怎样通过类名访问servlet，不用在web.xml配置

Question

以前记得好象可以，现在怎么不行了

Answer 1

那似乎就不是servlet了吧，就是个简单的class而已.

Answer 2

那也必须配置XML,起码把调用器打开,然后在你的XML中要写上<servlet></servlet>里的servlet-name和servlet-class
然后浏览器里地址/WEB应用/serlvet/包.类名

Answer 3

这个我有配啊
<servlet></servlet>里的servlet-name和servlet-class
也不行啊

Answer 4

<servlet>
    <servlet-name>RDepartmentTreeServlet</servlet-name>
    <servlet-class>com.clientRes.bo.RDepartmentTreeServlet</servlet-class>
  </servlet>
<servlet-mapping>
    <servlet-name>RDepartmentTreeServlet</servlet-name>
    <url-pattern>/RDepartmentTreeServlet</url-pattern>
  </servlet-mapping>

Answer 5

<?xml version="1.0" encoding="ISO-8859-1"?>
<web-app xmlns="http://java.sun.com/xml/ns/j2ee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd"
version="2.4">
<servlet>
<servlet-name>HelloWorldServlet</servlet-name>
<servlet-class>HelloWorldServlet</servlet-class>
</servlet>
</web-app>

Answer 6

<servlet>
    <servlet-name>RDepartmentTreeServlet</servlet-name>
    <servlet-class>com.clientRes.bo.RDepartmentTreeServlet</servlet-class>
  </servlet>
<servlet-mapping>
    <servlet-name>RDepartmentTreeServlet</servlet-name>
    <url-pattern>/RDepartmentTreeServlet</url-pattern>
  </servlet-mapping>

Answer 7

我的意思是通过类名这样访问
http://127.0.0.1:7001/web/com.cnmyth.bizaction.test.TestServlet

Answer 8

呵呵，不用跟全路径，只要用<url-pattern>/RDepartmentTreeServlet</url-pattern>的声明的servlet的名字就可以了