无法删除行：TransactionRequiredException:执行更新/删除查询

Question

关于这个问题有好几篇帖子，但仍然没有找到答案。这是父类Userr。在@OneToMany关系中，我想删除一个特定的子帐户。

现在，当我通过“删除”查询执行此操作时，我得到以下异常。

组织。springframework。刀。InvalidDataAccessApiUsageException：执行更新/删除查询；嵌套的异常是javax。坚持不懈TransactionRequiredException：执行更新/删除查询

@RooJavaBean
@RooToString
@RooJpaEntity
@RooJpaActiveRecord(finders = { "findUserrsByUserName"})
public class Userr {


@NotNull
@Column(unique = true)
private String userName;


@NotNull
private int userType;

@OneToMany(mappedBy = "user", fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true)
private List<Account> accounts = new ArrayList<Account>();
}

儿童班

@RooJavaBean
@RooToString
@RooJpaActiveRecord
@RooJpaEntity
public class Account {

@OneToMany(mappedBy="account", fetch=FetchType.LAZY, cascade = CascadeType.ALL)
List<Message> messages = new ArrayList<Message>();


/*@OneToMany(mappedBy="account", fetch=FetchType.LAZY, cascade = CascadeType.ALL)
List<PremiumPlayPositionCombination> premiumPlayPosition = new ArrayList<PremiumPlayPositionCombination>();*/


@OneToMany(mappedBy="account", fetch=FetchType.LAZY, cascade = CascadeType.ALL)
List<PositionCombinationArc> allPositionsArc = new ArrayList<PositionCombinationArc>();

@ManyToOne
@JoinColumn(name="user_id")
private Userr user;
}

这是我的删除查询

@Transactional
public static void deleteClientByClientId(Long clientId) {
    System.out.println("Delete query findUsersClientsByUser" + clientId);
    int deleteCount= entityManager().createQuery("DELETE FROM Account where id =:clientId").setParameter("clientId", clientId).executeUpdate();
    System.out.println("Delete query findUsersClientsByUser" + deleteCount);



}

我在ApplicationContext中添加了安全性。像这样的xml

 <?xml version="1.0" encoding="UTF-8" standalone="no"?>
<beans xmlns="http://www.springframework.org/schema/beans"      xmlns:aop="http://www.springframework.org/schema/aop" xmlns:context="http://www.springframework.org/schema/context" xmlns:jee="http://www.springframework.org/schema/jee" xmlns:tx="http://www.springframework.org/schema/tx" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd         http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd         http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd         http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd         http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">
<!--
    This will automatically locate any and all property files you have
    within your classpath, provided they fall under the META-INF/spring
    directory. The located property files are parsed and their values can
    then be used within application context files in the form of
    ${propertyKey}.
-->
<context:property-placeholder location="classpath*:META-INF/spring/*.properties"/>
<!--
    Turn on AspectJ @Configurable support. As a result, any time you
    instantiate an object, Spring will attempt to perform dependency
    injection on that object. This occurs for instantiation via the "new"
    keyword, as well as via reflection. This is possible because AspectJ
    is used to "weave" Roo-based applications at compile time. In effect
    this feature allows dependency injection of any object at all in your
    system, which is a very useful feature (without @Configurable you'd
    only be able to dependency inject objects acquired from Spring or
    subsequently presented to a specific Spring dependency injection
    method). Roo applications use this useful feature in a number of
    areas, such as @PersistenceContext injection into entities.
-->
<context:spring-configured/>
<bean class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close" id="dataSource">
    <property name="driverClassName" value="${database.driverClassName}"/>
    <property name="url" value="${database.url}"/>
    <property name="username" value="${database.username}"/>
    <property name="password" value="${database.password}"/>
</bean>
<bean class="org.springframework.orm.jpa.JpaTransactionManager" id="transactionManager">
    <property name="entityManagerFactory" ref="entityManagerFactory"/>
</bean>
<tx:annotation-driven mode="aspectj" transaction-manager="transactionManager"/>
<bean class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" id="entityManagerFactory">
    <property name="dataSource" ref="dataSource"/>
</bean>
<bean id="mailSender" class="org.springframework.mail.javamail.JavaMailSenderImpl">
    <property name="host" value="smtp.gmail.com"/>
    <property name="port" value="587"/>
    <property name="username" value="[e-mail]"/>
    <property name="password" value="noreply@123"/>
    <property name="javaMailProperties">
        <props>
            <prop key="mail.transport.protocol">smtp</prop>
            <prop key="mail.smtp.auth">true</prop>
            <prop key="mail.smtp.starttls.enable">true</prop>
            <prop key="mail.debug">true</prop>
        </props>
    </property>
</bean>
<!--
    This declaration will cause Spring to locate every @Component,
    @Repository and @Service in your application. In practical terms this
    allows you to write a POJO and then simply annotate the new POJO as an
    @Service and Spring will automatically detect, instantiate and
    dependency inject your service at startup time. Importantly, you can
    then also have your new service injected into any other class that
    requires it simply by declaring a field for your service inside the
    relying class and Spring will inject it. Note that two exclude filters
    are declared. The first ensures that Spring doesn't spend time
    introspecting Roo-specific ITD aspects. The second ensures Roo doesn't
    instantiate your @Controller classes, as these should be instantiated
    by a web tier application context. Refer to web.xml for more details
    about the web tier application context setup services.

    Furthermore, this turns on @Autowired, @PostConstruct etc support. These 
    annotations allow you to use common Spring and Java Enterprise Edition 
    annotations in your classes without needing to do any special configuration. 
    The most commonly used annotation is @Autowired, which instructs Spring to
    dependency inject an object into your class.
-->
<context:component-scan base-package="com.uforic.optionstrader">
    <context:exclude-filter expression=".*_Roo_.*" type="regex"/>
    <!--context:exclude-filter expression="org.springframework.stereotype.Controller" type="annotation"/-->
</context:component-scan>

Answer 1

当尝试使用HiberNate更新/删除时，最好使用一个事务初始化（）和提交（）示例来包围查询：

EntityTransaction tr=em.getTransaction();
tr.begin();

Query query = (Query) em.createQuery( "update etudiant set  email= :em, adresse= :adr,telephone= :tele,password= :pwd"
        + " where email= :mail");
        query.setParameter("em", email)
        .setParameter("adr", adresse)
        .setParameter("tele", tele)
        .setParameter("pwd", pass)
        .setParameter("mail", ancienEmail);
        
int a= query.executeUpdate();
tr.commit();

Answer 2

在Spring上下文文件中，您需要添加以下代码：

<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" 
xmlns:context="http://www.springframework.org/schema/context"
xmlns:tx="http://www.springframework.org/schema/tx"
xmlns:p="http://www.springframework.org/schema/p"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context
http://www.springframework.org/schema/context/spring-context-3.0.xsd 
http://www.springframework.org/schema/tx 
http://www.springframework.org/schema/tx/spring-tx.xsd">

<tx:annotation-driven />

另外，添加您的Spring Securitybean配置。

Answer 3

您可以在代码中检查以下内容：

>

我在代码中看到deleteClientByClientId方法是静态的@spring中的Transactional不支持静态方法。使您的方法成为非静态的，被注释为事务性的。您可以使用静态方法引用@Transactional

让我知道，哪个选项适合你。