我当前从 EntityManager 查询中收到连接超时错误。可以为这些设置超时吗?
持久性.xml
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">
<persistence-unit name="CallPU" transaction-type="RESOURCE_LOCAL">
<provider>org.eclipse.persistence.jpa.PersistenceProvider</provider>
<class>call.structure.Task</class>
<class>call.structure.Installation</class>
<class>call.structure.Contents</class>
<class>call.structure.Recipient</class>
<class>call.structure.CallTask</class>
<class>call.structure.SmsTask</class>
<class>call.structure.EmailTask</class>
<class>call.security.User</class>
<class>call.structure.content.Content</class>
<class>call.structure.content.RecordContent</class>
<class>call.structure.content.WaitContent</class>
<class>call.structure.content.TextContent</class>
<class>call.structure.content.VariableContent</class>
<class>call.structure.content.SoundContent</class>
<class>call.structure.content.SubjectContent</class>
<class>call.structure.content.FileContent</class>
<class>call.structure.Bounce</class>
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:oracle:thin:@127.0.0.1:1521:TEST"/>
<property name="javax.persistence.jdbc.password" value="userpassword"/>
<property name="javax.persistence.jdbc.driver" value="oracle.jdbc.OracleDriver"/>
<property name="javax.persistence.jdbc.user" value="username"/>
</properties>
</persistence-unit>
</persistence>
我的线程的 run 函数中代码超时:
private class TaskDB extends Thread {
private final long WAITING_TIME = 20000L;
@Override
public void run() {
Set<SmsTask> remove = SMSManager.this.getRemoveTask();
Set<SmsTask> normal = SMSManager.this.getNormalTask();
try {
while(true){
EntityManager em = DB.getEM(); //Calls EntityManagerFactory.createEntityManager()
em.getTransaction().begin();
Set<SmsTask> normalClone = new HashSet<SmsTask>(normal);
// Abort task in futur.
List<SmsTask> taskToRemove = new ArrayList<SmsTask>();
if (!remove.isEmpty()) {
String queryString = "SELECT t FROM SmsTask t WHERE t.id IN :remove ";
if (!normalClone.isEmpty())
queryString += "AND t.id NOT IN :normal ";
Query query = em.createQuery(queryString);
query.setParameter("remove", Utils.taskToIdList(remove));
if (!normalClone.isEmpty())
query.setParameter("normal", Utils.taskToIdList(normalClone));
taskToRemove = (List<SmsTask>) query.getResultList();
for (SmsTask task : taskToRemove) {
removedTask.add(task);
remove.remove(task);
}
}
String queryString = "SELECT t FROM SmsTask t WHERE (t.scheduleTime IS NULL OR t.scheduleTime < :dateNow) AND t.status = co.dium.call.structure.Task.StatusTask.NOT_START ";
if (!taskToRemove.isEmpty())
queryString += "AND t.id NOT IN :toRemove ";
Query query = em.createQuery(queryString);
query.setParameter("dateNow", Utils.obtainUniversalTime());
if (!taskToRemove.isEmpty())
query.setParameter("toRemove", Utils.taskToIdList(taskToRemove));
List<SmsTask> taskResults = (List<SmsTask>) query.getResultList();
em.getTransaction().commit();
for (SmsTask task : taskResults)
addTask(task);
SMSManager.TaskRemove.sleep(WAITING_TIME);
}
} catch (InterruptedException ex) {
Logger.getLogger(SMSManager.class.getName()).log(Level.SEVERE, null, ex);
}
System.out.println("Thread interrompu !");
Thread.currentThread().interrupt();
}
}
我得到的超时错误:
org.clipse.persistence.exceptions.DatabaseException
Internal Exception: java.sql.SQLRecoverableException: I/O error: Socket read timed out
Error Code: 17002
Call: [....sql query...]
[...]
at org.eclipse.persistence.internal.EJBQueryImpl.getResultList(EJBQueryImpl.java:742)
at call.manager.sms.SMSManager$TaskDB.run(SMSManager.java:367)
Caused by: java.sql.SQLRecoverableException: I/O Error: Scoket read timed out
[...]
是的,有javax.persistence.query.timeout。根据 JPA 2.0 规范,对此查询提示的支持是可选的:
便携式应用程序不应依赖此提示。取决于 正在使用的持久性提供程序和数据库,提示可能会也可能不会 观察到。
可以将所有查询的默认值(以毫秒为单位)设置为 persistence.xml:
<property name="javax.persistence.query.timeout" value="1000"/>
通过 Persistence 创建 EntityManagerFactory 时也可以给出相同的属性。createEntityManagerFactory。
它也可以根据查询被覆盖/设置:
query.setHint("javax.persistence.query.timeout", 2000);
可以通过 NamedQuery 中的属性 hints 获得相同的功能。
有两种方法可以使用 Hibernate 设置查询超时时间。
如果您本地引导 Hibernate 或者将 JPA
java.persistence.Query
解包为其 org.hibernate.query.Query
等效项,那么您只需使用 setTimeout
方法即可:
List<Post> posts = entityManager.createQuery("""
select p
from Post p
where lower(p.title) like lower(:titlePattern)
""", Post.class)
.setParameter("titlePattern", "%Hibernate%")
.unwrap(org.hibernate.query.Query.class)
.setTimeout(1)
.getResultList();
请注意,
方法采用setTimeout
参数,该参数指定以秒为单位的超时值。int
您还可以使用 JPA 查询提示,如以下示例所示:
List<Post> posts = entityManager.createQuery("""
select p
from Post p
where lower(p.title) like lower(:titlePattern)
""", Post.class)
.setParameter("titlePattern", "%Hibernate%")
.setHint("javax.persistence.query.timeout", 50)
.getResultList();
查询提示采用以毫秒为单位的超时值。javax.persistence.query.timeout
您还可以使用
org.hibernate.timeout
查询提示:
List<Post> posts = entityManager.createQuery("""
select p
from Post p
where lower(p.title) like lower(:titlePattern)
""", Post.class)
.setParameter("titlePattern", "%Hibernate%")
.setHint("org.hibernate.timeout", 1)
.getResultList();
查询提示采用以秒为单位的超时值。org.hibernate.timeout
您应该使用“
javax.persistence.query.timeout
”,而不是使用查询提示“javax.persistence.lock.timeout
”。
在后一种情况下,如果查询未在指定毫秒内返回,您将收到
CannotAcquireLockException
。在第一种情况下,如果事务没有在指定的毫秒内完成,您将收到 QueryTimeoutException
并且事务将被回滚。