基本的 Web 应用程序但出现 404 错误 Tomcat/Eclipse
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Basic web application but getting 404 error Tomcat/Eclipse
提问by Pradeep Simha
I have a simple dynamic web application in eclipse as shown in below image:
我在 Eclipse 中有一个简单的动态 Web 应用程序，如下图所示：
and my web.xml is as below (provided only relavent sections to reduce complexity):
我的 web.xml 如下（仅提供相关部分以降低复杂性）：
<web-app ...
<display-name>DemoRest</display-name>
<servlet>
<servlet-name>JerseyWebService</servlet-name>
<servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>JerseyWebService</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
<welcome-file-list>
<welcome-file>index.html</welcome-file>
</welcome-file-list>
But when I run I am getting "404 error" as shown below:
但是当我运行时，我收到“404 错误”，如下所示：
As you can see my Context-Path matches with my folder and actual files are there in correct location, still I am not able to run the application. I tried of cleaning, rebuilding, restarting of eclipse but still no luck. Can anyone help me, why this happens with tomcat?
正如您所看到的，我的 Context-Path 与我的文件夹匹配，并且实际文件位于正确的位置，但我仍然无法运行该应用程序。我尝试了清理，重建，重新启动 eclipse 但仍然没有运气。谁能帮助我，为什么tomcat会发生这种情况？
回答by Himanshu Bhardwaj
<servlet-mapping>
<servlet-name>JerseyWebService</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
This is not available because the url
这不可用，因为网址
http://localhost:8080/DemoRest/NewFile.jsp
will be intercepted by JerseyWebService, which I guess is not an intended action.
将被 JerseyWebService 拦截，我猜这不是预期的操作。
Why not host rest services on path like:
为什么不在路径上托管休息服务，例如：
<servlet-mapping>
<servlet-name>JerseyWebService</servlet-name>
<url-pattern>/rest/*</url-pattern>
</servlet-mapping>
