[英]How to use Criteria in JPA in Spring boot?
我正在使用 JPA 和弹簧靴。 我是 JPA 的新手。 我想通过传递函数的值来检索一个对象。 但在示例中,它是通过休眠配置完成的。 我的类路径中没有配置sessionFactory bean
。 我想使用 JPA 来检索对象。 在示例中,它是:
// Transactional for Hibernate
@Transactional
public class AccountDAOImpl implements AccountDAO {
@Autowired
private EntityManager entityManager;
@Override
public Account findAccount(String userName) {
Session session = sessionFactory.getCurrentSession();
Criteria crit = session.createCriteria(Account.class);
crit.add(Restrictions.eq("userName", userName));
return (Account) crit.uniqueResult();
}
}
但我在 JPA 中尝试过这样的:
// Transactional for Hibernate
@Transactional
public class AccountDAOImpl implements AccountDAO {
@Autowired
private EntityManager em;
@Override
public Account findAccount(String userName) {
CriteriaBuilder cm= em.getCriteriaBuilder();
cm.createQuery(Account.class);
cm.add(Restrictions.eq("userName", userName));
return (Account) crit.uniqueResult();
}
}
但它向我显示错误,如何使用 JPA 传递用户名?
我的 pom.xml 是:
<?xml version="1.0" encoding="UTF-8"?>
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd">
<modelVersion>4.0.0</modelVersion>
<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>2.2.1.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>
<groupId>com.ashwin</groupId>
<artifactId>vemployee</artifactId>
<version>0.0.1-SNAPSHOT</version>
<name>vemployee</name>
<description>Demo project for Spring Boot for offc</description>
<properties>
<java.version>1.8</java.version>
</properties>
<dependencies>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-jpa</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-web</artifactId>
</dependency>
<!-- Spring Security Core -->
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-core</artifactId>
</dependency>
<!-- Spring Security Config -->
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-config</artifactId>
</dependency>
<!-- Spring Security Web -->
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>spring-security-web</artifactId>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-devtools</artifactId>
<scope>runtime</scope>
<optional>true</optional>
</dependency>
<!-- https://mvnrepository.com/artifact/jstl/jstl -->
<dependency>
<groupId>jstl</groupId>
<artifactId>jstl</artifactId>
<version>1.2</version>
</dependency>
<!-- needed for jsp -->
<dependency>
<groupId>org.apache.tomcat.embed</groupId>
<artifactId>tomcat-embed-jasper</artifactId>
<version>9.0.27</version>
</dependency>
<!--bootsrap and jquery-->
<dependency>
<groupId>org.webjars</groupId>
<artifactId>bootstrap</artifactId>
<version>4.3.1</version>
</dependency>
<dependency>
<groupId>org.webjars</groupId>
<artifactId>jquery</artifactId>
<version>3.4.0</version>
</dependency>
<!-- https://mvnrepository.com/artifact/org.webjars/bootstrap-datepicker -->
<dependency>
<groupId>org.webjars</groupId>
<artifactId>bootstrap-datepicker</artifactId>
<version>1.7.1</version>
</dependency>
<dependency>
<groupId>commons-fileupload</groupId>
<artifactId>commons-fileupload</artifactId>
<version>1.4</version>
</dependency>
<dependency>
<groupId>mysql</groupId>
<artifactId>mysql-connector-java</artifactId>
<scope>runtime</scope>
</dependency>
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-test</artifactId>
<scope>test</scope>
<exclusions>
<exclusion>
<groupId>org.junit.vintage</groupId>
<artifactId>junit-vintage-engine</artifactId>
</exclusion>
</exclusions>
</dependency>
</dependencies>
<build>
<plugins>
<plugin>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-maven-plugin</artifactId>
</plugin>
</plugins>
</build>
</project>
我也试过如下:
TypedQuery<Account> query=em.createQuery("Select a from Account a where a.userName = :userName",Account.class);
query.setParameter(1,userName);
Account acc=query.getResultList().get(0);
return acc;
但是如何在这里设置用户名?
您可以像这样使用 JPQL
@Override
public Account findAccount(String userName) {
return em.createQuery("select account from Account account where account.username = :username", Account.class)
.setParameter("username", userName).getSingleResult();
}
试试这个方法
CriteriaBuilder criteriaB = entityManager.getCriteriaBuilder();
CriteriaQuery<Distance> criteriaQ = criteriaB.createQuery(Account.class);
Root<Account> element = criteriaQ.from(Account.class);
Predicate[]predicates = new Predicate[1];
predicates[0]= criteriaB.equal(element.get("userName"),userName);
criteriaQ.multiselect(element.get(Account)).where(predicates);
return entityManager.createQuery(criteriaQ).getSingleResult();
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