我有一个servlet,它基本上是hello world
种......
课程是:
package net.jgp.baseapp.validation.tomcat;
[...]
@WebServlet(name = "tv", description = "...", urlPatterns = {"/TomcatValidator", "/" })
public class TomcatValidator extends HttpServlet {
[...]
}
}
通过以下方式启动它时效果很好:
http://localhost:8081/net.jgp.baseapp.validation.tomcat/tv
http://localhost:8081/net.jgp.baseapp.validation.tomcat/qwe?test=89
等
但是,我可以将net.jgp.baseapp.validation.tomcat
更改为:
http://localhost:8081
一样运行?http://localhost:8081/toto
?到目前为止,我的web.xml实际上是空的(Eclipse生成的默认设置):
<?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"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"
id="WebApp_ID" version="3.0">
<welcome-file-list>
<welcome-file>index.html</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
<welcome-file>default.html</welcome-file>
<welcome-file>default.htm</welcome-file>
<welcome-file>default.jsp</welcome-file>
</welcome-file-list>
我看了很多东西,我必须承认我疯了:)...
谢谢!
PS:我知道:@WebServlet(name = "tv", description = "...", urlPatterns = {"/TomcatValidator", "/" })
在某种程度上等于:
@WebServlet(name = "tv", description = "...", urlPatterns = {"/" })
答案 0 :(得分:1)
这里有两件事。
如果您希望servlet提供根URL,则必须执行两项操作
1 - 您必须将您的webapp上下文设置为root,如此
2 - 然后你必须将你的servlet上下文设置为root
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
@WebServlet("/")
public class MyServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
public MyServlet() {
super();
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.getWriter().write("OK");
response.getWriter().flush();
}
}
那是因为您的servlet提供的URL就像这样
http://yourserver:port/yourapp/yourservlet
其中yourapp由#1
定义和yourservlet由#2
定义 像这样