Question

目前在我们的项目中，我们正在使用spring框架。由于某些项目要求，我们计划在项目中实现Spring Integration框架。

我要抛出Spring Integration Sample（Spring Integration Rest HTTP Path Usage Demo）应用程序

下面是applicationContext-http-int.xml文件

<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
    http://www.springframework.org/schema/integration http://www.springframework.org/schema/integration/spring-integration.xsd      
    http://www.springframework.org/schema/integration/http http://www.springframework.org/schema/integration/http/spring-integration-http.xsd   
    http://www.springframework.org/schema/oxm http://www.springframework.org/schema/oxm/spring-oxm-3.0.xsd"
xmlns:int="http://www.springframework.org/schema/integration"
xmlns:oxm="http://www.springframework.org/schema/oxm" 
xmlns:int-http="http://www.springframework.org/schema/integration/http">

<int:annotation-config/>

<!-- handler mapping implementation that is aware of inbound Spring Integration 
        http inbound gateway's and inbound adapter's with "path" attributes -->
<bean class="org.springframework.integration.http.inbound.UriPathHandlerMapping"/>

<!-- Inbound/Outbound Channels -->
<int:channel id="employeeSearchRequest" />
<int:channel id="employeeSearchResponse" />


<int-http:inbound-gateway id="inboundEmployeeSearchRequestGateway"      
    supported-methods="GET, POST" 
    request-channel="employeeSearchRequest"
    reply-channel="employeeSearchResponse"      
    mapped-response-headers="Return-Status, Return-Status-Msg, HTTP_RESPONSE_HEADERS"       
    view-name="/employee" 
    path="/services/employee/{id}/search"
    reply-timeout="50000">

    <int-http:header name="employeeId" expression="#pathVariables.id"/>

</int-http:inbound-gateway>


<!-- Note: The default parameter name for favorParameter is 'format'. For instance, when this flag is true, a request for /services/employee/{id}/search?format=json will result 
        in an MappingJacksonJsonView being resolved, while the Accept header can be the browser-defined text/html,application/xhtml+xml  -->

<bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">
    <property name="order" value="1" /> 
    <property name="defaultContentType" value="application/xml"/>
    <property name="favorParameter" value="true"/>  
    <property name="ignoreAcceptHeader" value="true" />     
    <property name="mediaTypes">
        <map>
            <entry key="json" value="application/json" />
            <entry key="xml" value="application/xml" />             
        </map>
    </property>
    <property name="defaultViews">
        <list>
            <bean
                class="org.springframework.integration.samples.rest.json.view.ExtendedMappingJacksonJsonView" >
                <property name="objectMapper" ref="jaxbJacksonObjectMapper"/>
            </bean> 
            <bean class="org.springframework.web.servlet.view.xml.MarshallingView">
                <constructor-arg ref="marshaller"/>                 
            </bean>             
        </list>
    </property>             
</bean>

<oxm:jaxb2-marshaller id="marshaller" contextPath="org.springframework.integration.samples.rest.domain" />

<int:service-activator id="employeeServiceActivator" 
                input-channel="employeeSearchRequest"
                output-channel="employeeSearchResponse" 
                ref="employeeSearchService" 
                method="getEmployee" 
                requires-reply="true"  
                send-timeout="60000"/>

<bean id="jaxbJacksonObjectMapper" class="org.springframework.integration.samples.rest.json.JaxbJacksonObjectMapper"/>

根据我的理解，流程就像输入渠道中有消息一样 employeeSearchService 将被激活。但是根据我们的项目要求，我们需要在运行时基于某些标头值激活服务，例如

如果服务名称= LoginService且方法名称=动作比 Service Activator应激活LoginService并调用action方法。基于网址格式
例如如果我的网址类似于http：// ipaddress：8080 / myapp / LoginService（是ServiceName）.action（方法名称）应该激活LoginService，并且应该调用action方法。

任何建议和帮助都会受到赞赏，因为SI对我来说是新的。

Answer 1

有几种方法可以回答这个问题。第一种是使用简单的标题值

<?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:int="http://www.springframework.org/schema/integration"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/integration http://www.springframework.org/schema/integration/spring-integration-2.2.xsd">

    <!-- input channel where the message starts -->
    <int:channel id="input.channel"/>

    <!-- routes to the different services based on the header value -->
    <int:header-value-router input-channel="input.channel" header-name="serviceName">
        <int:mapping value="a" channel="service.a.channel"/>
        <int:mapping value="b" channel="service.b.channel"/>
    </int:header-value-router>

    <!-- when serviceName header == 'a' -->
    <int:channel id="service.a.channel"/>

    <int:service-activator input-channel="service.a.channel" ref="serviceA"/>

    <!-- when serviceName == 'b' -->
    <int:channel id="service.b.channel"/>

    <int:service-activator input-channel="service.b.channel" ref="serviceB"/>
</beans>

此示例允许您根据可能需要的不同服务和多个选项进行扩展。

（input.channel与employeeSearchRequest相同）

另一个选项使用SpEL并假设只有两个服务

<?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:int="http://www.springframework.org/schema/integration"
    xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
        http://www.springframework.org/schema/integration http://www.springframework.org/schema/integration/spring-integration-2.2.xsd">

    <int:channel id="input.channel"/>

    <int:service-activator input-channel="input.channel"
        expression="headers['serviceName'] == 'a' ? @serviceA.process(payload) : @serviceB.process(payload)"/>

</beans>

如何在spring Integration中动态调用Service Activator

1 个答案: