如何设置 Servlet Path 使 HTML 正确调用 servlet 文件？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/13109898/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to set up the Servlet Path make HTML correctly invoke servlet file?
提问by Michael Lai
My goal: to access a .htm file, and pass the user input to the invoked servlet and display the content.
我的目标：访问 .htm 文件，并将用户输入传递给调用的 servlet 并显示内容。
What i did: I used eclipse Juno to create a dynamic project:ServeletTest. The structure of the project is as followed:
我做了什么：我使用 Eclipse Juno 创建了一个动态项目：ServeletTest。该项目的结构如下：
The servlet file is MyServlet.java and the related code is:
servlet 文件为 MyServlet.java，相关代码为：
package ylai.Servlet.test;
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;
import java.io.PrintWriter;
/**
* Servlet implementation class MyServlet
*/
@WebServlet(description = "test servlet", urlPatterns = { "/MyServlet" })
public class MyServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public MyServlet() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
String info = request.getParameter("info") ;
PrintWriter out = response.getWriter() ;
out.println("<html>") ;
out.println("<head><title>Hello Servlet</title></head>") ;
out.println("<body>") ;
out.println("<h1>" + info + "</h1>") ;
out.println("</body>") ;
out.println("</html>") ;
out.close() ;
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
this.doGet(request, response);
}
}
The html file is input.htm . And the detail code is:
html 文件是 input.htm 。详细代码是：
<html>
<head><title>This is html file</title></head>
<body>
<form action="myservlet" method="post">
Type something：<input type="text" name="info">
<input type="submit" value="submit">
</form>
</body>
</html>
And the web.xml is defined as :
web.xml 定义为：
<web-app xmlns="http://java.sun.com/xml/ns/javaee"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd"
version="2.5">
<display-name>Welcome to Tomcat</display-name>
<description>
Welcome to Tomcat
</description>
<servlet>
<servlet-name>myservlet</servlet-name>
<servlet-class>ylai.Servlet.test.MyServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/myservlet</url-pattern>
</servlet-mapping>
</web-app>
When i run the input.htm using Built-in Tomcat within the Eclipse, it works fine, and the input content in the input.htm can be displayed by MyServlet.java. The screen shot is as followed:
当我在Eclipse中使用内置Tomcat运行input.htm时，它工作正常，并且可以通过MyServlet.java显示input.htm中的输入内容。屏幕截图如下：
It seems works fine.
它似乎工作正常。
My Question:
我的问题：
If i want to modify the value of in the web.xml as
如果我想将 web.xml 中的值修改为
<servlet-mapping>
<servlet-name>myservlet</servlet-name>
<url-pattern>/myservletURL</url-pattern>
</servlet-mapping>
What i expected was once the input.htm is submitted, it will invoked the serlvet and the web page address should be:
我期望的是一旦input.htm被提交，它就会调用serlvet，网页地址应该是：
http://localhost:8080/ServeletTest/myservletURL
But the display page address is still, does not change :
但是显示页面地址依旧，没有变化：
http://localhost:8080/ServeletTest/myservletwith HTTP Status 404 error.
http://localhost:8080/ServeletTest/myservlet带有 HTTP 状态 404 错误。
It looks weird!!! The mechanism should be: When i submit the input.htm page, it will invoke servlet by servlet-name in the web.xml. In this case, servlet-name is myservlet. Tomcat will use servlet-name to find the actual location of servlet file: MyServlet.java and execute it. The redirect page address would be depends on what you define in . In this case, it should /ServeletTest/myservletURL But right now. Servlet file can not be invoked and the page address is not what i expect.
看起来很奇怪！！！机制应该是：当我提交 input.htm 页面时，它会通过 web.xml 中的 servlet-name 调用 servlet。在这种情况下，servlet-name 是 myservlet。Tomcat 将使用 servlet-name 查找 servlet 文件的实际位置：MyServlet.java 并执行它。重定向页面地址将取决于您在 . 在这种情况下，它应该是 /ServeletTest/myservletURL 但是现在。无法调用 Servlet 文件，页面地址不是我所期望的。
Do i have wrong understand on the servlet invoke mechanism or others?
我对 servlet 调用机制或其他机制有错误理解吗？
回答by user553180
If you changed the url-patternto myservletURL, you will also need to update the form actionto target this new url.
如果您将url-pattern更改为 myservletURL，您还需要更新表单操作以定位这个新的 url。
回答by sunleo
LifeCycleServlet--@WebServlet("/LifeCycleServlet")
MyServlet--@WebServlet(description = "test servlet", urlPatterns = { "/MyServlet" })
delete these lines because here you have mentioned url as MyServlet
删除这些行，因为在这里你提到了 url 作为 MyServlet
or
或者
Change this urlpattern { "/MyServlet" } also
还要更改此 urlpattern { "/MyServlet" }
