小编kma*_*oor的帖子

在网上快速搜索显示3或4个变体人们如何指定xmlns和xsi:schemaLocation进入persistence.xml.

指定JPA 2.1版的"正确"方式是什么？

我正在使用:

<persistence version="2.1"
             xmlns="http://xmlns.jcp.org/xml/ns/persistence"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence
             http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">

我遇到了一个简单的问题; 奋力如何调用order by上join编辑实体.基本上我正在努力实现以下目标JPA Criteria:

select distinct d from Department d 
left join fetch d.children c 
left join fetch c.appointments a
where d.parent is null 
order by d.name, c.name

我有以下内容:

CriteriaBuilder cb = getEntityManager().getCriteriaBuilder();
CriteriaQuery<Department> c = cb.createQuery(Department.class);
Root<Department> root = c.from(Department.class);
Fetch<Department, Department> childrenFetch = root.fetch(
    Department_.children, JoinType.LEFT);
childrenFetch.fetch(Department_.appointments, JoinType.LEFT);

c.orderBy(cb.asc(root.get(Department_.name)));
c.distinct(true);
c.select(root);
c.where(cb.isNull(root.get(Department_.parent)));

如何实现order by d.name, c.name用Criteria API？我尝试使用Expression,Path但是没有用.任何指针将不胜感激.

我正在尝试为Android项目添加布局:

右键单击res文件夹New | Directory.该目录已创建,但未显示在app/res.也许是因为默认情况下隐藏了空目录？转到Windows资源管理器并将文件添加到新创建的目录(隐藏在Android Studio中).回到工作室,如何"刷新"项目结构以显示新创建的目录及其下的文件？

基本上我的问题是,Eclipse的刷新菜单选项的替代方案是什么？

我正在尝试更改以下HQL以使用JPA Criteria:

select distinct d from Department d 
left join fetch d.children c 
where d.parent is null 
and (
    d.name like :term 
    or c.name like :term
    ) 
order by d.name

Department有一个Set<Department>孩子.

标准:

CriteriaBuilder cb = getEntityManager().getCriteriaBuilder();
CriteriaQuery<Department> c = cb.createQuery(Department.class);
Root<Department> root = c.from(Department.class);
root.fetch("children", JoinType.LEFT);
Path<Department> children = root.join("children", JoinType.LEFT);
c.orderBy(cb.asc(root.get("name")));
c.distinct(true);
c.where(cb.isNull(root.get("parent")));
String param = "%" + "term" + "%";
cb.and(cb.like(root.<String> get("name"), param));
cb.or(cb.like(children.<String> get("name"), param));

TypedQuery<Department> tq = getEntityManager().createQuery(c);
departments = tq.getResultList();

我知道它可能有点简洁,但是,HQL返回24和Criteria版本28.我想我不处理:

and …

对stop或undeploy/redeploy一的Spring framework 3.0.5基于Web的应用程序中记录以下错误Tomcat7's catalina.out:

SEVERE: The web application [/nomination##1.0-qa] created a ThreadLocal with key of type [java.lang.ThreadLocal] (value [java.lang.ThreadLocal@4f43af8f]) and a value of type [org.springframework.security.core.context.SecurityContextImpl] (value [org.springframework.security.core.context.SecurityContextImpl@ffffffff: Null authentication]) but failed to remove it when the web application was stopped. Threads are going to be renewed over time to try and avoid a probable memory leak.

我最初想到实现ServletContextListener和close()那里的背景.但是,发现ContextLoaderListener哪个工具ServletContextListener设置如下web.xml:

<listener>
<listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
</listener>

来自Javadocs:

**contextDestroyed** …

我有六次自定义注销过滤器.我尝试访问应用程序时两次,当我输入用户名/密码并单击"登录"时再两次,然后当我点击"注销"时再次两次.

