后台主题是here
只是为了明确目标 - 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。但是文件很大，需要时间从控制器的 InputStream 中读取。所以我不情愿地决定派一个新线程来处理这个I / O.代码如下：
控制器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(...)");
}
}
Runnable
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/怀疑：
当前代码给出了一个非常明显的异常 - 在run（）方法完成之前doPost（...）方法返回时，流不可访问：
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)
请记住第1点，Executor框架的使用是否对我有帮助？
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止并且流仍然可访问以便可以从中读取（上传的）文件？
答案 0 :(得分：2)
实际上它并没有。你试图产生线程并读取POST请求的内容，并且你还试图将用户转发到具有相同请求对象的另一个页面。这会混淆servlet容器。
你可以