小编fat*_*zzy的帖子

我正在尝试创建自定义AccessDecisionVoter,并在调用它时调试它.

我已经在每种方法中都提出了一个问题,但没有发生任何事情.

弹簧security.xml文件:

<bean id="accessDecisionManager" class="org.springframework.security.access.vote.UnanimousBased">
  <property name="decisionVoters">
     <list>
          <bean class="com.affiliates.server.security.voters.VoterTest">
              <property name="brandsApi"  ref="brandsApi"/>
          </bean>
        </list>
  </property>

IBrandsApi.java

    public interface IBrandsApi {

    IHibernateBean getByPK(Integer id);

    @Secured({ "ROLE_BRAND_ADMIN" })    
    IHibernateBean update(IHibernateBean brand);

    @Secured({ "ROLE_BRAND_ADMIN" })    
    IHibernateBean insert(IHibernateBean brand);

    @Secured({ "ROLE_BRAND_ADMIN" })    
    ResultContainer getAll(IFilter filter);

    @Secured({ "ROLE_ADMIN" })  
    Integer delete(IFilter filter); 
}

VoterTest.java(带有断点的空文件)

    public class VoterTest implements AccessDecisionVoter {
private IBrandsApi brandsApi;

    public IBrandsApi getBrandsApi() {
        return brandsApi;
    }

    public void setBrandsApi(IBrandsApi brandsApi) {
        this.brandsApi = brandsApi;
    }

        @Override
        public boolean supports(ConfigAttribute …

我试图在java中设置一个计划任务,一天运行一次.
问题是它只在第一天运行.
有什么想法吗？
谢谢

log.info("Schdualing midnight task");
    Timer timer = new Timer();
    Calendar date = Calendar.getInstance();

    date.set(Calendar.HOUR_OF_DAY, 23);
    date.set(Calendar.MINUTE, 30);
    date.set(Calendar.SECOND, 0);

    timer.schedule(new EndOfDayRatesTimerTask(new MidnightQuotesEvent()),
            date.getTime());

我有以下UserDetailsS​​ervice实现.
到目前为止,身份验证过程非常有效.
如何在"会话"中存储我的"MyUser bean"(已成功登录),以便我可以在我的应用程序的其他区域中访问它

谢谢.

@Transactional(readOnly = true)
public class CustomUserDetailsService implements UserDetailsService {


    private EmployeesApi employeesApi = new EmployeesApi();

    /**
     * Retrieves a user record containing the user's credentials and access. 
     */
    public UserDetails loadUserByUsername(String userName)
            throws UsernameNotFoundException, DataAccessException {

        // Declare a null Spring User
        UserDetails user = null;

        try {


            MyUser employee = employeesApi.getByUserName(userName);



            user =  new User(
                    employee.getUserName(), 
                    employee.getPassword().toLowerCase(),
                    true,
                    true,
                    true,
                    true,
                    getAuthorities(1) );

        } catch (Exception e) {
            logger.error("Error in retrieving user"); …

我构建了一个Async任务服务执行器,它通过外部请求执行任务.

每个任务都包含函数void,run(),因此任何想要向系统添加任务的程序员都需要继承From BaseTask.

    interface ITask{
       void run();
    }
    abstract BaseTask : ITask{
       //force "run()" to set Result
       public ResultContainer Result {set; get;}
       void run();
    }

    class SomeTask : BaseTask {
       void run(){
           ////run the operation here, in the end, set the result.
           //force the programmer to set the Result;
           this.Result = new ResultContainer("task ok");
       }
    }

由于内部原因,run()必须无效.

有没有什么办法,我可以强迫谁想要添加一个任务来调用一个程序员Result中BaseTask,并将其值？ 你认为这是一种不好的做法吗？

谢谢

我在将数据转换为json时遇到问题.

Session session = sessionFactory.openSession();       
Affiliate affiliate = (affiliate) session.get( Affiliate , pk );
session.close();
JSONArray.fromObject(affiliate);

调试器显示该行已被提取.
但我尝试转换为json字符串时得到此异常:

Exception in thread "main" net.sf.json.JSONException: org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.affiliates.hibernate.Affiliate.employees, no session or session was closed
at net.sf.json.JSONObject._fromBean(JSONObject.java:959)    ...

