Question

我使用以下方式启动服务器：

python -m SimpleHTTPServer

在浏览器中，我可以通过转到'localhost'来查看当前目录列表。但是，它按字母顺序列出它们。

我可以在Windows shell样式中使用它，在文件之前列出子目录，但是这两个组是按字母顺序单独排序的吗？

Answer 1

我们需要在SimpleHTTPServer.py文件中进行更改。我在下面的功能中做了一些改动。您可以替换并检查

def list_directory(self, path):
    """Helper to produce a directory listing (absent index.html).

    Return value is either a file object, or None (indicating an
    error).  In either case, the headers are sent, making the
    interface the same as for send_head().

    """
    lst1=[]
    lst2=[]
    try:
        list = os.listdir(path)
        for i in list:
            if os.path.isdir(os.path.join(path,i)):
                lst1.append(i)
            elif os.path.isfile(os.path.join(path,i)):
                lst2.append(i)
    except os.error:
        self.send_error(404, "No permission to list directory")
        return None
    lst1.sort(key=lambda a: a.lower())
    #print lst1
    #print list
    lst2.sort(key=lambda a: a.lower())
    f = StringIO()
    displaypath = cgi.escape(urllib.unquote(self.path))
    f.write('<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">')
    f.write("<html>\n<title>Directory listing for %s</title>\n" % displaypath)
    f.write("<body>\n<h2>Directory listing for %s</h2>\n" % displaypath)
    f.write("<hr>\n<ul>\n")
    for name in lst1:
        fullname = os.path.join(path, name)
        displayname = linkname = name
        # Append / for directories or @ for symbolic links
        if os.path.isdir(fullname):
            displayname = name + "/"
            linkname = name + "/"
        f.write('<li><a href="%s">%s</a>\n'
                % (urllib.quote(linkname), cgi.escape(displayname)))
    for name in lst2:
        fullname = os.path.join(path, name)
        displayname = linkname = name
        # Append / for directories or @ for symbolic links
        if os.path.islink(fullname):
            displayname = name + "@"
            # Note: a link to a directory displays with @ and links with /
        f.write('<li><a href="%s">%s</a>\n'
                % (urllib.quote(linkname), cgi.escape(displayname)))

    f.write("</ul>\n<hr>\n</body>\n</html>\n")
    length = f.tell()
    f.seek(0)
    self.send_response(200)
    self.send_header("Content-type", "text/html")
    self.send_header("Content-Length", str(length))
    self.end_headers()
    return f

在顶部创建python SimpleHTTPServer列表子目录

1 个答案: