您好我正在尝试在我的Spring MVC Web应用程序中配置消息源.

我目前使用ReloadableResourceBundleMessageSource运行它,但我无法使用ResourceBundleMessageSource运行它.我宁愿使用ResourceBundleMessageSource,因为我不需要Reload功能,而ResourceBundleMessageSource的效率稍高.

在我的rootApplicationContext中,我已经定义了bean如下.

        <bean id="messageSource"
        class="org.springframework.context.support.ReloadableResourceBundleMessageSource">
        <property name="basename" value="/resources/locale/messages" />
        <property name="defaultEncoding" value="UTF-8"/>
    </bean>

    <bean id="localeChangeInterceptor"
        class="org.springframework.web.servlet.i18n.LocaleChangeInterceptor">
        <property name="paramName" value="lang" />
    </bean>

    <bean id="localeResolver"
        class="org.springframework.web.servlet.i18n.SessionLocaleResolver" />

    <bean id="handlerMapping"
        class="org.springframework.web.servlet.mvc.annotation.DefaultAnnotationHandlerMapping">
        <property name="interceptors">
            <ref bean="localeChangeInterceptor" />
        </property>
    </bean>

这工作很精细..

但是一改变到

<bean id="messageSource"
        class="org.springframework.context.support.ResourceBundleMessageSource">
        <property name="basename" value="/resources/locale/messages" />
        <property name="defaultEncoding" value="UTF-8"/>
</bean>

该应用程序打破了例外:

12:35:57,433 ERROR [org.apache.catalina.core.ContainerBase.[jboss.web].[default-host].[/ SpringJAXWS].[jsp]](http-localhost-127.0.0.1-8080-1 Servlet jsp的Servlet.service()抛出异常:org.apache.tiles.util.TilesIOException:JSPException包括路径'/jsp/views/layout/top.jsp'.at org.apache.tiles.servlet.context.ServletUtil.wrapServletException(ServletUtil.java:241)[tiles-servlet-2.2.2.jar:2.2.2] at org.apache.tiles.jsp.context.JspTilesRequestContext.include (JspTilesRequestContext.java:105)[tiles-jsp-2.2.2.jar:2.2.2]

救命 !

整个项目代码可在GITHUB获得

https://github.com/localghost8080/JaxWS

这是所有感兴趣的人的整个堆栈跟踪.

https://github.com/localghost8080/JaxWS/blob/master/ResourceBundleException.txt

我正在尝试Drools规则引擎,我是一个初学者.

我在单个规则文件中有以下规则:

rule "A stand alone rule" 
salience 2
no-loop
when
    $account : Account()
    Account($account.balance>100)
then
    System.out.println("balance>100");
    System.out.println($account.getBalance());
    System.out.println($account.getCustomer().getName());    
end

rule "A second Rule"
salience 1
no-loop
when
    $account : Account()
    Account($account.balance<100)
then
    System.out.println("balance<100");
    System.out.println($account.getBalance());
    System.out.println($account.getCustomer().getName());
end 

在StatefulKnowledgeSession中,我通过了两个账户,一个账户余额为15000,另一个账户余额为15,

Account account=new Account(7l,15000l);
        Account account1=new Account(5l,15l);

        Customer customer = new Customer("Samrat", 28, "Sector51", account);
        Customer customer1 = new Customer("Alexi", 28, "Sector50", account1);
        account.setCustomer(customer);
        account1.setCustomer(customer1);
        session.insert(account);
        session.insert(account1);

        session.fireAllRules();

根据我的预期结果应该是每个规则应该只被触发一次,并且应该打印相应的对象.

但我得到的结果是:

balance>100
15000
Samrat
balance>100
15000
Samrat
balance<100
15
Alexi
balance<100
15
Alexi

我无法理解为什么每条规则都运行两次????