本文介绍了属性是对象时的命名查询?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想在JPA中进行以下查询:

SELECT * FROM `happyDB`.`users` U WHERE U.party_as_user =1 AND U.party_party_id =2

这很好,但是我的问题是我只将Party作为对象,而没有作为id,并且我无法使其工作.

在我要执行命名查询的Users实体中,我具有以下内容:

@JoinColumn(name = "pary_party_id", referencedColumnName = "party_id")
@ManyToOne
private Party partyId;

@Column(name = "party_as_user")
private Boolean partyAsUser;

我试图像带点符号的对象一样实现它,但这不起作用:

@NamedQuery(name = "Users.findByPartyAsUser", query = "SELECT u FROM Users u WHERE u.partyAsUser = :partyAsUser AND u.partyId.partyId = :partyId")

Party对象中有一个名为partyId的属性,但是它不起作用.对此有什么解决方案吗?还是每次向Users中插入新的Party时,都必须向Users-bean添加一个属性,如private int partyID并填充它?

感谢您的帮助!萨米·努尔米(Sami Nurmi)

解决方案

通常,您可以在JPA中将对象用作参数,

SELECT u FROM Users u WHERE u.partyAsUser = :partyAsUser AND u.party  = :party

Party party = new Party(id);
query.setParameter("party", party);

但是,如果您使用正确的变量名,那么您应该可以使用

SELECT u FROM Users u WHERE u.partyAsUser = :partyAsUser AND u.party.id  = :id

如果您比对象更了解SQL,那么您始终可以使用本机SQL查询.

I would like to make this query in JPA:

SELECT * FROM `happyDB`.`users` U WHERE U.party_as_user =1 AND U.party_party_id =2

This is working fine, but my problem is that I have Party only as an object, not as an id and I can't make it work.

In the Users-entity where I am trying to do the named query I have the following:

@JoinColumn(name = "pary_party_id", referencedColumnName = "party_id")
@ManyToOne
private Party partyId;

@Column(name = "party_as_user")
private Boolean partyAsUser;

I tried to implement it like an object with dot notation, but that's not working:

@NamedQuery(name = "Users.findByPartyAsUser", query = "SELECT u FROM Users u WHERE u.partyAsUser = :partyAsUser AND u.partyId.partyId = :partyId")

There is a property called partyId inside the Party-object, but it's not working. Is there any solution for that or do I have to add one property to the Users-bean like private int partyID and populate it every time when a new Party is inserted into Users?

Thanks for helping!Sami Nurmi

解决方案

In general you can use an object as a parameter in JPA,

SELECT u FROM Users u WHERE u.partyAsUser = :partyAsUser AND u.party  = :party

Party party = new Party(id);
query.setParameter("party", party);

But what you have should work if you use the correct variable names, my guess is,

SELECT u FROM Users u WHERE u.partyAsUser = :partyAsUser AND u.party.id  = :id

And you can always use a native SQL query, if you understand SQL better than objects.

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10-21 14:08