我想问一下,我这个无包的servlet怎么总是在tomcat上出不来,总是提示404错误 我的classpath是这样的 C:\jdk1.5.0_16\jre\lib\rt.jar;.;C:\Tomcat 5.5\common\lib\servlet-api.jar 同时我也将servlet-api.jar文件放在了C:\jdk1.5.0_16\jre\lib\ext下 javac可以编译成功,也把class文件放在了C:\Tomcat 5.5\webapps\ROOT\WEB-INF\classes下,但是是上了tomcat就不行了,先谢谢各位了。。
<?xml version="1.0" encoding="ISO-8859-1"?>
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distributed under the License is distributed on an "AS IS" BASIS,
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limitations under the License.
--><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"> <display-name>Welcome to Tomcat</display-name>
<description>
Welcome to Tomcat
</description><!-- JSPC servlet mappings start --> <servlet>
<servlet-name>org.apache.jsp.index_jsp</servlet-name>
<servlet-class>org.apache.jsp.index_jsp</servlet-class>
</servlet> <servlet-mapping>
<servlet-name>org.apache.jsp.index_jsp</servlet-name>
<url-pattern>/index.jsp</url-pattern>
</servlet-mapping><!-- JSPC servlet mappings end --></web-app>
<servlet>
<servlet-name>你的servlet名字</servlet-name>
<servlet-class>你的servlet class的package路径</servlet-class>
</servlet> <servlet-mapping>
<servlet-name>对应上面的servlet名字</servlet-name>
<url-pattern>你自己定义的url</url-pattern>
</servlet-mapping>
<servlet-name>HelloServlet</servlet-name>
<servlet-class>C:\Tomcat 5.5\webapps\ROOT\WEB-INF\classes</servlet-class>
</servlet> <servlet-mapping>
<servlet-name>HelloServlet</servlet-name>
<url-pattern>http://localhost:8080/servlet</url-pattern>
</servlet-mapping>这样加?钢材这样加了以后,服务器启动不了了
这里指的是package路径喔
import java.io.*;
import javax.servlet.*;
import javax.servlet.http.*;
那就不应该放到ROOT\WEB-INF\classes
应该放到webapps/下你自己建立的工程的demo\WEB-INF\classes下
理解吧
没有包名 那就直接写你的类名唉