Jetty - Servlet无法找到 [英] Jetty - Servlet cannot be found
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问题描述
HTTP错误：404
访问/ ProjectServlet时遇到问题。原因：
NOT_FOUND
我认为我正确配置了所有文件，无法解释Jetty为什么返回此错误。感谢您帮助我！
索引.html
<！DOCTYPE html>
< html>
< head>
< meta charset =ISO-8859-1>
< title> TEST< / title>
< / head>
< body>
< h1> Servlet呼叫测试：< / h1>
< form action =/ ProjectServletmethod =GET>
< input type =textname =namemaxlength =20value =Nameonfocus =this.select（）/>< br>
< input type =submitname =callservletvalue =Call Servlet。/>
< / form>
< / body>
< / html>
ProjectServlet.java
包servlets;
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;
/ **
* Servlet实现类ProjectServlet
* /
@WebServlet（description =使用项目列表返回JSON响应的Servlet，urlPatterns = { / ProjectServlet}）
public class ProjectServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/ **
*默认构造函数。
* /
public ProjectServlet（）{
super（）;
}
/ **
* @see HttpServlet＃doGet（HttpServletRequest请求，HttpServletResponse响应）
* /
protected void doGet（HttpServletRequest request， HttpServletResponse响应）throws ServletException，IOException {
//调用Servlet按钮
if（request.getParameter（callservlet）！= null）{
响应。的setContentType（ 应用程序/ JSON）;
String name = request.getParameter（name）;
String jsonexample =hi+ name;
response.getWriter（）。write（jsonexample）;
}
}
/ **
* @see HttpServlet＃doPost（HttpServletRequest请求，HttpServletResponse响应）
* /
protected void doPost（HttpServletRequest请求，HttpServletResponse响应）抛出ServletException，IOException {
// TODO自动生成的方法存根
}
}
web.xml
<？xml version =1.0encoding =UTF-8？>
< web-app xmlns：xsi =http://www.w3.org/2001/XMLSchema-instancexmlns =http://java.sun.com/xml/ns/javaeexsi ：schemaLocation =http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsdid =WebApp_IDversion =3.0 >
< display-name> Test< / display-name>
< 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>
< servlet>
< servlet-name> ProjectServlet< / servlet-name>
< servlet-class> servlets.ProjectServlet< / servlet-class>
< / servlet>
< servlet-mapping>
< servlet-name> ProjectServlet< / servlet-name>
< url-pattern> / ProjectServlet< / url-pattern>
< / servlet-mapping>
< / web-app>
解决方案
这应该如下所示添加
< form action =$ {pageContext.servletContext.contextPath} / ProjectServlet
method =GET>
而不是
code>< form action =/ ProjectServletmethod =GET>
I am new to Servlets and wanted to call a simple Servlet by using Eclipse's Jetty plugin. I can call the index.html, but when trying to access the Servlet I get:
HTTP ERROR: 404
Problem accessing /ProjectServlet. Reason:
NOT_FOUND
I think I configured all my files correctly and cannot explain why Jetty returns this error.
Thanks for helping me!
index.html
<!DOCTYPE html>
<html>
<head>
<meta charset="ISO-8859-1">
<title>TEST</title>
</head>
<body>
<h1>Servlet Call Test:</h1>
<form action="/ProjectServlet" method="GET">
<input type="text" name="name" maxlength="20" value="Name" onfocus="this.select()"/><br>
<input type="submit" name="callservlet" value="Call Servlet."/>
</form>
</body>
</html>
ProjectServlet.java
package servlets;
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;
/**
* Servlet implementation class ProjectServlet
*/
@WebServlet(description = "Servlet to return JSON response with project list", urlPatterns = { "/ProjectServlet" })
public class ProjectServlet extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* Default constructor.
*/
public ProjectServlet() {
super();
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
//Call Servlet Button
if (request.getParameter("callservlet") != null) {
response.setContentType("application/json");
String name = request.getParameter("name");
String jsonexample = "hi " + name;
response.getWriter().write(jsonexample);
}
}
/**
* @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse response)
*/
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
// TODO Auto-generated method stub
}
}
web.xml
<?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">
<display-name>Test</display-name>
<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>
<servlet>
<servlet-name>ProjectServlet</servlet-name>
<servlet-class>servlets.ProjectServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>ProjectServlet</servlet-name>
<url-pattern>/ProjectServlet</url-pattern>
</servlet-mapping>
</web-app>
解决方案
This should be as shown below to add the context path before Servlet url.
<form action="${pageContext.servletContext.contextPath}/ProjectServlet"
method="GET">
instead of
<form action="/ProjectServlet" method="GET">
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