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">
Related
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.
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
I am using some third-part jars that PARTLY migrated to spring. The pain point is that there are lots of initialization module that mot migrated to spring. There initialization module need to be executed first before lots' of beans creation.
I also read Spring 3 bean instantiation sequence, the problem here is, the third part library uses #Component for bean creation (which depends on initialization module that not migrated to spring, wired right...?).
Now I can write a spring bean to wrap all initialization modules. And create the bean before beans that need it.
So is there a way to specify the bean creation sequence to create the initialization bean first?
Also I checked some documents, spring bean creation is in single thread so this would work.
You can use BeanPostProcessor and add your initizaling module as the dependency
<?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"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.2.xsd">
<bean class="com.foo.CustomBeanPostProcessor" depends-on="com.foo.InitModuleBean"/>
<bean class="com.foo.BarBean" />
<bean id="com.foo.InitModuleBean" class="com.foo.InitModuleBean" />
</beans>
you put the wraperBean that initialize as the first Bean in your xmlConfig file then spring will process the initialization by initializing that Wraper An other solution is to add depends-on attribute to all your beans that depend on that Wrapper and you specify the id of that wrapper bean inside the attribute depends-on like this
<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.xsd">
<bean id="modules" class="com.mycompany.ModulesWrapper" />
<bean class="com.foo.ClientBean" depends-on="modules" />
</beans>
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 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.