【发布时间】:2010-02-12 23:59:32
【问题描述】:
我希望我的 Jython servlet 实现 HttpServlet.contextInitialized 方法,但我不确定如何在 web.xml 中表达这一点。我目前拥有的是:
from javax.servlet import ServletContextListener;
from javax.servlet.http import HttpServlet
class JythonServlet1 ( HttpServlet, ServletContextListener ):
def contextInitialized( self, event ):
print "contextInitialized"
context = event.getServletContext()
def contextDestroyed( self, event ):
print "contextDestroyed"
context = event.getServletContext()
def doGet( self, request, response ):
print "doGet"
def doPost( self, request, response ):
print "doPost"
我的 web.xml 看起来像这样:
<?xml version="1.0" encoding="UTF-8"?>
<web-app
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
id="WebApp_ID" version="2.5">
<display-name>JythonTest</display-name>
<servlet>
<servlet-name>PyServlet</servlet-name>
<servlet-class>org.python.util.PyServlet</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>PyServlet</servlet-name>
<url-pattern>*.py</url-pattern>
</servlet-mapping>
<servlet>
<description></description>
<display-name>JythonServlet1</display-name>
<servlet-name>JythonServlet1</servlet-name>
<servlet-class>JythonServlet1</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
</web-app>
如您所见,在最后一个<servlet> 条目中,我想使用上下文(我可以在其中启动调度程序)初始化servlet,但它似乎与Java servlet 的工作方式不同。
【问题讨论】: