为什么后台任务会阻止 SimpleHTTPServer 中的响应？

发布于 2024-09-28 15:49:35 字数 1468 浏览 2 评论 0原文

我正在编写一个简单的基于浏览器的前端，它应该能够启动后台任务，然后从中获取进度。我希望浏览器收到一个响应，说明任务是否成功启动，然后轮询以确定任务何时完成。但是，后台任务的存在似乎会阻止立即发送 XMLHttpRequest 响应，因此我无法报告启动该进程的成功。考虑以下（简化的）代码：

import SocketServer
import SimpleHTTPServer
import multiprocessing
import time

class MyProc(multiprocessing.Process):
    def run(self):
        print 'Starting long process..'
        for i in range(100): time.sleep(1)
        print 'Done long process'

class Page(SimpleHTTPServer.SimpleHTTPRequestHandler):
    def do_GET(self):
        if self.path == '/':
            print >>self.wfile, "<html><body><a href='/run'>Run</a></body></html>"
        if self.path == '/run':
            self.proc = MyProc()
            print 'Starting..'
            self.proc.start()
            print 'After start.'
            print >>self.wfile, "Process started."

httpd = SocketServer.TCPServer(('', 8000), Page)
httpd.serve_forever()

当我运行此代码并浏览到 http://localhost:8000 时，我得到一个名为“运行”的按钮。当我单击它时，终端显示：

Starting..
After start.

但是浏览器视图不会改变..实际上光标在旋转。仅当我在终端中按 Ctrl-C 来中断程序时，浏览器才会更新并显示消息 Process started.

消息 After start 清楚地被打印出来。因此我可以假设 do_GET 在启动进程后返回。然而，直到我中断长时间运行的进程后，浏览器才会得到响应。我必须得出结论，do_GET 和正在发送的响应（位于 SimpleHTTPServer 内部）之间存在某些阻塞。

我也尝试过使用线程和 subprocess.Popen 但遇到了类似的问题。有什么想法吗？

原文

I'm writing a simple browser-based front end that should be able to launch a background task and then get progress from it. I want the browser to receive a response saying whether the task launched successfully, and then poll to determine when it is done. However, the presence of a background task seems to be stopping the XMLHttpRequest response from being sent immediately, so I can't report the success of launching the process. Consider the following (simplified) code:

import SocketServer
import SimpleHTTPServer
import multiprocessing
import time

class MyProc(multiprocessing.Process):
    def run(self):
        print 'Starting long process..'
        for i in range(100): time.sleep(1)
        print 'Done long process'

class Page(SimpleHTTPServer.SimpleHTTPRequestHandler):
    def do_GET(self):
        if self.path == '/':
            print >>self.wfile, "<html><body><a href='/run'>Run</a></body></html>"
        if self.path == '/run':
            self.proc = MyProc()
            print 'Starting..'
            self.proc.start()
            print 'After start.'
            print >>self.wfile, "Process started."

httpd = SocketServer.TCPServer(('', 8000), Page)
httpd.serve_forever()

When I run this, and browse to http://localhost:8000, I get a button named "Run". When I click on it, the terminal displays:

Starting..
After start.

However the browser view does not change.. in fact the cursor is spinning. Only when I press Ctrl-C in the terminal to interrupt the program, then the browser is update with the message Process started.

The message After start is clearly being printed. Therefore I can assume that do_GET is returning after starting the process. Yet, the browser doesn't get a response until after I interrupt the long-running process. I have to conclude there is something blocking between do_GET and the response being sent, which is inside SimpleHTTPServer.

I've also tried this with threads and subprocess.Popen but ran into similar problems. Any ideas?

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那小子欠揍 2024-10-05 15:49:35

除了上面史蒂夫和我的评论之外，这里还有一个适合我的解决方案。

确定内容长度的方法有点难看。如果您不指定，尽管显示了内容，浏览器仍可能显示旋转的光标。关闭 self.wfile 也可以。

from cStringIO import StringIO

class Page(SimpleHTTPServer.SimpleHTTPRequestHandler):
    def do_GET(self):
        out = StringIO()
        self.send_response(200)
        self.send_header("Content-type", "text/html")
        if self.path == '/':
            out.write("<html><body><a href='/run'>Run</a></body></html>\n")
        elif self.path == '/run':
            self.proc = MyProc()
            print 'Starting..'
            self.proc.start()
            print 'After start.'
            out.write("<html><body><h1>Process started</h1></body></html>\n")
        text = out.getvalue()
        self.send_header("Content-Length", str(len(text)))
        self.end_headers()
        self.wfile.write(text)

In addition to Steve's and my comments above, here is a solution that works for me.

The method to determine a content-length is a bit ugly. If you don't specify one, the browser may still show a spinning cursor although the content is shown. Closing the self.wfile instead could also work.

from cStringIO import StringIO

class Page(SimpleHTTPServer.SimpleHTTPRequestHandler):
    def do_GET(self):
        out = StringIO()
        self.send_response(200)
        self.send_header("Content-type", "text/html")
        if self.path == '/':
            out.write("<html><body><a href='/run'>Run</a></body></html>\n")
        elif self.path == '/run':
            self.proc = MyProc()
            print 'Starting..'
            self.proc.start()
            print 'After start.'
            out.write("<html><body><h1>Process started</h1></body></html>\n")
        text = out.getvalue()
        self.send_header("Content-Length", str(len(text)))
        self.end_headers()
        self.wfile.write(text)

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惜醉颜 2024-10-05 15:49:35

我使用此代码片段来运行 SimpleHTTPServer 的线程版本。

例如，我将此文件保存为 ThreadedHTTPServer.py ，然后像这样运行：

$ python -m /path/to/ThreadedHTTPServer PORT

所以它将受到单独威胁线程，现在您可以并行下载并正确导航。

from BaseHTTPServer import HTTPServer, BaseHTTPRequestHandler
from SocketServer import ThreadingMixIn
import threading
import SimpleHTTPServer
import sys

PORT = int(sys.argv[1])

Handler = SimpleHTTPServer.SimpleHTTPRequestHandler

class ThreadedHTTPServer(ThreadingMixIn, HTTPServer):
    """Handle requests in a separate thread."""

if __name__ == '__main__':
    server = ThreadedHTTPServer(('0.0.0.0', PORT), Handler)
    print 'Starting server, use <Ctrl-C> to stop'
    server.serve_forever()

I use this snippet to run Threaded Version of SimpleHTTPServer.

I save this file as ThreadedHTTPServer.py for example and then I run like that:

$ python -m /path/to/ThreadedHTTPServer PORT

So it'll be threated in separated threads and now you can download in paralell and navigate properly.

from BaseHTTPServer import HTTPServer, BaseHTTPRequestHandler
from SocketServer import ThreadingMixIn
import threading
import SimpleHTTPServer
import sys

PORT = int(sys.argv[1])

Handler = SimpleHTTPServer.SimpleHTTPRequestHandler

class ThreadedHTTPServer(ThreadingMixIn, HTTPServer):
    """Handle requests in a separate thread."""

if __name__ == '__main__':
    server = ThreadedHTTPServer(('0.0.0.0', PORT), Handler)
    print 'Starting server, use <Ctrl-C> to stop'
    server.serve_forever()

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