Question

(从中文译出)

Hey, I ve got a problem. I have an html file hosted on github pages who looks like this: (It can be accessed at this site)

<head>
  <meta http-equiv="refresh" content="0; url=https://gi-b716.github.io/modrinth-api-tools/api/files/1.2.0.zip">
</head>

Ideally, it should redirect to this 1.2.0.zip after accessing the above html file, and when I accessed it using Google Chrome, it did work that way. However, I wanted to get this file in Python, so I wrote this code:

download_url =  https://gi-b716.github.io/modrinth-api-tools/api/version/d/lasted 
r = requests.get(download_url, allow_redirects=True)
with open("1.2.0.zip", "wb") as f:
    f.write(r.content)
    f.close()

但是,我总是能够拿到档案,尽管请求发现的是转头。我找不到有关返回时档案的任何信息。

What can I do to capture the correct file?

Answer 1

这些类型的转头设计成超文本,而不是基于头盔。你们将不得不使用reg或配对法拉。你们可以这样做:

def requests_redirects():
    download_url =  https://gi-b716.github.io/modrinth-api-tools/api/version/d/lasted 
    r = requests.get(download_url, allow_redirects=True)

    if b content="0; url=  in r.content:
        # regex
        pattern = re.compile(r content="0; url=(.*?)" )
        redirect_url = pattern.findall(r.content.decode())[0]
        
        r = requests.get(redirect_url, allow_redirects=True)
        

    with open("1.2.0.zip", "wb") as f:
        f.write(r.content)
        f.close()

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