getting error No qualifying bean of type [javax.persistence.EntityManagerFactory] is defined: expected single matching bean but found 2

I had the same issue today. Solved it doing the following:

First I've added the parameter unitName to @PersistenceContext to both entity manager properties:

@PersistenceContext(unitName="appPU")
@Qualifier(value = "appEntityManagerFactory")
private EntityManager appEntityManager;

@PersistenceContext(unitName="managerPU")
@Qualifier(value = "managerEntityManagerFactory")
private EntityManager managerEntityManager;

And in my configuration file I've added a property persistenceUnitName to my bean definitions:

<bean id="appEntityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="dataSource" ref="dataSource1" />
    <property name="persistenceUnitName" value="appPU" />
    <property name="packagesToScan" value="br.com.app.domain" />
    ...
</bean>

<bean id="managerEntityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="dataSource" ref="dataSource2" />
    <property name="persistenceUnitName" value="managerPU" />
    <property name="packagesToScan" value="br.com.app.domain" />
    ...
</bean>

Also I'd like to add once more useful comment: you need to extend the section in the 'web.xml' file of your web-app. Since now you have 2 Entity Managers, you need 2 OpenEntityManagerInViewFilters. Look the example:

<filter>
  <filter-name>OpenEntityManagerInViewFilter1</filter-name>
  <filter-class>org.springframework.orm.jpa.support.OpenEntityManagerInViewFilter</filter-class>
     <init-param>
         <param-name>entityManagerFactoryBeanName</param-name>
         <param-value>appEntityManagerFactory</param-value>
    </init-param>
</filter>

<filter-mapping>
    <filter-name>OpenEntityManagerInViewFilter1</filter-name>
    <url-pattern>/*</url-pattern>
    </filter-mapping>


<filter>
  <filter-name>OpenEntityManagerInViewFilter2</filter-name>
  <filter-class>org.springframework.orm.jpa.support.OpenEntityManagerInViewFilter</filter-class>
    <init-param>
        <param-name>entityManagerFactoryBeanName</param-name>
        <param-value>managerEntityManagerFactory</param-value>
    </init-param>
  </filter>

<filter-mapping>

<filter-name>OpenEntityManagerInViewFilter2</filter-name>
 <url-pattern>/*</url-pattern>
</filter-mapping>

Pay attention on fact the name 'appEntityManagerFactory' in < param-value >appEntityManagerFactory< / param-value > = 'appEntityManagerFactory' in < bean id="appEntityManagerFactory".

I also faced such problems and solved it. Please do the following to solve this error:

Add the following line to all your entity classes of both schema.

@PersistenceContext(unitName="<persistenceUnit>")
transient EntityManager entityManager;

<persistenceUnit> is the name of the persistence unit you defined in the persistence.xml file.