I use ActiveMQ as Embedded with Spring Boot.
It seems the Broker is created trough an ActiveMQConnectionFactory.
I understand that the way to configure the broker is to set parameters in the query with broker. as described here : http://activemq.apache.org/how-do-i-embed-a-broker-inside-a-connection.html
I would like to setup some features about the DLQ, so it's in the destinationPolicy attribute, but the attribute type is not a simple type but a complex type, how can I write the query parameter to disable DLQ, please ?
Complementing #Petter and #April answers, below the same solutions but with more complete samples:
1. Petter solution, import activemq.xml at connnection factory url
build.gradle
ext {
springBootVersion = "1.5.3.RELEASE"
activeMQVersion = "5.14.5"
}
dependencies {
compile("org.springframework.boot:spring-boot-starter-activemq:${springBootVersion}")
compile("org.apache.activemq:activemq-broker:${activeMQVersion}")
testCompile("org.springframework.boot:spring-boot-starter-test:${springBootVersion}")
testCompile group: 'org.apache.activemq', name: 'activemq-spring', version: "${activeMQVersion}"
testCompile("junit:junit:4.12")
}
src/main/resources/activemq.xml
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:amq="http://activemq.apache.org/schema/core"
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://activemq.apache.org/schema/core
http://activemq.apache.org/schema/core/activemq-core-5.4.0.xsd
">
<broker xmlns="http://activemq.apache.org/schema/core" brokerName="broker1" persistent="false" >
<transportConnectors>
<transportConnector name="vm" uri="vm://broker1"/>
</transportConnectors>
</broker>
</beans>
Config.java
#EnableJms
#SpringBootApplication
#EnableAutoConfiguration
#Configuration
public class Config {}
application.properties
spring.activemq.broker-url=vm://broker1?brokerConfig=xbean:activemq.xml
2. April solution, import activemq.xml at Spring Configuration
Just remove application.properties then add #ImportResource("classpath:activemq.xml") entry to Config.java
Config.java
#EnableJms
#SpringBootApplication
#EnableAutoConfiguration
#Configuration
#ImportResource("classpath:activemq.xml")
public class Config {}
Good question. The properties on the vm-transport for auto-broker creation are great, but only up to a point which I think you have hit.
My suggestion is that you define the broker configuration as you normally would have done in XML and then just refer to this xml in the URI. Destination policies are indeed a complex structure and I don't see how it would be a good idea to define them with simple query params even if it was possible.
vm://localhost?brokerConfig=xbean:activemq.xml
I had this problem and solved it by using a spring configuration file. In my case, I wanted to configure my broker to persist.
I added the needed libs in my pom: including activemq-broker, activemq-spring, spring-jms (and in my case, activemq-leveldb-store).
My spring xml file looked like this:
<beans
xmlns="http://www.springframework.org/schema/beans"
xmlns:amq="http://activemq.apache.org/schema/core"
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://activemq.apache.org/schema/core http://activemq.apache.org/schema/core/activemq-core.xsd">
<broker xmlns="http://activemq.apache.org/schema/core" brokerName="xyz">
<persistenceAdapter>
<levelDB directory="activemq-data"/>
</persistenceAdapter>
<transportConnectors>
<transportConnector uri="vm:localhost?persistent=true" />
</transportConnectors>
</broker>
</beans>
And I registered the spring file in one of my configuration classes:
#ImportResource("activemq-spring.xml")
That did the job.
I tried the xbeans solution first, but I got stuck because I was missing some xbeans classes, and I didn't know if it was a version thing or what. I'm using activemq 5.12.1
Related
I have a project in grails 3, that has a project spring dependency, in the spring project, xml beans are configured. How should import the bens in grails architecture?
build.gradle
dependencies {
compile (project(':spring-project')) { transitive = false }
}
settings.gradle
includeFlat 'spring-project'
I tried the following ways:
in the resources.groovy load the beans:
beans = {
importBeans('path/to/beans-definition.xml')
}
in the resources.xml:
<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:task="http://www.springframework.org/schema/task"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-4.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-4.0.xsd http://www.springframework.org/schema/task http://www.springframework.org/schema/task/spring-task.xsd">
<import resource="path/to/beans-definition.xml" />
</beans>
Besides that, in the spring project, beans using java annotation are configured. The beans is not working in grails app, even by setting the spring project packages in the conponent scan.
#ComponentScan(basePackages = ["package.spring.project.beans"])
class Application extends GrailsAutoConfiguration {
static void main(String[] args) {
GrailsApp.run(Application, args)
}
}
But, nothing works. Any help would be appreciated..
Beans can also be configured using a grails-app/conf/spring/resources.xml. In earlier versions of Grails this file was automatically generated for you by the run-app script, but the DSL in resources.groovy is the preferred approach now so it isn’t automatically generated now. But it is still supported - you just need to create it yourself.