这是我的会员实体

    @Entity(name="AFFILIATE")
    @Inheritance(strategy=InheritanceType.SINGLE_TABLE)
    public class Affiliate extends HibernateBean{ 

        @Id
        @GeneratedValue(strategy=GenerationType.IDENTITY) 
        @Column(name="AFFILIATE_ID")
        private long id;

        @ManyToOne(targetEntity = Affiliate.class)
        @JoinColumn(name="PARENT_ID")
        private Affiliate parent;



        @ManyToMany(cascade=CascadeType.ALL,fetch=FetchType.LAZY)
        @JoinTable(name="EMPLOYEES_AFFILIATES"  , joinColumns = {@JoinColumn(name="AFFILIATE_ID")},inverseJoinColumns={@JoinColumn(name="EMPLOYEE_ID")})
        private Set<Employee> employees = new HashSet<Employee>(0);

 getters and setters...

    }

谢谢

我正在使用Spring Security + MVC.
注释@Secured({ "ROLE_ADMIN" })仅在控制器层中正常工作.
如果我尝试在更深层/其他层中使用它,我就不会遇到安全性错误.
或者,如果我尝试在"无mvc映射"方法上使用它,我没有得到任何安全错误.
关注我的xml配置文件:
web.xml:

  <?xml version="1.0" encoding="UTF-8"?>
<web-app id="WebApp_ID" version="2.4"
    xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">
    <filter>
        <filter-name>springSecurityFilterChain</filter-name>
        <filter-class>org.springframework.web.filter.DelegatingFilterProxy</filter-class>
    </filter>
    <filter-mapping>
        <filter-name>springSecurityFilterChain</filter-name>
        <url-pattern>/*</url-pattern>
    </filter-mapping>
    <context-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>
        /WEB-INF/spring-security.xml
        /WEB-INF/applicationContext.xml
        </param-value>
    </context-param>
    <context-param>
        <param-name>log4jConfigLocation</param-name>
        <param-value>/WEB-INF/classes/log4j-myapp.properties</param-value>
    </context-param>
    <servlet>
        <servlet-name>spring</servlet-name>
        <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
        <load-on-startup>1</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>spring</servlet-name>
        <url-pattern>/Management/*</url-pattern>
    </servlet-mapping>
    <listener>
        <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
    </listener>
</web-app>

为spring-servlet.xml

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p"
    xmlns:context="http://www.springframework.org/schema/context"
    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">

    <!-- Declare a view resolver -->
    <bean id="viewResolver"
    class="org.springframework.web.servlet.view.InternalResourceViewResolver"
    p:prefix="/WEB-INF/pages/" p:suffix=".jsp" /> …

我的spring-security.xml有两个问题

1. 当我有多个角色时,access="ROLE_ADMIN,ROLE_EMPLOYEE"
   我得到了例外: Caused by: java.lang.IllegalArgumentException: Failed to parse expression 'ROLE_ADMIN,ROLE_EMPLOYEE'
   但如果我有一个角色:access="ROLE_ADMIN"它会正常工作

2. 如果我直接登陆/Management/main/admin我不会被规则重定向:security:form-login login-page="/Management/auth/login/",这意味着我可以在没有角色管理员的情况下进入应用程序.

这是我的spring-security.xml

<?xml version="1.0" encoding="UTF-8"?> 
<beans xmlns="http://www.springframework.org/schema/beans"
       xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
       xmlns:lang="http://www.springframework.org/schema/lang"
       xmlns:context="http://www.springframework.org/schema/context"
       xmlns:tx="http://www.springframework.org/schema/tx"
       xmlns:sec="http://www.springframework.org/schema/security" 
       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-3.0.xsd
        http://www.springframework.org/schema/security 
        http://www.springframework.org/schema/security/spring-security-3.0.xsd">

    <sec:global-method-security secured-annotations="enabled"  jsr250-annotations="enabled" />
    <sec:http auto-config="true" use-expressions="true"
        access-denied-page="/Management/auth/denied">

