如何使用 Python 请求获取 pdf 文件名？
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How to get pdf filename with Python requests?
提问by kramer65
I'm using the Python requests libto get a PDF file from the web. This works fine, but I now also want the original filename. If I go to a PDF file in Firefox and click downloadit already has a filename defined to save the pdf. How do I get this filename?
我正在使用 Python请求库从网络获取 PDF 文件。这工作正常，但我现在也想要原始文件名。如果我在 Firefox 中转到 PDF 文件并单击download它已经定义了一个文件名来保存 pdf。我如何获得这个文件名？
For example:
例如：
import requests
r = requests.get('http://www.researchgate.net/profile/M_Gotic/publication/260197848_Mater_Sci_Eng_B47_%281997%29_33/links/0c9605301e48beda0f000000.pdf')
print r.headers['content-type'] # prints 'application/pdf'
I checked the r.headersfor anything interesting, but there's no filename in there. I was actually hoping for something like r.filename..
我检查了r.headers任何有趣的东西，但那里没有文件名。我实际上希望像r.filename..
Does anybody know how I can get the filename of a downloaded PDF file with requests library?
有人知道如何使用请求库获取下载的 PDF 文件的文件名吗？
采纳答案by user3255354
It is specified in an http header content-disposition. So to extract the name you would do:
它在 http 标头中指定content-disposition。因此，要提取名称，您将执行以下操作：
import re
d = r.headers['content-disposition']
fname = re.findall("filename=(.+)", d)[0]
Name extracted from the string via regular expression (remodule).
通过正则表达式（re模块）从字符串中提取的名称。
回答by Maksim Solovjov
Apparently, for this particular resource it is in:
显然，对于这个特定的资源，它位于：
r.headers['content-disposition']
Don't know if it is always the case, though.
不过不知道是不是一直这样。
回答by Nilpo
Building on some of the other answers, here's how I do it. If there isn't a Content-Dispositionheader, I parse it from the download URL:
基于其他一些答案，这就是我的做法。如果没有Content-Disposition标题，我从下载 URL 解析它：
import re
import requests
from requests.exceptions import RequestException
url = 'http://www.example.com/downloads/sample.pdf'
try:
with requests.get(url) as r:
fname = ''
if "Content-Disposition" in r.headers.keys():
fname = re.findall("filename=(.+)", r.headers["Content-Disposition"])[0]
else:
fname = url.split("/")[-1]
print(fname)
except RequestException as e:
print(e)
There are arguably better ways of parsing the URL string, but for simplicity I didn't want to involve any more libraries.
可以说有更好的方法来解析 URL 字符串，但为了简单起见，我不想涉及更多的库。
回答by myildirim
You can use werkzeugfor options headers https://werkzeug.palletsprojects.com/en/0.15.x/http/#werkzeug.http.parse_options_header
您可以使用werkzeug选项标题https://werkzeug.palletsprojects.com/en/0.15.x/http/#werkzeug.http.parse_options_header
>>> import werkzeug
>>> werkzeug.parse_options_header('text/html; charset=utf8')
('text/html', {'charset': 'utf8'})