<bean id="myBean" class="my.company.MyBeanImpl">
<property name="someProperty" value="42" />
<property name="otherProperty" value="blue" />
further check this link
I am interested to know the classes used in spring integration tag so that I can get more details of the tags by going through the javadoc of the classes.
I have two basic questions:
Do the spring integration xml tags (for example stdin-channel-adapter) convert to <bean class=".." /> tags?
how to figure out the bean class associated with the spring integration tags?
Here is a simple example of spring integration xml context file:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:int="http://www.springframework.org/schema/integration"
xmlns:int-stream="http://www.springframework.org/schema/integration/stream"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/integration http://www.springframework.org/schema/integration/spring-integration-4.0.xsd
http://www.springframework.org/schema/integration/stream http://www.springframework.org/schema/integration/stream/spring-integration-stream-4.0.xsd">
<int-stream:stdin-channel-adapter id="producer" channel="messageChannel" />
<int:poller id="defaultPoller" default="true" max-messages-per-poll="2" fixed-rate="100" />
<int-stream:stdout-channel-adapter id="consumer" channel="messageChannel" append-newline="true" />
<int:channel id="messageChannel" />
</beans>
Thanks
I think you have some mistake in your first question. See Andreas' comment.
Anyway the answer for you is like.
Any custom tags in Spring are handler by the particular NamespaceHandler. Typically you can find the target impl in file like META-INF/spring.handlers in the particular Spring jar, e.g.:
http\://www.springframework.org/schema/integration/stream=org.springframework.integration.stream.config.StreamNamespaceHandler
With that in hands you can find the code like:
this.registerBeanDefinitionParser("stdin-channel-adapter", new ConsoleInboundChannelAdapterParser());
Where you can determine that a ConsoleInboundChannelAdapterParser is responsible for parsing and instantiation beans for <stdin-channel-adapter> tag.
And there you can find the code like:
BeanDefinitionBuilder builder = BeanDefinitionBuilder.genericBeanDefinition(
"org.springframework.integration.stream.CharacterStreamReadingMessageSource");
So, the real class for target bean instance is CharacterStreamReadingMessageSource. But that's not all.
Please, look here for the design and model: http://docs.spring.io/spring-integration/docs/4.3.0.RELEASE/reference/html/overview.html#programming-tips
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.
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">
for my spring-mvc application I created several types of configuration (unittest, integration, qa, production). All the configs are in one war-file, so there is only one type of application I create. Which configuration to take should be decided by the server, where the application is running.
To decide what kind of configuration should be used, I have to look into a file. After that I can decide which configuration should be used by spring mvc.
For now by convention there is always the -servlet.xml used. Is there a way how to decide dynamically which config to take?
Regards,
Michael
Here is a solution that I use. It works very well:
Put the configuration differences in property files.
Keep a single Spring xml with placeholders.
Use PropertyPlaceholderConfigurer to set the properties.
PropertyPlaceholderConfigurer can use system properties to resolve the name of the property file to load.
Set a system property with the name of your environment before initilizing the PropertyPlaceholderConfigurer (you can do this in a bean that reads the value out of your file).
And there you go! The environment will be cleanly detected, and the relevant properties will be loaded!
No need to wait for Spring 3.1, you can use this solution today with 3.0.
I have the same setup, but I use maven to build the WARs differently. I use a PropertyPlaceholderConfigurer in the context:
<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:context="http://www.springframework.org/schema/context"
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">
<context:property-placeholder location="classpath:datasource.properties" ignore-unresolvable="true" />
<bean id="dataSource" class="org.springframework.jdbc.datasource.DriverManagerDataSource"
p:driverClassName="${jdbc.driver}"
p:url="${jdbc.url}"
p:username="${jdbc.username}"
p:password="${jdbc.password}" />
<!--other beans-->
</beans>
then I setup an environments folder:
src
--main
----environments
------dev
--------datasource.properties
------cert
--------datasource.properties
------prod
--------datasource.properties
Then in my Maven pom, I use a build profile to copy anything in the environment folder based on a parameter flag in the maven command:
<profiles>
<profile>
<id>environment</id>
<activation>
<property>
<name>environment</name>
</property>
</activation>
<build>
<resources>
<resource>
<directory>
src/main/environments/${environment}
</directory>
</resource>
</resources>
<!-- other build config and plugins -->
so the following command:
mvn clean -Denvironment=dev install
would copy the dev datasource.properties to the war
After all I'm using PropertyPlaceholderConfigurer but slightly differnt than Axel mentioned: I load just one property from my configuration and use it to determine which import to use. Because of https://jira.springframework.org/browse/SPR-1332 I cant use a file to store the instance-type, but have to use environment-variables.
<bean id="propertyConfigurerOne" class="org.springframework.beans.factory.config.PropertyPlaceholderConfigurer"/>
<import resource="classpath:/web${vabse.Environment}.xml"/>