查询:参数中的属性名称 [英] Query : property name in parameter
问题描述
通过此查询,我成功检索了数据库中的电话号码:
With this query, I succeed to retrieve a phone number in database:
import java.util.List;
import org.springframework.data.jpa.repository.JpaReposit ory;
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.query.Param;
import com.mc.appcontacts.domain.hibernate.Contact;
public interface ContactRepository extends JpaRepository<Contact, Integer> {
@Query("SELECT c.phoneNumber from Contact c WHERE LOWER(c.name) = LOWER(:name)")
String find(@Param("name") String name);
但是可以在参数中动态指定要检索的属性的名称吗?
But is it possible to specify dynamically name of the property i want to retrieve in parameter?
在网上阅读的所有tuto中,我了解到我们可以在参数中传递属性的值(在我的示例中:@Param("name")字符串名) 但是我要传递的参数是属性的名称而不是值!
In all tuto i've read on the net, i learn we can pass the value of the property in parameter (In my exemple : @Param("name") String name ) but what i want to pass in parameter is the name of the property not the value !
我知道下面的例子是不正确的,但这是为了给出大致的想法:
I know the exemple below is not correct but it's to give the general idea :
@Query(从联系人c WHERE LOWER(c.name)= LOWER(:name)中选择c.(属性)") 字符串查找(@Param("name")字符串名称,@ Param("property")字符串属性);
@Query("SELECT c.(property) from Contact c WHERE LOWER(c.name) = LOWER(:name)") String find(@Param("name") String name, @Param("property") String property);
具有property = phoneNumber(或我的表格的其他属性).
With property = phoneNumber (or an other property of my table).
谢谢您的帮助!
我不知道该怎么做(一切对我来说都是新的):
I don't understand how to do that (everything is new for me):
我已经阅读(并尝试过)像这样定义jpql:
I have read (and try) that jpql is defined like this :
import com.mysema.query.jpa.impl.JPAQuery;
import com.mc.appcontacts.repository.ContactRepository; // managed by spring data
//jpa repository
public class ServicesC {
@Autowired
private ContactRepository repository;
@PersistenceContext // try
private EntityManager em; // try
JPAQuery query = new JPAQuery(em); // try
public Contact getOne(Integer id) {
return repository.findOne(id);
}
public String findAtt(String att, String key){
String jpql = "SELECT c." + att + " from Contact c WHERE LOWER(c.name) = LOWER(:key)"; // try
List<Contact> l = (List<Contact>) em.createQuery(jpql); // try
return "test";
}
}
但是它不起作用(我不感到惊讶...):
But it doesn't work (i'm not surprised...) :
2014-02-24 18:18:34.567:WARN::Nested in org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'appMapping': Injection of autowired dependencies failed; nested exception is org.springframework.beans.factory.BeanCreationException: Could not autowire field: private com.mc.appcontacts.service.ServiceC com.mc.appcontacts.mvc.MappingService.service; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'Service' defined in file [C:\Professional\Workspaces\Eclipse\ContactMain\ContactCore\target\classes\com\mc\appcontacts\service\ServiceC.class]: Instantiation of bean failed; nested exception is org.springframework.beans.BeanInstantiationException: Could not instantiate bean class [com.mc.appcontacts.service.ServiceC]: Constructor threw exception; nested exception is java.lang.NullPointerException:
java.lang.NullPointerException
at com.mysema.query.jpa.impl.JPAProvider.get(JPAProvider.java:72)
at com.mysema.query.jpa.impl.JPAProvider.getTemplates(JPAProvider.java:80)
at com.mysema.query.jpa.impl.JPAQuery.<init>(JPAQuery.java:46)
我必须仅为jpql定义第二个EntityManager吗? (有可能吗?这是正确的方法吗?我不这样认为...) 我已经在xml文件中为Spring-data定义了EntityManager:
Must i define a second EntityManager only for jpql ? (Is it possible ? is it the right way ? I don't think so...) I have already a EntityManager defin for Spring-data in xml file :
<tx:annotation-driven transaction-manager="transactionManager" />
<!-- Activate Spring Data JPA repository support -->
<jpa:repositories base-package="com.mc.appcontacts.repository" />
<!-- Declare a JPA entityManagerFactory -->
<bean id="entityManagerFactory" class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
<property name="persistenceXmlLocation" value="classpath:META-INF/contacts/hibernate/persistence.xml" />
<property name="persistenceUnitName" value="hibernatePersistenceUnit" />
<!-- <property name="dataSource" ref="dataSource" /> -->
<property name="jpaVendorAdapter" ref="hibernateVendor" />
</bean>
<!-- Specify our ORM vendor -->
<bean id="hibernateVendor" class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
<property name="showSql" value="${hibernate.showSql}" />
</bean>
<!-- Declare a transaction manager-->
<bean id="transactionManager" class="org.springframework.orm.jpa.JpaTransactionManager">
<property name="entityManagerFactory" ref="entityManagerFactory" />
</bean>
请帮助我...它是如何工作的?
Please help me ... how does it work ?
推荐答案
对于这种动态构建查询的用例,我认为最好的选择是探索Criteria API,它非常适合此类情况. http://docs.oracle.com/javaee/6/tutorial/doc/gjitv.html
I think for this use case of building queries dynamically your best bet would be to explore Criteria API, which is very suitable for such things. http://docs.oracle.com/javaee/6/tutorial/doc/gjitv.html
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