How to pass the -D parameter when multiple WAR files are running? - java

When multiple WAR files are running under tomcat,
each WAR is expecting a
-Dconfig-path=/path/app.conf.ini
Is it possible to pass a unique -D Parameter value to each of the running applications?
tomcat
webapps
APPLICATION_1.war -Dconfig-path=/path/app.conf1.ini
APPLICATION_2.war -Dconfig-path=/path/app.conf2.ini
APPLICATION_3.war -Dconfig-path=/path/app.conf3.ini

You can add all the configuration properties in one file and pass it as command line parameter while starting tomcat. All those properties will be available to all .war files.
But if all the war files are using same property name then you have to modify the property name in config file and your code.
For example: If you are using app.version=1.1 for 1st war and 2.1 for 2nd war then you have to add them like
Firstwarname.app.version=1.1
SecondwarName.app.version=2.1
Accordingly, your code needs to be modified to access properties.

I found a way
In the web.xml added this:
<context-param>
<param-name>config-path</param-name>
<param-value>/path/app.conf1.ini</param-value>
</context-param>

Related

Set a path to an outside-the-app file in web.xml

I'm working in a java webapp that uses Picketlink package. This package needs a config file ( picketlink.xml ) that is located in WEB-INF directory by default. This app will be hosted in 3 different linux jboss servers/environments (development, testing and production).
This configuration file is different for each env.
It's a project requirement that I have only one single build (single artifact) for all envs, so env-specific stuff must be in the server (can't use profiles). So, I need to store this file in the server itself and reference it from inside the app.
The documentation recommends that I use a context-param inside my web.xml to point to this external file:
<context-param>
<param-name>CONFIG_FILE</param-name>
<param-value>/path/to/picketlink.xml</param-value>
</context-param>
I just can't make it work as the app does not find the file.
What would be the right way to write this path? I wanted something like this:
<context-param>
<param-name>CONFIG_FILE</param-name>
<param-value>home/user/picketlink_config/picketlink.xml</param-value>
</context-param>

How to set up Tomcat to use the config file in conf/Catalina/localhost?

so my question is how can I set up a Tomcat Server to use this configuration file conf/Catalina/localhost/MyApp.xml?
It works like a charm if my application is named like this: MyApp.war so tomcat will extract the archive to MyApp.
But I want to use a name with the version inside like MyApp##1.0-SNAPSHOT.war.
Is it possible to configure the tomcat so it will use the MyApp.xml anyway?
You can set path and docBase attributes of Context element as described in the docs.
Create a directory outside $CATALINA_HOME/webapps and place war files there, let's say /opt/tomcat/wars/. Then set docBaseattribute to the full path to your war. Make user tomcat user has read permissions on that directory. If war is inside webapps directory you could end up with a double deployment depending on deployOnStartup and autoDeploy attributes of Host element.
<Context path="/MyApp" docBase="/opt/tomcat/wars/MyApp##1.0-SNAPSHOT.war">
...
</Context>
You can try also naming the xml the same as the war file.

Different properties files on the same server

We have a database library that gets the connection information (user, host, etc) from a properties file. This file can be a config.properties file located in the classpath, or next to the execution jar or can be passed as an argument -Dproperties=/path/to/myConfig.properties.
We also have several applications that use this library, so each one has its own config.properties file used in its own execution.
But now I'm creating two web applications that use the same library. So, if I deploy them in Tomcat (war file), I have two options (to my knowledge):
1.- Include each config inside the WAR file. But with this, every time I need to tweak something in the config.properties I'll have to repack the war.
2.- Pass the -Dproperties parameter as an execution argument of Tomcat. But different war deployments will have to share the same properties file.
Is there a way around this?
Can I pass the -D argument to a specific deployment in Tomcat (or any other server)?
PS: This is one of the scenarios we have, but is not constraint to database connection info. We have other libraries that get parameters through config.properties file.
EDIT: I want to be able to have different config.properties file for each deployment. Not the same properties shared among them.
I think I found a way around using self contained webserver inside the application, like Jetty.
We've a similar requirement in which we share a common property file between different applications deployed into JBoss EAP server.
In $JBOSS_HOME/bin/standalone.conf file you can add configuration file path as below:
JAVA_OPTS="$JAVA_OPTS -DCONFIG_LOCATION=/external/config/configuration.properties"
Start the server with above specified property and within your application you can read this property file with apache commons-configuration api as below:
try {
props = new PropertiesConfiguration(System.getProperty("CONFIG_LOCATION"));
FileChangedReloadingStrategy strategy = new FileChangedReloadingStrategy();
// Delay 30s
// strategy.setRefreshDelay(30000);
props.setReloadingStrategy(strategy);
} catch (ConfigurationException e) {
e.printStackTrace();
}
With this reload strategy you can change your properties while your server is running. Also you can specify the interval after which all properties specified in the external file needs to be refreshed within your application without bouncing it. Hope this helps!
You can create an environment variable whose value will be the path where the properties file are located. Later use this environment variable will creating bean for property placeholder config.
for UNIX, you can add in your bash profile file
export CONF_DIR=/path/to/conf
And in spring context file, add this
<context:property-placeholder
location="file:///${CONF_DIR}/path/myConfig1.properties,
file:///${CONF_DIR}/path/myConfig2.properties"
properties-ref="applicationDefaultProperties" ignore-resource-not-found="false"
ignore-unresolvable="false"/>
So, when you want to change any thing in the properties file, you can change at one location, and then restart the application to load the new values in your app.
So, if your config file is this
db.user=username
db.password=password
Inside java class, you can use the keys as like this
#Value("${db.user")
private String username;
#Value("${db.password")
private String password;
The solution I found for my problem is using an embedded web server in my application. In my case, I'm using Jetty.
Now I pack my application as an executable jar and pass the system parameters as -D arguments and they live inside the instance of the application.
Like this:
java -Dproperties=config.properties -jar java_app_with_embedded_server.jar

Tomcat: Get short URL for Jersey based Rest Service

I have a Jersey based Rest service running on a tomcat server. There is no UI, just a server that offers some rest services. Now, to access this service the URL that i have to type in is pretty long. Something like localhost:8080/MyApp/url_pattern/classPath/method where MyApp is the webapp that i deployed, url_pattern is the pattern that i defined in the servlet-mapping in web.xml, classPath and method being the #Path annotations for the Class and method respectively. Is it possible to shorten it such that I get rid of the MyApp and url_pattern part of this URL. Something like localhost:8080/classPath/method.
PS: There is just one webApp running on this server, so no point having the MyApp part
I don't think you can remove all what you desire from the url but you can definitely remove the MyApp part by making it the root application for tomcat.
Answer on this related link describes it pretty well, how to set your application as the root application. So you can access your REST services without having the app name in url:
Setting default application in tomcat 7
Content copied from the above link:
First Method:
first shutdown your tomcat [from the bin directory (sh shutdown.sh)]
then you must delete all the content of your tomcat webapps folder (rm
-fr *) then rename your WAR file to ROOT.war finally start your tomcat [from the bin directory (sh startup.sh)]
Second Method:
leave your war file in CATALINA_BASE/webapps, under its original name
- turn off autoDeploy and deployOnStartup in your Host element in the server.xml file. explicitly define all application Contexts in
server.xml, specifying both path and docBase. You must do this,
because you have disabled all the Tomcat auto-deploy mechanisms, and
Tomcat will not deploy your applications anymore unless it finds their
Context in the server.xml.
Note:
that this last method also implies that in order to make any change to
any application, you will have to stop and restart Tomcat.
Third Method:
Place your war file outside of CATALINA_BASE/webapps (it must be
outside to prevent double deployment). - Place a context file named
ROOT.xml in CATALINA_BASE/conf//. The single element in this context
file MUST have a docBase attribute pointing to the location of your
war file. The path element should not be set - it is derived from the
name of the .xml file, in this case ROOT.xml. See the Context
Container above for details.

do we need to deploy war after changing property file

Do we need to deploy war after changing the property file in java?
At least you may have to restart the application so that it reads the new property file. So re-deploying the entire application may not be necessary.
you do need war to deploy if property file is not loaded each time it is needed
Yes, unless you make your context reloadable. To do that update your conf/context.xml, adding the "reloadable" attribute to the root node like so:
<Context reloadable="true">
Take a look at the Tomcat Config Docs.

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