поскольку мое приложение работает на платформе Spring-MVC 3.0, в чем проблема с контроллером ниже, который получает ошибку «Http Status 400»?
package com.maskkk.controller;
import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import org.springframework.stereotype.Controller;
import org.springframework.ui.Model;
import org.springframework.validation.BindingResult;
import org.springframework.web.bind.annotation.ModelAttribute;
import org.springframework.web.bind.annotation.PathVariable;
import org.springframework.web.bind.annotation.RequestHeader;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.multipart.MultipartFile;
@Controller
public class FileDownloadController
{
@RequestMapping(value="/spring/download")
public void downloadPDFResource( HttpServletRequest request,
HttpServletResponse response,
@PathVariable("fileName") String fileName
)
{
//If user is not authorized - he should be thrown out from here itself
//Authorized user will download the file
//String dataDirectory = request.getServletContext().getRealPath("/WEB-
INF/downloads/pdf/");
String dataDirectory =
request.getServletContext().getRealPath("fileupload");
Path file = Paths.get(dataDirectory, fileName);
if (Files.exists(file))
{
response.setContentType("application/pdf");
response.addHeader("Content-Disposition", "attachment;
filename="+fileName);
try
{
Files.copy(file, response.getOutputStream());
response.getOutputStream().flush();
}
catch (IOException ex) {
ex.printStackTrace();
}
}
}
}
А мой web.xml ниже:
<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_3_0.xsd"
version="3.0">
<welcome-file-list>
<welcome-file>dashboard.jsp</welcome-file>
<welcome-file>form.jsp</welcome-file>
<welcome-file>index.htm</welcome-file>
<welcome-file>index.jsp</welcome-file>
</welcome-file-list>
<!-- log4j2-begin -->
<listener>
<listener-
class>org.apache.logging.log4j.web.Log4jServletContextListener</listener-
class>
</listener>
<filter>
<filter-name>log4jServletFilter</filter-name>
<filter-class>org.apache.logging.log4j.web.Log4jServletFilter</filter-
class>
</filter>
<filter-mapping>
<filter-name>log4jServletFilter</filter-name>
<url-pattern>/*</url-pattern>
<dispatcher>REQUEST</dispatcher>
<dispatcher>FORWARD</dispatcher>
<dispatcher>INCLUDE</dispatcher>
<dispatcher>ERROR</dispatcher>
</filter-mapping>
<!-- log4j2-end -->
<!-- 编码处理过滤器 -->
<filter>
<filter-name>encodingFilter</filter-name>
<filter-
class>org.springframework.web.filter.CharacterEncodingFilter</filter-class>
<init-param>
<param-name>encoding</param-name>
<param-value>utf-8</param-value>
</init-param>
<init-param>
<param-name>forceEncoding</param-name>
<param-value>true</param-value>
</init-param>
</filter>
<filter-mapping>
<filter-name>encodingFilter</filter-name>
<url-pattern>*.do</url-pattern>
</filter-mapping>
<servlet>
<servlet-name>spring</servlet-name>
<servlet-
class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
<init-param>
<param-name>contextConfigLocation</param-name>
<param-value>/WEB-INF/spring-servlet.xml</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<!-- 这里的servlet-name和上面的要一致. -->
<servlet-name>spring</servlet-name>
<!-- 这里就是url的匹配规则, / 就是匹配所有 -->
<url-pattern>*.do</url-pattern>
</servlet-mapping>
</web-app>
Данный контроллер предназначен для загрузки файла с веб-сайта. И этот контроллер не активен, когда я набираю "http://localhost:8888/Upload/spring/download.do" на URL и нажимаю ввод.
Так в чем проблема?
Любые предложения будут очень признательны!