring.io/spring/docs/current/javadoc-api/org/springframework/beans/factory/config/PropertyPlaceholderConfigurer.html
<bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="location" value="${config.file.location}" />
</bean>
If I understand your question, then you can use Class.getResourceAsStream(String) the linked Javadoc says (in part)
This method delegates to this object's class loader. If this object was loaded by the bootstrap class loader, the method delegates to ClassLoader.getSystemResourceAsStream(java.lang.String).
The better way to externalize env specific properties is to use "user.home" or "user.dir".
Thanks #Martin F..
Resolved:This is the final one i used and its working fine in dev,stage Env.
<bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="ignoreUnresolvablePlaceholders" value="false"/>
<property name="order" value="1"/>
<property name="locations">
<list>
<value>classpath:conf-${cisco.life}.properties</value>
<value>file:///${openshift.home}/data/conf-${cisco.life}.properties</value>
<value>file:${openshift.home}/data/conf-${cisco.life}.properties</value>
</list>
</property>
</bean>.
and i used script action hook in openshift to set the lifecycle on system level.
appname=echo $OPENSHIFT_APP_NAME
case "$appname" in
*dev)export JAVA_OPTS_EXT="${JAVA_OPTS_EXT} -Dcisco.life=dev";
echo "setting up env life dev for " $appname
;;
*stage)export JAVA_OPTS_EXT="${JAVA_OPTS_EXT} -Dcisco.life=stg;
echo "setting up env life as stg for " $appname.
I have property place holder in my spring context.xml file
<bean id="propertyConfigurer" class="com.techpleiad.poc.RMCPropertyUtil">
<property name="basenames" value="file:${config.file.dir}/prop_application" />
<property name="defaultEncoding" value="UTF-8" />
<property name="cacheSeconds" value="30"></property>
</bean>
and this property 'config.file.dir' is not getting resolved.
'config.file.dir' is the environment variable and when i debug the code and check for the basename the file path comes as it is.. '{config.file.dir}/prop_application'
I need to know what spring code/classes are involved in resolving such properties.
How i could debug and resolve this problem?
You'll need to register a PropertySourcesPlaceholderConfigurer with a reference to your property sources (or not since this is an environment property which are implicitly added).
With XML you can do that with
<context:property-placeholder location="classpath:spring.properties" />
With Java config, simply define a static #Bean annotated method which returns a PropertySourcesPlaceholderConfigurer.
You can try with Spring SpEL to get the system properties
#{systemProperties['config.file.dir']}
To read environment variable use
#{systemEnvironment['config.file.dir']}
The systemEnvironment property contains all the environment variables on the machine where the program is running. Meanwhile, the systemProperties contains all the properties that we set in Java when the application started, using the -D argument.
I'm using my context.xml file to set init parameters for my java application, for example:
<Parameter
name="Environment"
description="The environment in which this code is running (e.g. Production, Staging, Development)."
value="Production"/>
I would like to be able to create a parameter who's value attribute is loading from a file. Is there anyway to do this? Should I be using a < Resource > element instead? If so, how do I setup a resource to load the contents of a file? I've tried Google, but I my not understand the context.xml file well enough to know what to look for. Any help is much appreciated!
In application-context.xml add this element
<context:property-placeholder ignore-unresolvable="true" location="classpath*:application.properties"/>
In the file application.properties you can define parameters like this
userName=root
password=123
And then you can use the parameters by this way
<property name="username" value="${userName}"/>
<property name="password" value="${password}"/>
I'm using Spring in a Console java application. I'm using PropertyPlaceholderConfigurer to load database details, and it works flawlessly:
<bean id="propertyConfigurer" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer">
<property name="location" value="file:/appdata/configs/myapplication/connection.properties" />
</bean>
So the root dir will be relative to where the jar is located. I need the same functionality with log4j.properties
log4j.appender.file.File=/appdata/configs/myapplication/myapplication-log.log
How can I achieve this without defining environment variables?
Just tried it like this:
log4j.appender.file.File=\\appdata\\configs\\myapplication\\myapplication-log.log
And it works fine.
I have a properties file which sits on APPSERVERS HOME directory(JBOSS_HOME/PROJECT_PROPERTIES/abc.properties).PROJECT_PROPERTIES is the directory where we are keeping all the project related property files.I need to read this properties file from spring config.Earlier i was using the following approach.
<bean id="propertyOverrideConfigurer"
class="org.springframework.beans.factory.config.PropertyOverrideConfigurer">
<property name="location" value="classpath:abc.properties" />
Now we moved all our properties file to JBOSS_HOME/PROJECT_PROPERTIES directory.
Please provide me some pointers how to access the properties file using spring.
This should help you out:
How to read properties in spring
You can pass the properties file path as a command line argument
<property name="location" value="${propertiesFile}" />
Then for the jvm parameters pass
-DpropertiesFile=my/configuration/file.properties