多请求 pycurl 永远运行（无限循环）

Question

我想使用 Pycurl 执行多请求。代码是： m.add_handle（句柄） requests.append((句柄，响应))

    # Perform multi-request.
    SELECT_TIMEOUT = 1.0
    num_handles = len(requests)
    while num_handles:
        ret = m.select(SELECT_TIMEOUT)
        if ret == -1: continue
        while 1:
            ret, num_handles = m.perform()
            print "In while loop of multicurl"
            if ret != pycurl.E_CALL_MULTI_PERFORM: break

问题是，这个循环需要永远运行。它没有终止。 谁能告诉我，它的作用是什么以及可能出现的问题是什么？

Answer 1

你看过 PyCurl 官方代码了吗？下面的代码实现了多种东西，我尝试执行它，我能够在 300 秒内并行抓取大约 10,000 个 url。我认为这正是您想要实现的目标？请纠正我，如果我错了。

#! /usr/bin/env python
# -*- coding: iso-8859-1 -*-
# vi:ts=4:et
# $Id: retriever-multi.py,v 1.29 2005/07/28 11:04:13 mfx Exp $

#
# Usage: python retriever-multi.py <file with URLs to fetch> [<# of
#          concurrent connections>]
#

import sys
import pycurl

# We should ignore SIGPIPE when using pycurl.NOSIGNAL - see
# the libcurl tutorial for more info.
try:
    import signal
    from signal import SIGPIPE, SIG_IGN
    signal.signal(signal.SIGPIPE, signal.SIG_IGN)
except ImportError:
    pass


# Get args
num_conn = 10
try:
    if sys.argv[1] == "-":
        urls = sys.stdin.readlines()
    else:
        urls = open(sys.argv[1]).readlines()
    if len(sys.argv) >= 3:
        num_conn = int(sys.argv[2])
except:
    print "Usage: %s <file with URLs to fetch> [<# of concurrent connections>]" % sys.argv[0]
    raise SystemExit


# Make a queue with (url, filename) tuples
queue = []
for url in urls:
    url = url.strip()
    if not url or url[0] == "#":
        continue
    filename = "doc_%03d.dat" % (len(queue) + 1)
    queue.append((url, filename))


# Check args
assert queue, "no URLs given"
num_urls = len(queue)
num_conn = min(num_conn, num_urls)
assert 1 <= num_conn <= 10000, "invalid number of concurrent connections"
print "PycURL %s (compiled against 0x%x)" % (pycurl.version, pycurl.COMPILE_LIBCURL_VERSION_NUM)
print "----- Getting", num_urls, "URLs using", num_conn, "connections -----"


# Pre-allocate a list of curl objects
m = pycurl.CurlMulti()
m.handles = []
for i in range(num_conn):
    c = pycurl.Curl()
    c.fp = None
    c.setopt(pycurl.FOLLOWLOCATION, 1)
    c.setopt(pycurl.MAXREDIRS, 5)
    c.setopt(pycurl.CONNECTTIMEOUT, 30)
    c.setopt(pycurl.TIMEOUT, 300)
    c.setopt(pycurl.NOSIGNAL, 1)
    m.handles.append(c)


# Main loop
freelist = m.handles[:]
num_processed = 0
while num_processed < num_urls:
    # If there is an url to process and a free curl object, add to multi stack
    while queue and freelist:
        url, filename = queue.pop(0)
        c = freelist.pop()
        c.fp = open(filename, "wb")
        c.setopt(pycurl.URL, url)
        c.setopt(pycurl.WRITEDATA, c.fp)
        m.add_handle(c)
        # store some info
        c.filename = filename
        c.url = url
    # Run the internal curl state machine for the multi stack
    while 1:
        ret, num_handles = m.perform()
        if ret != pycurl.E_CALL_MULTI_PERFORM:
            break
    # Check for curl objects which have terminated, and add them to the freelist
    while 1:
        num_q, ok_list, err_list = m.info_read()
        for c in ok_list:
            c.fp.close()
            c.fp = None
            m.remove_handle(c)
            print "Success:", c.filename, c.url, c.getinfo(pycurl.EFFECTIVE_URL)
            freelist.append(c)
        for c, errno, errmsg in err_list:
            c.fp.close()
            c.fp = None
            m.remove_handle(c)
            print "Failed: ", c.filename, c.url, errno, errmsg
            freelist.append(c)
        num_processed = num_processed + len(ok_list) + len(err_list)
        if num_q == 0:
            break
    # Currently no more I/O is pending, could do something in the meantime
    # (display a progress bar, etc.).
    # We just call select() to sleep until some more data is available.
    m.select(1.0)


# Cleanup
for c in m.handles:
    if c.fp is not None:
        c.fp.close()
        c.fp = None
    c.close()
m.close()

Answer 2

我认为这是因为您只跳出第一个 while 循环

# Perform multi-request.
SELECT_TIMEOUT = 1.0
num_handles = len(requests)
while num_handles:                           #  while nr.1
    ret = m.select(SELECT_TIMEOUT)
    if ret == -1: continue
    while 1:                                 #  while nr.2
        ret, num_handles = m.perform()
        print "In while loop of multicurl"
        if ret != pycurl.E_CALL_MULTI_PERFORM: break
    '**'

所以如果你使用'break'会发生什么，你将跳出当前的while循环（当你使用break时你处于第二个while循环中。） 程序的下一步将在此处输入写为 '**' 的行，因为它是它跳回的最后一行。 （到 while num_handles 的第一行） 然后再走 3 行，它会运行到 'while 1:' 等等......这就是你获得 inf 循环的方式。

所以要解决这个问题：

# Perform multi-request.
SELECT_TIMEOUT = 1.0
num_handles    = len(requests)
while num_handles:                           #  while nr.1
    ret = m.select(SELECT_TIMEOUT)
    if ret == -1: continue
    while 1:                                 #  while nr.2
        ret, num_handles = m.perform()
        print "In while loop of multicurl"
        if ret != pycurl.E_CALL_MULTI_PERFORM: 
            break
    break

所以这里发生的事情是，一旦它从嵌套的 while 循环中中断，它也会自动从第一个循环中中断。 （否则它永远不会到达这条线，因为while，以及之前使用的continue