Java 休眠一对一映射更新,而不是插入和抛出StaleStateException
我试图使用hibernate进行一对一的映射,将一些信息插入数据库,但每次尝试时都会出现此错误Java 休眠一对一映射更新,而不是插入和抛出StaleStateException,java,hibernate,one-to-one,Java,Hibernate,One To One,我试图使用hibernate进行一对一的映射,将一些信息插入数据库,但每次尝试时都会出现此错误 Caused by: org.hibernate.StaleStateException: Batch update returned unexpected row count from update [0]; actual row count: 0; expected: 1 at org.hibernate.jdbc.Expectations$BasicExpectation.checkBatche
Caused by: org.hibernate.StaleStateException: Batch update returned unexpected row count from update [0]; actual row count: 0; expected: 1
at org.hibernate.jdbc.Expectations$BasicExpectation.checkBatched(Expectations.java:85)
at org.hibernate.jdbc.Expectations$BasicExpectation.verifyOutcome(Expectations.java:70)
at org.hibernate.jdbc.BatchingBatcher.checkRowCounts(BatchingBatcher.java:90)
at org.hibernate.jdbc.BatchingBatcher.doExecuteBatch(BatchingBatcher.java:70)
at org.hibernate.jdbc.AbstractBatcher.executeBatch(AbstractBatcher.java:268)
at org.hibernate.engine.ActionQueue.executeActions(ActionQueue.java:268)
at org.hibernate.engine.ActionQueue.executeActions(ActionQueue.java:185)
at org.hibernate.event.def.AbstractFlushingEventListener.performExecutions(AbstractFlushingEventListener.java:321)
at org.hibernate.event.def.DefaultFlushEventListener.onFlush(DefaultFlushEventListener.java:51)
at org.hibernate.impl.SessionImpl.flush(SessionImpl.java:1217)
at org.springframework.orm.hibernate3.SpringSessionSynchronization.beforeCommit(SpringSessionSynchronization.java:145)
... 12 more
这是我的client.hbm.xml文件
<hibernate-mapping>
<class name="register.Client" table="client" lazy="false">
<id name="clientId" type="string" column="CLIENT_ID"
length="255">
<generator class="increment"></generator>
</id>
<property name="address" type="string" column="ADDRESS"
length="255" />
<property name="email" type="string" column="EMAIL"
length="255" />
<property name="username" type="string" column="USERNAME"
unique="true">
</property>
<!-- Associations -->
<!-- bi-directional one-to-one association to Login -->
<one-to-one name="login" class="login.UsrPwd" lazy="false"
outer-join="auto" />
</hibernate-mapping>
我不认为第二个应该是更新的,因为行还不存在,这就是为什么它抛出我得到的异常。我不知道它为什么这么做,如果能帮上忙,我将不胜感激
如果您需要更多信息,请告诉我“我可以提供”看起来您需要在登录之前创建并提交客户端对象。您是否正在创建一个与客户端关联的登录对象,但只提交/保存该登录对象?我找到了答案。我有时间的时候会把答案贴出来。这只是我代码中的一个愚蠢的小错误。
<hibernate-mapping>
<class name="login.UsrPwd" table="login">
<id name="username" type="string" column="USER_NAME"> </id>
<property name="password" column="PASSWORD" type="string"/>
<property name="type" column="TYPE_ID" type="int"></property>
<one-to-one name="consultantInfo" class="register.ConsultantInfo"
cascade="all" />
<!-- bi-directional one-to-one association to Client -->
<one-to-one name="client" class="register.Client" cascade="all"/>
</class>
</hibernate-mapping>
Hibernate: insert into login (PASSWORD, TYPE_ID, USER_NAME) values (?, ?, ?)
Hibernate: update client set ADDRESS=?, EMAIL=?, USERNAME=? where CLIENT_ID=?