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Spring instance is not starting if one of the route-builder instance is failed

We are initializing the 2 different camel routes via spring context xml,
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.0.xsd
http://camel.apache.org/schema/spring http://camel.apache.org/schema/spring/camel-spring.xsd">
<camelContext id="camelContext" xmlns="http://camel.apache.org/schema/spring">
<routeBuilder ref="DescriptiveInfoRoute"/>
<routeBuilder ref="DescriptiveInfoRouteV2"/>
</camelContext>
</beans>
Both the routes are separate each will having it's input properties files to read the input values. (which makes 2 input files should be compulsory in order for both the routes to get initialized and starts the application). Now we got a new requirement one input file may or may not present. So is there any way in the spring to successfully start the application even if one the route was successfully initialized.
Thanks in advance,
Raghavan

How to serialize/deserialize spring xml config

I am going to change ldap config of another program (its called openkm) which includes editing a spring xml file like this:
<beans:beans xmlns="http://www.springframework.org/schema/security"
xmlns:b="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security.xsd">
<security:ldap-server id="ldapServer"
url="ldap://192.168.0.6:389/DC=ldap,dc=weyler,dc=local"
manager-dn="CN=Administrator,cn=users,dc=weyler,dc=local"
manager-password="password"/>
<security:authentication-manager alias="authenticationManager">
<security:ldap-authentication-provider
server-ref="ldapServer"
user-search-base="cn=Users"
user-search-filter="(sAMAccountName={0})"
group-search-base="cn=Users"
group-search-filter="(member={0})"
group-role-attribute="cn"
role-prefix="none">
</security:ldap-authentication-provider>
</security:authentication-manager>
</beans:beans>
The configuration just concerns to replace existing values (for example to change ip, port or baseDn).
Using replace all with regex is not reliable and Dom xml parser is a mess for this big xml file. How else could this be done?

Use Spring PropertyPlaceHolderConfigurer to externalize these configurations into a property file and access the same from your XML. An example here.
And then you can update your property file programmetically with apache common properties or any other property writer.
This way it will be far efficient and cleaner than to manipulate XMLs.

spring java xml configuration

im using spring for my java web app. the site has got bigger and i would like to set some configurations.
i have been researching and came across things like document builder factory, replacing spring xml with java config and others. i dunno where to start.
im thinking of implementing the configurations in xml (WEB/newConfig.xml) and have it read by the java beans. basically i wanna input my cofiguration values into xml and have it load by a java bean so that i can use it in controllers and jstl.
im just giving some examples here. for example xml configurations:
<property name="numberOfCars" value="3" />
<property name="webSiteName" value="New Spring Web App" />
....
and i read it in my java class:
class Config {
public getNumberOfCars() {
return numOfCars;
}
public getWebSiteName() {
return webSiteName;
}
}
where should i start and what online materials can i read?
==============================
update
here is what i have created.
applicationContext.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:aop="http://www.springframework.org/schema/aop" xmlns:context="http://www.springframework.org/schema/context"
xmlns:jee="http://www.springframework.org/schema/jee" xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">
<context:property-placeholder location="/WEB-INF/your_prop_file.properties" />
<bean id="ConfigMgr" class="org.domain.class.ConfigMgr">
<property name="username" value="${username}">
</bean>
</beans>
you_prop_file.properties
username=hello world name
ConfigMgr.java
public class ConfigMgr {
private String username;
...getter
...setter
}
in my controller, here is what i did:
ConfigMgr config = new ConfigMgr();
sysout.out.println(config.getUsername());
i am getting null and i am sure im missing something here. where should i set the username value to the ConfigMgr class?

Spring Java configuration is a newer feature that allows you to configure your Spring application using Java classes instead of XML files. Its just an alternative for XML configuration. XML way is equally feature rich.
From what I could figure out from your problem, you want to move the hardcoded values of params (numberOfCars,webSiteName.. ) outisde your configuration file.
If that is the case, you don't have to go that far.
Just use :-
<context:property-placeholder location="classpath:your_prop_file.properties" />
in your spring xml file and replace the param values like:-
<property name="webSiteName" value="${website.name}" />
You need to have a your_prop_file.properties file in your classpath with enteries like:-
website.name=New Spring Web App

You are not injecting the ConfigMgr bean that you created in XML file.
What you are doing is you are creating a new Object in controller which does not have a clue about properties file.
Now you can inject it using either #Autowired inside your controller or through xml configuration.
There are plenty of examples available in google on basic spring dependency injection.

Validating spring config xml Files in Java

i want to validate some files looking like this:
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/task http://www.springframework.org/schema/task/spring-task-3.0.xsd />
...
<bean>...</bean>
...
</beans>
Some of them have more xmlns and locations and some less. Someone knows if there is any api where u just give in such files and it will validate it in java?
greetings

Spring will validate spring configuration files. You can write a junit that loads your configuration to test if the configuration files are formatted correctly.

There you go.
http://download.oracle.com/javase/1,5,0/docs/api/javax/xml/validation/package-summary.html
You can wrap it and make parser that will first extract schemas, then use them for validation.

Getting null value from applicationcontext.xml

I am working on struts2 application with spring for back end.
We are using database.properties file and the entries are as follows:
jdbc.url=jdbc:mysql://localhost:3306/myDb
jdbc.username=root
jdbc.password=rooooot
jdbc.csvlocation=C:\myCSV
I added the following new entry in database.properties
enhancePerf.Flag=true
In applicationcontext.xml I am fetching the value like this :-
<bean id="userLogin" scope="prototype"
class="com.hello.something.actions.UserLoginAction">
<property name="perfEnhance" value="${enhancePerf.Flag}"/>
</bean>
After declaring a global variable perfEnhance in UserLoginAction, and forming the setters and getters method of the same, I'm still not getting the value.
I followed the following link:-
http://www.roseindia.net/tutorial/spring/spring3/web/applicationcontext.xml-properties-file.html
Please advise.

Instead of your PropertyPlaceholderConfigurer bean, put:
<context:property-placeholder location="classpath:path/to/database.properties"
ignore-unresolvable="false"/>
This way if the property is not found, it will complain. Otherwise it seems that you may have another "database.properties" file in your classpath, that simply does not have such a property.
Make sure that "path/to/database.properties" is in your classpath. If database.properties itself is your class path, then no "path/to" is needed => just classpath:database.properties
You also have to configure Spring to manage your Actions as beans, using the ContextLoaderPlugin, as well as you have to use bean names in Struts config. If you have the following in your struts-config.xml file:
<action path="/users" .../>
You must define that Action's bean with the "/users" name in action-servlet.xml:
<bean name="/users" .../>
Please take a look at Spring Struts Integration from official Spring's docs.
EDIT to answer the comment:
context is an XML namespace that should be defined in the XML file where it is used:
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xmlns:util="http://www.springframework.org/schema/util"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context.xsd
http://www.springframework.org/schema/util
http://www.springframework.org/schema/util/spring-util.xsd">