        <sec:intercept-url pattern="/Management/auth/login" filters="none" access="permitAll"/>
        <sec:intercept-url pattern="/Management/main/admin" filters="none" access="ROLE_ADMIN,ROLE_EMPLOYEE" />
        <sec:intercept-url pattern="/Management/api/affiliates/**" filters="none" access="ROLE_ADMIN,ROLE_EMPLOYEE" />

        <sec:form-login  login-page="/Management/auth/login/"
             authentication-success-handler-ref="loginAuthenticationSuccessHandler"
            authentication-failure-url="/Management/auth/login?error=true"
            login-processing-url="/Management/auth/j_spring_security_check"
            default-target-url="/Management/auth/login?error=false" />
        <sec:logout invalidate-session="true"
            logout-success-url="/Management/auth/login/" logout-url="/Management/auth/logout" />
    </sec:http>

    <sec:authentication-manager>
        <sec:authentication-provider
            user-service-ref="customUserDetailsService">
            <sec:password-encoder ref="passwordEncoder" …

虽然很有趣,但大多数优秀的MySQL编辑器都是基于Windows的.我正在为Ubuntu寻找一个工具(400美元以上),它可以:

• 自动填写表格和字段名称+保留关键字.
• 语法着色.
• 内联行内容编辑.
• 将表/数据库从一个主机复制到另一个主机.

我认为最好的基于Windows的是SQLyog,但这是一个关于Wine的噩梦.

我有一个网格,我需要/必须(无论如何)向它发送多级json.
例:

{ "root" : [ { 
        "affiliateId" : 8,
        "name" : "Affiliate Name",
        "email" : "affiliate@gmail.co.il",
        "manager" : { 
            "name"  : "I am the manager",
            "email" : "manager@gmail.co.il"
          },
      } ],
  "totalCount" : 1
}

现在,当我构建网格字段时,我想使用更深层的项目,例如:

{
        name:'manager_email',
        header: "Manager",
        dataIndex: 'manager.email',/******access a deep level******/
        width: 100,
        sortable: true,
        type:'text'
    }

我没有错误,只是网格中的空单元格.

谢谢

我有一个拥有很多汽车的用户实体。
我无法使用“ IN”语句按汽车列表获取用户。

List<Car> cars = getCarsList();
String hql = "From User WHERE user.cars in(:cars)";
Query query = session.createQuery(hql)
query.setParameterList("cars",cars);
//query.setParameterList("cars",cars.toArray());//not working also
//query.setParameter("cars","1,4,8,30");//not working also
query.setMaxResults(1);

这样，该参数将忽略汽车。
我也尝试发送vars.toArray（）;
而且我还尝试发送String 1,4,8,300（如果是汽车，则为ID）。

我有多个品牌的应用程序。
单个用户可以拥有多个品牌，几乎应用程序中的每个查询都通过其品牌来限制用户。

用户有 ManyToMany 和 Brand(s) 。

例如，我想List<Employee>从数据库中提取。

s.createQuery("FROM User user INNER JOIN user.brands brands WHERE brands IN(1,2)");//(1,2) = :brands

在这种情况下，我将不止一次获得同一个用户，因为该用户拥有多个与他相关的品牌。

如何在不需要不同的情况下查询品牌而不会影响我的结果？

这是用户实体：

    public class User  {
        @Id
        @Column(name="USER_ID")
        @GeneratedValue(strategy=GenerationType.IDENTITY)
        private Integer userId;

        @ManyToMany(fetch=FetchType.EAGER)
        @JoinTable(name="USERS_BRANDS" , joinColumns = {@JoinColumn(name="USER_ID")},inverseJoinColumns={@JoinColumn(name="BRAND_ID")})
        private Collection<Brand> brands;
        ...other members
        ...getters and setters
}

谢谢

有没有人设法在drupal 7上安装CCK？我收到很多错误.
我想要的原因是因为CCK不为每个字段创建2个表,它每个内容类型只创建一个表,这样更有效.

谢谢