java - 以非阻塞方式处理文件上传

Question

后台线程是here

为了明确目标 - 用户将上传一个大文件，并且必须立即重定向到另一个页面以进行不同的操作。但文件很大，从 Controller 的InputStream读取需要时间。所以我不情愿地决定fork一个新的Thread来处理这个I/O。代码如下:

Controller servlet

/**
     * @see HttpServlet#doPost(HttpServletRequest request, HttpServletResponse
     *      response)
     */
    protected void doPost(HttpServletRequest request,
            HttpServletResponse response) throws ServletException, IOException {
        // TODO Auto-generated method stub

        System.out.println("In Controller.doPost(...)");

        TempModel tempModel = new TempModel();
        tempModel.uploadSegYFile(request, response);

        System.out.println("Forwarding to Accepted.jsp");

        /*try {
            Thread.sleep(1000 * 60);
        } catch (InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }*/

        request.getRequestDispatcher("/jsp/Accepted.jsp").forward(request,
                response);
    }

模型类

package com.model;

import java.io.IOException;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.Future;

import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

import com.utils.ProcessUtils;

public class TempModel {

    public void uploadSegYFile(HttpServletRequest request,
            HttpServletResponse response) {
        // TODO Auto-generated method stub

        System.out.println("In TempModel.uploadSegYFile(...)");

        /*
         * Trigger the upload/processing code in a thread, return immediately
         * and notify when the thread completes
         */
        try {
            FileUploaderRunnable fileUploadRunnable = new FileUploaderRunnable(
                    request.getInputStream());

            /*
             * Future<FileUploaderRunnable> future = ProcessUtils.submitTask(
             * fileUploadRunnable, fileUploadRunnable);
             * 
             * FileUploaderRunnable processed = future.get();
             * 
             * System.out.println("Is file uploaded : " +
             * processed.isFileUploaded());
             */

            Thread uploadThread = new Thread(fileUploadRunnable);
            uploadThread.start();

        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } /*
         * catch (InterruptedException e) { // TODO Auto-generated catch block
         * e.printStackTrace(); } catch (ExecutionException e) { // TODO
         * Auto-generated catch block e.printStackTrace(); }
         */

        System.out.println("Returning from TempModel.uploadSegYFile(...)");
    }

}

可运行

package com.model;

import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.nio.ByteBuffer;
import java.nio.channels.Channels;
import java.nio.channels.ReadableByteChannel;

public class FileUploaderRunnable implements Runnable {

    private boolean isFileUploaded = false;
    private InputStream inputStream = null;

    public FileUploaderRunnable(InputStream inputStream) {
        // TODO Auto-generated constructor stub
        this.inputStream = inputStream;
    }

    public void run() {
        // TODO Auto-generated method stub

        /* Read from InputStream. If success, set isFileUploaded = true */
        System.out.println("Starting upload in a thread");

        File outputFile = new File("D:/06c01_output.seg");/*
                                                         * This will be changed
                                                         * later
                                                         */
        FileOutputStream fos;
        ReadableByteChannel readable = Channels.newChannel(inputStream);
        ByteBuffer buffer = ByteBuffer.allocate(1000000);

        try {

            fos = new FileOutputStream(outputFile);

            while (readable.read(buffer) != -1) {
                fos.write(buffer.array());
                buffer.clear();
            }

            fos.flush();
            fos.close();

            readable.close();
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        System.out.println("File upload thread completed");
    }

    public boolean isFileUploaded() {
        return isFileUploaded;
    }

}

我的疑问/疑问:

1. 从 Servlet 手动生成线程在逻辑上对我来说是有意义的，但在编码方面却让我害怕 - 容器毕竟不知道这些线程(我认为是这样!)

2. 当前代码给出了一个非常明显的异常 - 由于 doPost(...) 方法在 run() 方法完成之前返回，因此无法访问流:

   In Controller.doPost(...)
   In TempModel.uploadSegYFile(...)
   Returning from TempModel.uploadSegYFile(...)
   Forwarding to Accepted.jsp
   Starting upload in a thread
   Exception in thread "Thread-4" java.lang.NullPointerException
       at org.apache.coyote.http11.InternalInputBuffer.fill(InternalInputBuffer.java:512)
       at org.apache.coyote.http11.InternalInputBuffer.fill(InternalInputBuffer.java:497)
       at org.apache.coyote.http11.InternalInputBuffer$InputStreamInputBuffer.doRead(InternalInputBuffer.java:559)
       at org.apache.coyote.http11.AbstractInputBuffer.doRead(AbstractInputBuffer.java:324)
       at org.apache.coyote.Request.doRead(Request.java:422)
       at org.apache.catalina.connector.InputBuffer.realReadBytes(InputBuffer.java:287)
       at org.apache.tomcat.util.buf.ByteChunk.substract(ByteChunk.java:407)
       at org.apache.catalina.connector.InputBuffer.read(InputBuffer.java:310)
       at org.apache.catalina.connector.CoyoteInputStream.read(CoyoteInputStream.java:202)
       at java.nio.channels.Channels$ReadableByteChannelImpl.read(Unknown Source)
       at com.model.FileUploaderRunnable.run(FileUploaderRunnable.java:39)
       at java.lang.Thread.run(Unknown Source)
   

3. 记住第一点，使用 Executor 框架对我有帮助吗？

   package com.utils;
   
   import java.util.concurrent.Future;
   import java.util.concurrent.ScheduledThreadPoolExecutor;
   
   public final class ProcessUtils {
   
       /* Ensure that no more than 2 uploads,processing req. are allowed */
       private static final ScheduledThreadPoolExecutor threadPoolExec = new ScheduledThreadPoolExecutor(
               2);
   
       public static <T> Future<T> submitTask(Runnable task, T result) {
   
           return threadPoolExec.submit(task, result);
       }
   }
   

那么我应该如何确保用户不会阻塞并且流仍然可访问，以便可以从中读取(上传的)文件？

Answer 1

实际上并非如此。您正在尝试生成线程并读取 POST 请求的内容，并且您还尝试将用户转发到具有相同请求对象的另一个页面。这会让 servlet 容器感到困惑。

你可以

• 使用单独的框架来上传表单和 Controller
• 使用带有上传表单和单独 Controller 的弹出窗口
• 使用 AJAX 就地加载下一页，同时仍然上传当前页面中的内容(让浏览器为您处理)