Is there a built-in Spring environment variable fo

问题:

I'm using Spring Web Services to expose a service as a web service. In my spring configuration xml file, I have a bean which is an instance of DefaultWsdl11Definition. One of the properties that needs to be set is the locationUri. This needs to be a fully qualified Uri, but I don't want to have to change this value when the application gets promoted from dev to uat and to production. Spring knows what the web application context root is, so is there a variable that I can specify in my configuration file to access it?

Something like:

<bean id="myWebServices"
    class="org.springframework.ws.wsdl.wsdl11.DefaultWsdl11Definition">
    <property name="schemaCollection">
        <bean
            class="org.springframework.xml.xsd.commons.CommonsXsdSchemaCollection">
            <property name="xsds" ref="xsdList"/>
            <property name="inline" value="true" />
        </bean>
    </property>
    <property name="portTypeName" value="myWebServices" />
    <property name="locationUri" value="${webContextRoot}/webServices" />
</bean>

回答1:

With Spring 3.0, you should be able to access the servlet context through the servletContext bean in the web application context:

<property name="locationUri" value="#{servletContext.contextPath}/webServices" />

If you're using pre-Spring-EL (before 3.0), you should be able to do

<bean name="servletContextBean" class="org.springframework.web.context.support.ServletContextFactoryBean" />
<bean name="contextPath" factory-bean="servletContextBean" factory-method="getContextPath" />
<bean name="locationUri" factory-bean="contextPath" factory-method="concat">
   <constructor-arg value="/webServices" />
</bean>

and inside your myWebservices bean

<property name="locationUri" ref="locationUri" />

EDIT:

I don't think getting the server name and port from the ServletContext, as depending on the setup the web container may not know the hostname (i.e. a HTTP server may be in front of the web container, e.g. tomcat may be behind an Apache web server or depending on the Websphere configuration).

However, the following may be part of a solution to get the hostname. With Spring 3.0, you could do the following:

<property name="host" 
          class="java.net.InetAddress" 
          factory-method="getLocalHost"/>

<property name="locationUri" 
          value="http://#{host.canonicalHostName}:8080/#{servletContext.contextPath}/webServices" />

回答2:

I had a similar problem that you described, I use property files to do this

• ws_dev.properties
• ws_prod.properties

I configured my property file like this, The deployment property is java vm argument like

-Ddeployment=dev

<context:property-placeholder location="ws_${deployment}.properties"/>

回答3:

May be late, but may some other need a solution too:

set property in servlet:

web.xml

<servlet>
<servlet-name>spring-ws</servlet-name>
    <servlet-class>org.springframework.ws.transport.http.MessageDispatcherServlet</servlet-class>
    <init-param>
        <param-name>contextConfigLocation</param-name>
        <param-value>classpath:/spring-ws-context.xml</param-value>
    </init-param>
    <init-param>
        <param-name>transformWsdlLocations</param-name>
        <param-value>true</param-value>
    </init-param>
</servlet>

The bean declaration in spring-ws-context.xml:

<bean id="WebService"
    class="org.springframework.ws.wsdl.wsdl11.DefaultWsdl11Definition"
    p:portTypeName="App" p:locationUri="/WebServices" p:requestSuffix="Request"
    p:responseSuffix="Response">
    <property name="schema">
        <bean class="org.springframework.xml.xsd.SimpleXsdSchema" p:xsd="classpath:/requestTypes.xsd" />
    </property>
</bean>

回答4:

You can add ApplicationContextAware interface to your bean, cast it to WebApplicationContext and then get ServletContext. Also see class org.springframework.web.context.ContextLoader