我究竟做错了什么？

组态:

<http auto-config="true" use-expressions="true">
    <intercept-url pattern="/admin/**" access="hasRole('ROLE_ADMIN_FUNCTIONS')" />      
    <intercept-url pattern="/**" access="hasRole('ROLE_USER')" />

    <form-login login-page="/login"
        authentication-success-handler-ref="customAuthenticationSuccessHandlerBean"
        authentication-failure-handler-ref="customAuthenticationFailureHandlerBean" />
    <logout invalidate-session="true" success-handler-ref="logoutHandlerBean" />
    <session-management session-fixation-protection="migrateSession">
        <concurrency-control max-sessions="1"
            expired-url="/login_sessionexpired" />
    </session-management>

    <custom-filter before="LOGOUT_FILTER" ref="customLogoutFilter" />
</http>

<beans:bean id="customLogoutFilter" class="com.hurontg.libms.security.CustomLogoutFilter" />

过滤器:

public class CustomLogoutFilter extends OncePerRequestFilter {
/**
 * 
 */
private XLogger logger = XLoggerFactory
        .getXLogger(CustomLogoutFilter.class.getName());

@Override
protected void doFilterInternal(HttpServletRequest req,
        HttpServletResponse res, FilterChain chain)
        throws ServletException, IOException {

    logger.error("========================================================================================");
    logger.error("$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Custom Logout Filter $$$$$$$$$$$$$$$$$$$$$$$$$$$$$");
    logger.error("========================================================================================");

    chain.doFilter(req, res);
}

}

Spring版本:4.1.1 …

Hibernate允许使用来通过@ManyToOne映射添加索引@org.hibernate.annotations.Index。

有没有一种方法可以为@ManyToMany关系中的连接表指定索引？

如果Entity A和Entity 的拥有方B为@ManyToManywith A，则联接表将具有复合索引（a1，b1）。

我想知道这是否足够，还是需要创建另一个索引（b1，a1）？

关于停止/重启web-app Tomcat 7.0.26,我有一个非常简单的'Hello world'类型的Web应用程序(Spring 3.2.1,Hibernate 4.1.9)

The following web applications were stopped (reloaded, undeployed), but their
classes from previous runs are still loaded in memory, thus causing a memory
leak (use a profiler to confirm):
/myapp

我采取了以下步骤:启动JVisualVM右键单击Tomcat并选择"堆转储"单击[heapdump]上的"OQL控制台"然后执行此查询:

select x from org.apache.catalina.loader.WebappClassLoader x

找到4个实例:

org.apache.catalina.loader.WebappClassLoader

选中的"已启动"字段为"false"右键单击"this"引用并单击"Show Nearest GC Root"将显示"未找到GC根"的对话框.

我错过了什么？任何帮助将不胜感激.谢谢.

从Hibernate 3.6文档中：

您可以使用带有关键字的HQL提供额外的加入条件。

from Cat as cat
left join cat.kittens as kitten
    with kitten.bodyWeight > 10.0

该with子句允许对JOIN条件（ON子句）添加限制。JPQL中有这样的事情吗？

当我运行以下JPQL时：

select c from ContainerDef c left join fetch c.displayState ds where c.id = 1 and ds.user.id = 2

生成以下SQL：

select
        ...
    from
        CONTAINER_DEF containerd0_ 
    left outer join
        USER_CONTAINERDEF displaysta1_ 
            on containerd0_.CONTAINERDEF_ID=displaysta1_.CONTAINERDEF_ID 
    where
        containerd0_.CONTAINERDEF_ID=? 
        and displaysta1_.AUTHUSER_ID=?

真正应该生成的是：

select
        ...
    from
        CONTAINER_DEF containerd0_ 
    left outer join
        USER_CONTAINERDEF displaysta1_ 
            on containerd0_.CONTAINERDEF_ID=displaysta1_.CONTAINERDEF_ID 
            and displaysta1_.AUTHUSER_ID=?
    where
        containerd0_.CONTAINERDEF_ID=? 

我确定我缺少适用于HQL的JPQL子句with。

HTML:

<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>TEsT</title>    
</head>
<body>
<div id="wizard">
    <h1>First Step</h1>
    <div>First Content</div>

    <h1>Second Step</h1>
    <div>Second Content</div>
</div>

    <script src="jquery-1.11.0.min.js"></script>
<script src="jquery.steps.min.js"></script>
<script src="homepage.js"></script>
    </body>
</html>

jQuery的:

$(document).ready(function(){
$("#wizard").steps();
});

(非特别)演绎: