Implementación de un servlet de descarga de archivos simple

Solución:

Suponiendo que tenga acceso al servlet como se muestra a continuación

http://localhost:8080/myapp/download?id=7

Necesito crear un servlet y registrarlo en web.xml

web.xml

<servlet>
     <servlet-name>DownloadServlet</servlet-name>
     <servlet-class>com.myapp.servlet.DownloadServlet</servlet-class>
</servlet>
<servlet-mapping>
     <servlet-name>DownloadServlet</servlet-name>
     <url-pattern>/download</url-pattern>
</servlet-mapping>

DownloadServlet.java

public class DownloadServlet extends HttpServlet {


    protected void doGet( HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

         String id = request.getParameter("id");

         String fileName = "";
         String fileType = "";
         // Find this file id in database to get file name, and file type

         // You must tell the browser the file type you are going to send
         // for example application/pdf, text/plain, text/html, image/jpg
         response.setContentType(fileType);

         // Make sure to show the download dialog
         response.setHeader("Content-disposition","attachment; filename=yourcustomfilename.pdf");

         // Assume file name is retrieved from database
         // For example D:\file\test.pdf

         File my_file = new File(fileName);

         // This should send the file to browser
         OutputStream out = response.getOutputStream();
         FileInputStream in = new FileInputStream(my_file);
         byte[] buffer = new byte[4096];
         int length;
         while ((length = in.read(buffer)) > 0){
            out.write(buffer, 0, length);
         }
         in.close();
         out.flush();
    }
}

Eso depende. Si dicho archivo está disponible públicamente a través de su servidor HTTP o contenedor de servlets, simplemente puede redirigirlo a través de response.sendRedirect().

Si no es así, deberá copiarlo manualmente en el flujo de salida de respuesta:

OutputStream out = response.getOutputStream();
FileInputStream in = new FileInputStream(my_file);
byte[] buffer = new byte[4096];
int length;
while ((length = in.read(buffer)) > 0){
    out.write(buffer, 0, length);
}
in.close();
out.flush();

Deberá manejar las excepciones apropiadas, por supuesto.

Prueba con Resource

File file = new File("Foo.txt");
try (PrintStream ps = new PrintStream(file)) {
   ps.println("Bar");
}
response.setContentType("application/octet-stream");
response.setContentLength((int) file.length());
response.setHeader( "Content-Disposition",
         String.format("attachment; filename="%s"", file.getName()));

OutputStream out = response.getOutputStream();
try (FileInputStream in = new FileInputStream(file)) {
    byte[] buffer = new byte[4096];
    int length;
    while ((length = in.read(buffer)) > 0) {
        out.write(buffer, 0, length);
    }
}
out.flush();

---

---

Utiliza Nuestro Buscador