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Spring Batch - more than one writer based on field value

I am working on spring batch, for writer currently using FlatFileItemWriter.
I would like to write my input file content to more than one flat file based on some field value. Is Spring batch support any kind of functionality by default.[something similar to CompositeItemWriter]
For example, my input file content is something like this.
john,35,retail,10000
joe,34,homeloan,20000
Amy,23,retail,2000
Now i would like to write two different files based on third column, it means row 1 and row 3 should go to file1 and row 2 should go to file2.
My writer configuration is:
<bean id="fileWriter" class="org.springframework.batch.item.file.FlatFileItemWriter">
<property name="resource" value="file:C:/output.dat"/>
<property name="lineAggregator">
<bean class="org.springframework.batch.item.file.transform.DelimitedLineAggregator">
<property name="delimiter" value="|" />
<property name="fieldExtractor">
<bean class="org.springframework.batch.item.file.transform.BeanWrapperFieldExtractor">
<property name="names" value="field1,field2...." />
</bean>
</property>
</bean>
</property>
</bean>

Take a look at the ClassifierCompositeItemWriter. This ItemWriter implementation allows you to define a Classifier that chooses which of the defined delegate ItemWriter instances to delegate to. In your case, you'd create a Classifier that decided based on field4 and delegate the writing to the appropriate instance of the FlatFileItemWriter.
You can read more about the ClassifierCompositeItemWriter in the documentation here: http://docs.spring.io/spring-batch/trunk/apidocs/org/springframework/batch/item/support/ClassifierCompositeItemWriter.html

Use a ClassifierCompositeItemWriter
Calls one of a collection of ItemWriters for each item, based on a
router pattern implemented through the provided Classifier.
Router pattern is based on bean content

How to use property file in util:properties?

I must use spring util:properties, however, I need something like this:
if propertyFile x exists, use x, otherwise, use y.
Can you please advise how I can obtain this?

May be this is half solution! Spring can load the wildcard resources. Please see spring <util:properties /> with wildcards
In this way you can name your files like: x-config.properties and y-config.properties:
<bean id = "config"
class = "org.springframework.beans.factory.config.PropertiesFactoryBean">
<property name = "locations"
value = "classpath*:somefolder/*-config.properties" />
</bean>
If both x and y exist both of them are loaded.

Actually, there is an option ignoreResourceNotFound, but it isn't available for namespace component. You have to use PropertiesFactoryBean directly:
<bean id="props" class="org.springframework.beans.factory.config.PropertiesFactoryBean">
<property name="locations" value="y.properties, x.properties"/>
<property name="ignoreResourceNotFound" value="true"/>
</bean>
If your x.properties doesn't exist, it will be ignored and the properties from y.properties will remain.
If x.properties exists and it contains the same keys as y.properties, they override those from y.properties, because all locations are loaded one by one respectively.

Insert external data into persistence.xml

I want my persistence.xml to set some of its properties dynamically, to be specific:
<property name="hibernate.connection.password" value="password"/>
<property name="hibernate.connection.username" value="username"/>
I can build a class that could provide me the data I need, but I don't know how to set the class up in a way that it works like this:
<property name="hibernate.connection.password" value="${my.clazz.pass}"/>
<property name="hibernate.connection.username" value="${my.clazz.user}"/>
I have tried to set the class up like this
public class clazz{
String pass;
String user;
public clazz(){
//do stuff to set pass and user
}
//getter/setter
}
But that does not work. I haven't found a way here or in google, but I have seen the ${my.clazz.smth}-way several times.
So, how can I set that up? :)
Thanks in advance!

So, resolved this a while ago, but I still didn't answer:
Anthony Accioly pointed me to the right direction:
I added this to my applicationContext.xml's entityManagerFactory
<bean id="entityManagerFactory"
class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="persistenceUnitPostProcessors">
<bean class="my.package.SetupDatabase">
</bean>
</property>
//the other stuff
</bean>
The corresponding class, in this case I use hibernate:
package my.package;
public class SetupDatabase implements PersistenceUnitPostProcessor {
private String username;
private String password;
private String dbserver;
public void SetupDatabase(){
//do stuff to obtain needed information
}
public void postProcessPersistenceUnitInfo(MutablePersistenceUnitInfo pui) {
pui.getProperties().setProperty("hibernate.connection.username", username );
pui.getProperties().setProperty("hibernate.connection.password", password);
pui.getProperties().setProperty("hibernate.connection.url", dbserver );
}
}
This way the setup is done just once when starting the whole thing up, but the required data may be 'outsourced'.
Thanks again for pointing me to the right direction!

