问题描述
为什么 config.getInitParameter（String）总是在以下代码示例中返回 null ？
Why does config.getInitParameter(String) always return null in the following code example?
public void init(ServletConfig config) throws ServletException
{
super.init(config);
filename = config.getInitParameter("addressfile");
这是web.xml文件
This is web.xml file
<servlet>
<servlet-name>ListManagerServlet</servlet-name>
<servlet-class>savva.listmanagerservlet.ListManagerServlet</servlet-class>
<init-param>
<param-name>addressfile</param-name>
<param-value>d:\temp\demo.txt</param-value>
</init-param>
</servlet>
<servlet-mapping>
<servlet-name>ListManagerServlet</servlet-name>
<url-pattern>/ListManagerServlet</url-pattern>
</servlet-mapping>
UPD：Eclipse EE Indigo，Java 1.6，Tomcat 7.0
UPD: Eclipse EE Indigo, Java 1.6, Tomcat 7.0
推荐答案
规范的方法是使用继承的在无参数方法（以及删除任何 init（config）方法）。
The canonical way is to just use the inherited GenericServlet#getInitParameter() in the argumentless init() method (and remove any init(config) method).
@Override
public void init() throws ServletException {
filename = getInitParameter("addressfile");
}
如果仍然无效，那么 web.xml 未正确部署，或者参数名称中有拼写错误，或者您实际访问的是与 filename 不同的实例变量使用/测试它。
If that still doesn't work, then your web.xml is not properly been deployed, or you have a typo in the parameter name, or you actually accessed a different instance variable than filename to use/test it.
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