如何在 Python 3 中实现 urllib.urlretrieve 的等价物？答案

【问题标题】：How to implement the equivalent of urllib.urlretrieve in Python 3?如何在 Python 3 中实现 urllib.urlretrieve 的等价物？
【发布时间】：2017-05-19 14:09:06
【问题描述】：

我想使用不同的用户代理从 http://www.apkmirror.com 下载 APK 文件。以下适用于 Python 2：

import urllib

class ApkURLopener(urllib.FancyURLopener):
    version = 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.81 Safari/537.36'

urllib._urlopener = ApkURLopener()

download_link = 'https://www.apkmirror.com/wp-content/themes/APKMirror/download.php?id=215041'
download_file = '/tmp/apkmirror_test/youtube.apk'

if __name__ == "__main__":
    urllib.urlretrieve(url=download_link, filename=download_file)

在不使用urllib.request.urlretrieve 的情况下，我很难找到在 Python 3 中执行相同操作的代码，这在未来可能会被弃用。到目前为止，我想出了

#!/usr/bin/python3
import urllib.request

download_link = 'https://www.apkmirror.com/wp-content/themes/APKMirror/download.php?id=215041'
download_file = '/tmp/apkmirror_test/youtube.apk'

USER_AGENT = 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.81 Safari/537.36'

request = urllib.request.Request(url=download_link, headers={'User-Agent': USER_AGENT})

if __name__ == "__main__":
    response = urllib.request.urlopen(url=request)

我怀疑我应该使用response 对象的write 方法，但我不确定如何使用，或者是否确实是这样做的方法。有什么建议吗？

【问题讨论】：

标签： python urllib

【解决方案1】：

基于Alternative of urllib.urlretrieve in Python 3.5，我完成了Python 3脚本如下：

#!/usr/bin/python3
import urllib.request
import contextlib

download_link = 'https://www.apkmirror.com/wp-content/themes/APKMirror/download.php?id=215041'
download_file = '/tmp/apkmirror_test/youtube2.apk'

USER_AGENT = 'Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/58.0.3029.81 Safari/537.36'

request = urllib.request.Request(url=download_link, headers={'User-Agent': USER_AGENT})

if __name__ == "__main__":
    response = urllib.request.urlopen(url=request)

    with open(download_file, 'wb') as out_file:
        with contextlib.closing(response) as fp:
            block_size = 1024 * 8
            while True:
                block = fp.read(block_size)
                if not block:
                    break
                out_file.write(block)

【讨论】：

【解决方案2】：

这有点老了，但是下面的工作吗？

import urllib.request as rq, os
from urllib.parse import urlparse


url = "your_url"
result = urllib.request.urlopen(url)
result_url = result.url
result_url_parse = urlparse(result_url)
result_path = result_url_parse.path
filename = os.path.basename(result_path)
download = rq.urlretrieve(url,filename)

【讨论】：

【解决方案3】：

您可以使用shutil.copyfileobj() 神奇地从 url 字节流复制到文件。

import urllib.request
import shutil
    
url = "https://www.apkmirror.com/wp-content/themes/APKMirror/download.php?id=215041"    
output_file = "/tmp/apkmirror_test/youtube.apk"
user_agent = 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36'

with urllib.request.urlopen(url, headers={'User-Agent': user_agent}) as response, open(output_file, 'wb') as out_file:
  shutil.copyfileobj(response, out_file)

来源：https://stackoverflow.com/a/48691447/1174102

【讨论】：