The value placeholder ${my.clazz.smth} that you refer to is usually read from a properties file as opposed to a class directly.
This is done using Spring's PropertyPlaceholderConfigurer.
Here is an example of a project which combined the Hibernate & Spring.

If you really need to delay the configuration until runtime (e.g., to obtain the database credentials from an external source such as a Web service) you can do it with Hibernate API Programatic Configuration, particularly Ejb3Configuration for legacy versions of Hibernate or ServiceRegistryBuilder (v 4.X)...
But be warned that, to the best of my knowledge, there is no way to dynamically update the username and password of a PersintenceUnit. You will have to build another EntityManagerFactory from a new Configuration instance (a quite expensive operation) every time that you need to change it's properties.
Anyway, unless you have a really good reason to, do not manage database credentials from your application, delegate it to a JNDI-Bound DataSource instead.
Ejb3Configuration cfg = new Ejb3Configuration()
// Add classes, other properties, etc
.setProperty("hibernate.connection.password", "xxxx")
.setProperty("hibernate.connection.username", "yyyy");
EntityManagerFactory emf= cfg.buildEntityManagerFactory();

Concatenate strings within a Spring XML configuration file?

I have a String value in a Spring configuration file that comes to be as the result of a JNDI lookup -- it happens to be a path name:
<jee:jndi-lookup id="myAppHomeDir" jndi-name="myAppHomeDir" />
Now I need to concatenate on to the end of this path another string and hand it off to another Spring bean as follows (which of course doesn't work):
<bean id="LogPath" class="org.mystuff.initBean">
<property name="logDirectory">
<jee:jndi-lookup id="myAppHomeDir"
jndi-name="myAppHomeDir" /> + "/logs"
</property>
</bean>
Is there a simple way to do this without me having to write a utility class in Java?

Try using Spring EL (expression language). I would try the following (not tested):
<jee:jndi-lookup id="myAppHomeDir" jndi-name="myAppHomeDir" />
<bean id="LogPath" class="org.mystuff.initBean">
<property name="logDirectory" value="#{myAppHomeDir+'/logs'}"/>
</bean>
Not quite sure if it would work. The thing that troubles me is the cast from File (I guess) to String when concatenating. So if the previous one didn't work, I would try:
#{myAppHomeDir.canonicalPath+'/logs'}
Let us know if it works.

Spring Security 3.0 with jdbc

I read the "Spring Security 3 database authentication with Hibernate"! But I don't know how I should implementate it into my project!
In which document I have to set the password/username/drivers/url for the database?
I have different column titles like OPERATOR_ID/USR_ID/PASSWORD
OPERATOR_ID should be the login name, USR_ID the role and the password for the login
Please, maybe you could post an example which implements my questions? Maybe for a checkout or a *.war file?

I dont think there is any configuration as such for doing this. You have to implement the UserDetailsService which has only one method loadUserByUsername to load the user and you have to implement the same to load your user information from your database using hibernate.
See here

You would need to configure a JDBCDaoImpl bean which takes a Datasource as a parameter. How you retrieve the Datasource is up to you, you may grab it from the app server or use something like Spring's DriverManagerDatasource Here is some (pseudo) config
<bean id="datasource" class="org.springframework.jdbc.datasource.DriverManagerDataSource">
<property name="driverClassName"><value>your.driver.classname</value></property>
<property name="url"><value>yourDatabaseUrl</value></property>
<property name="username"><value>yourUsername</value></property>
<property name="password"><value>yourPassword</value></property>
</bean>
<bean id="dao" class="org.springframework.security.core.userdetails.jdbc.JdbcDaoImpl">
<property name="DataSource" ref="datasource" />
...
</bean>