我创建了一个自定义数据类型,用于存储有效且规范化的电子邮件地址:
public class Email implements Serializable {
private final String value;
public Email(String emailAddress) {
this.value = normalize(emailAddress);
if (!isValid(value)) {
throw new IllegalArgumentException("Email address format is not valid: " + value);
}
}
...
}
以及相应的JPA 2.1转换器,用于自动存储到数据库和从数据库中检索:
@Converter(autoApply = true)
public class EmailConverter implements AttributeConverter<Email, String> {
@Override
public String convertToDatabaseColumn(Email email) {
return email == null ? null : email.toString();
}
@Override
public Email convertToEntityAttribute(String email) {
return email == null ? null : new Email(email);
}
}
现在,我可以将其用作个人实体的财产:
public class Person extends BaseEntity {
private String name;
private LocalDate birthDate;
@QueryType(PropertyType.STRING)
private Email email;
...
}
这非常有效。但当我试图使用Querydsl通过电子邮件地址的开头找到所有人时,我遇到了一个问题。我用注释了电子邮件属性
@QueryType
作为字符串。通过这种方式,创建了Querydsl元模型,这样我就可以进行查询(使用
.startsWith()
)像这样:
private static final QPerson person = QPerson.person;
public List<Person> getEmailStartWith(String pattern) {
pattern = Email.normalize(pattern);
return new JPAQuery(getEntityManager())
.from(person)
.where(person.email.startsWith(pattern))
.orderBy(person.email.asc())
.list(person);
}
但当我运行它时,我会得到这个异常:
java.lang.IllegalArgumentException: You have attempted to set a value of type class java.lang.String for parameter 1 with expected type of class xxx.Email from query string select person
from Person person
where person.email like ?1 escape '!'
order by person.email asc.
at org.eclipse.persistence.internal.jpa.QueryImpl.setParameterInternal(QueryImpl.java:932) ~[eclipselink-2.5.0.jar:2.5.0.v20130507-3faac2b]
at org.eclipse.persistence.internal.jpa.QueryImpl.setParameterInternal(QueryImpl.java:906) ~[eclipselink-2.5.0.jar:2.5.0.v20130507-3faac2b]
at org.eclipse.persistence.internal.jpa.EJBQueryImpl.setParameter(EJBQueryImpl.java:469) ~[eclipselink-2.5.0.jar:2.5.0.v20130507-3faac2b]
at org.eclipse.persistence.internal.jpa.EJBQueryImpl.setParameter(EJBQueryImpl.java:1) ~[eclipselink-2.5.0.jar:2.5.0.v20130507-3faac2b]
at com.mysema.query.jpa.impl.JPAUtil.setConstants(JPAUtil.java:44) ~[querydsl-jpa-3.2.1.jar:na]
at com.mysema.query.jpa.impl.AbstractJPAQuery.createQuery(AbstractJPAQuery.java:130) ~[querydsl-jpa-3.2.1.jar:na]
at com.mysema.query.jpa.impl.AbstractJPAQuery.createQuery(AbstractJPAQuery.java:97) ~[querydsl-jpa-3.2.1.jar:na]
at com.mysema.query.jpa.impl.AbstractJPAQuery.list(AbstractJPAQuery.java:240) ~[querydsl-jpa-3.2.1.jar:na]
...
我可以使用JPQL得到正确的结果(还有一个不雅的技巧,
concat
):
public List<Person> getEmailStartWith(String pattern) {
pattern = Email.normalize(pattern);
return getEntityManager()
.createQuery("select p from Person p where p.email like concat(?1, '%')", Person.class)
.setParameter(1, pattern)
.getResultList();
}
但我当然更喜欢Querydsl的类型安全性。是否可以使用此库创建此查询?
