I am extending my model class to WCMUse class to write my business logic. I have osgi configuration which has couple of properties. I want to read one of the property in my model class. I am not sure how to get the handle of osgi configuration in WCMUse class. Any pointers will be highly appreciated.
Adding the answer for future references.
You can use the #getSlingScriptHelper() method of the WCMUse class to get an handle of the SlingScriptHelper. This is the same sling object that is available through inclusion of global.jsp in traditional JSP scripts.
You can then call #getService() method of the SlingScriptHelper to look up the desired OSGi Service.
The following code snippet can be used in the model to get the service configuration
getSlingScriptHelper().getService(<<Configuration Service>>.class)
Related
I'm trying to override this bean which is provided by standard Hybris (OOTB) framework. I would love to override it so it uses my own custom class. Is this possible?
You can create new extended abstract class but it is not used by existing classes. You can use customize functionality to overwrite class and/or spring config. Bu you must careful during patching or updating system.
How to override hybris platform files other than customize
Do you want to inject a custom implementation for one of the properties mapped to the abstract bean definition or do you want the abstract bean to use a custom class that you provide?
The latter just doesnt work, because you would also break all configurations of implementing beans.
You should seek for all beans that use this abstract bean as a parent and think, which of those you want to change and change that. Most likely, those beans will have an alias defined that you can use to change it. :-)
Best Bechte
Im trying to upgrade my project from spring-boot-1.4.3.RELEASE to spring-boot-1.5.9.RELEASE.
In the 1.4.3.RELEASE the way I have used my custom implementation of repositories is as follow:
Made an interface MyCustomRepositroy that extends JpaRepository
Had a class MyCustomRepositoryImpl that implements MyCustomRepositroy and SimpleJpaRepository. In that class I have changed the behavior of
the of save, find and delete methods, because I needed a
ceratin behavior for entitys of certain type (lets say that I need a
custom save for all entitys that implements Special interface)
I made a MyCustomJpaRepositoryFactoryBean that extends JpaRepositoryFactoryBean. In that factory, I've overridden createRepositoryFactory and gave it my MyRepositoryFactory implementation.
In MyRepositoryFactory implementation I've overridden the getTargetRepository, and getRepositoryBaseClass.In these methods, I check if the entity is of type Special, and if so, I return MyCustomRepositoryImpl, otherwise I return SimpleJpaRepository.
Also, I can get the beanFactory in my MyCustomRepositoryImpl class because I call my own constructor of MyCustomRepositoryImpl that has also a beanFactory parameter via getTargetRepositoryViaReflection.
Now, with the new version (that uses spring-data-commons-1.13.9.RELEASE), I cannot override the Factory class, and hence can't decide for each entitiy which implementation to give, and have no way to get the beanFactory.
Is there any way I can get what I want?
Sorry for the mess, but I cannot post my code here.
P.S - my project is a spring based library, so I can't do anything to the entitys because my clients declare them, all I know that some entitys implement the Special interface and some don't
I am trying to figure out what happened in 1.5.9.RELEASE, but in the meantime, just as a fyi - 1.5.6.RELEASE works ok.
I am having the same issue when trying to update from 1.5.6.RELEASE to 1.5.9.RELEASE
I have a Spring app and I created a Rest service with a couple of GET methods.
Now that web service needs to be moved to a library module to be used by other modules. How can I do that? Where to place it in the project I want to use it? Extend it in the project? I don't have anything to add to it or override (it is a kind of utility web service). I just need its name as a component. Would I just have some child class that is just extending it with empty body and give it a component name? Is that ok?
import xxx.yyy.lib.UtilityWebService;
#Component("utilityWS")
public class MyWebService extends UtilityWebService{
}
What would be the best practice for this situation?
Right now, I bootstrap Camel using Spring. Using Spring, I can point my Camel contexts to their respective properties files which are then injected at boot time. My issues is that I now want to move my properties from a file to a database, yet still be able to use the property placeholders as I was before. What's the best way to go about doing this?
I've noticed that there's a PropertiesResolver interface that I could implement, but I wouldn't know how to tell Camel about my implementation. Camel's documentation is very lacking in this area.
I also wouldn't be opposed to having Spring get the properties from the database for me, although I don't see that happening.
The PropertiesResolver was designed to help Camel to locate the properties files from OSGi bundle or normal class path.
If you want to setup your owner PropertiesResolver, you can try to use org.apache.camel.spring.spi.BridgePropertyPlaceholderConfigurer.
You can inject your customer implementation of PropertiesResolver there.
I thought I'd update this with what I found to work. This probably isn't the best method way to do it, but it works and I didn't have to modify the Camel source code. Basically, I converted all of my classes from inheriting from RouteBuilder to inherit from GJKRouteBuilder (which inherits from RouteBuilder). Then in there, I did this:
public class GJKRouteBuilder extends RouteBuilder {
#Override
protected void checkInitialized() throws Exception {
//Get properties from CamelContext using getContext()
//Lookup properties from DB based on CamelContext
//Get the properties component from the context (or create one)
//call setOverrideProperties() on properties component
super.checkInitialized();
}
}
Again, probably not the best method, but it works. Now, any route that inherits from GJKRouteBuilder and has the proper values wired up through Spring will have the properties injected into the properties component as if they were coming right from a properties file.
I want to clear some moments about integrating spring and struts. I have only one action class per application extended from MappingDispatchAction. So, actually my app when doing something uses not Action objects, but methods from my action. All I want from spring is to initialize this action and all for now. Just simply set DAO object. I looked through documentation, but I don't understand following:
We use action path from struts-config.xml as a name of bean in action-servlet.xml. Okay, but am I supposed to write beans in action-servlet.xml for every path name and set this poor DAO ref or what ?
The Struts 1 config file will use the DelegatingActionProxy class as the type attribute for all action configurations.
The Spring config file will contain the bean definitions of each action implementation. I don't know what DAO you're talking about, but actions that require DAO or service injection need to have them listed, yes--that's what Spring configuration is.
You may also be able to use annotations if you're not interested in using XML configuration, or use bean inheritance if many beans share the same DAO/service/etc. property values.