Java hibernate中一对一关联的延迟加载问题_Java_Hibernate_Lazy Loading_One To One

Java hibernate中一对一关联的延迟加载问题

java hibernate

Java hibernate中一对一关联的延迟加载问题,java,hibernate,lazy-loading,one-to-one,Java,Hibernate,Lazy Loading,One To One,我有一个关于延迟加载OneToOne关联映射的问题 +-----------------+ +------------------+ | USER | | Address | | |1 1| | | ADDRESS_ID +-------------+ | |

我有一个关于延迟加载

OneToOne

关联映射的问题

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

案例1外键位于子表（地址）中

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

在上面的地址中，延迟加载不起作用

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

情况2外键在父表中（用户）

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

在上面的地址中，延迟加载工作

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

有人能解释一下为什么一对一延迟加载在第一种情况下不起作用，但在第二种情况下起作用吗？

@OneToOne

有点棘手

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

这完全取决于您使用的持久性提供程序

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

某些提供程序不尊重

FetchType.LAZY

提示

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

您可以尝试指定（在关系的两端）

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

为了了解这里发生了什么，让我们看一下级别：

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

第一种情况：

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
|                 +-------------+  USER_ID (FK)    |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

加载时，用户Hibernate必须知道地址是否存在

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

因此Hibernate发出一个类似于以下内容的SQL请求：

SELECT * FROM USER WHERE ID = ?
SELECT * FROM ADDRESS WHERE user_id = ?

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

当获得结果时，实体已经加载，因此没有必要为地址分配LazyProxy。Hibernate分配获取的对象

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

第二种情况：

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

Hibernate不需要检查地址是否存在。这就是创建代理的原因。

@OneToOne

处理起来有点棘手

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

这完全取决于您使用的持久性提供程序

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

某些提供程序不尊重

FetchType.LAZY

提示

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

您可以尝试指定（在关系的两端）

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

为了了解这里发生了什么，让我们看一下级别：

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

第一种情况：

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
|                 +-------------+  USER_ID (FK)    |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

加载时，用户Hibernate必须知道地址是否存在

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

因此Hibernate发出一个类似于以下内容的SQL请求：

SELECT * FROM USER WHERE ID = ?
SELECT * FROM ADDRESS WHERE user_id = ?

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

当获得结果时，实体已经加载，因此没有必要为地址分配LazyProxy。Hibernate分配获取的对象

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

第二种情况：

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

Hibernate不需要检查地址是否存在。这就是创建代理的原因。

阅读本文。它将使您了解为什么反向侧（映射方式）属性上的一对一无法按预期工作

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

实际上，反过来hibernate需要知道映射的值是什么，因为用户可以立即请求值

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

阅读这篇文章，你就会清楚地了解事情是如何运作的

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

感谢阅读本文。它将使您了解为什么反向侧（映射方式）属性上的一对一无法按预期工作

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

实际上，反过来hibernate需要知道映射的值是什么，因为用户可以立即请求值

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

阅读这篇文章，你就会清楚地了解事情是如何运作的

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

谢谢

我没能让它工作，所以我使用了一个准备好的语句，在那里我加入了FETCH，无论如何Hibernate在事后查询的东西

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

添加

optional=true

或伪造

OneToMany

关系在我看来是错误的解决方法。

我一直无法让它工作，所以我使用了一个准备好的语句，其中我

JOIN-FETCH

无论如何Hibernate在事后查询的东西

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?

添加

optional=true

或伪造

OneToMany

关系，依我看，是错误的解决方法。

你所说的引号是什么意思：“延迟加载不起作用”？它只意味着在获取用户时，不应获取地址。你所说的引号是什么意思：“延迟加载不起作用”？它只意味着在获取用户时，地址不能取，你能给我解释一下这两种情况的区别吗？你能给我解释一下这两种情况的区别吗？

+-----------------+             +------------------+
| USER            |             |  Address         |
|                 |1           1|                  |
| ADDRESS_ID      +-------------+                  |
|                 |             |                  |
|                 |             |                  |
+-----------------+             +------------------+

SELECT * FROM USER WHERE ID = ?