我创建了一个自定义数据类型来存储有效和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);
        }
    }
    ...
}

及其相应的 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，我可以将它用作个人实体的属性:

public class Person extends BaseEntity {

    private String name;

    private LocalDate birthDate;

    @QueryType(PropertyType.STRING)
    private Email email;

    ...
}

这非常有效。但是当我尝试使用 Querydsl 通过电子邮件地址的开头来查找所有人时，我遇到了问题。我用 @QueryType 作为字符串注释了电子邮件属性。这样，就创建了 Querydsl 元模型，这样我就可以像这样进行查询(使用 .startsWith()):

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);
}

但是当我运行它时出现这个异常:

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(和一个不雅的技巧，concat)得到正确的结果:

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。是否可以使用此库创建此查询？

最佳答案

根据 Querydsl 创建者，Timo Westkämper , 这个issue与 EclipseLink 更相关。

关于java - 使用 EclipseLink 和 Querydsl 的自定义数据类型，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/19017907/