是否有从文件名中提取扩展名的功能?


当前回答

# try this, it works for anything, any length of extension
# e.g www.google.com/downloads/file1.gz.rs -> .gz.rs

import os.path

class LinkChecker:

    @staticmethod
    def get_link_extension(link: str)->str:
        if link is None or link == "":
            return ""
        else:
            paths = os.path.splitext(link)
            ext = paths[1]
            new_link = paths[0]
            if ext != "":
                return LinkChecker.get_link_extension(new_link) + ext
            else:
                return ""

其他回答

filename='ext.tar.gz'
extension = filename[filename.rfind('.'):]

此方法需要字典、列表或集合。您可以使用内置字符串方法使用“.endswitch”。这将在文件末尾的列表中搜索名称,只需使用str.endswith(fileName[index])即可完成。这更多用于获取和比较扩展名。

https://docs.python.org/3/library/stdtypes.html#string-方法

示例1:

dictonary = {0:".tar.gz", 1:".txt", 2:".exe", 3:".js", 4:".java", 5:".python", 6:".ruby",7:".c", 8:".bash", 9:".ps1", 10:".html", 11:".html5", 12:".css", 13:".json", 14:".abc"} 
for x in dictonary.values():
    str = "file" + x
    str.endswith(x, str.index("."), len(str))

示例2:

set1 = {".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"}
for x in set1:
   str = "file" + x
   str.endswith(x, str.index("."), len(str))

示例3:

fileName = [".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"];
for x in range(0, len(fileName)):
    str = "file" + fileName[x]
    str.endswith(fileName[x], str.index("."), len(str))

示例4

fileName = [".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"];
str = "file.txt"
str.endswith(fileName[1], str.index("."), len(str))

具有输出的示例5、6、7

示例8

fileName = [".tar.gz", ".txt", ".exe", ".js", ".java", ".python", ".ruby", ".c", ".bash", ".ps1", ".html", ".html5", ".css", ".json", ".abc"];
exts = []
str = "file.txt"
for x in range(0, len(x)):
    if str.endswith(fileName[1]) == 1:
         exts += [x]
     

对于简单的用例,一个选项可能是从点拆分:

>>> filename = "example.jpeg"
>>> filename.split(".")[-1]
'jpeg'

文件没有扩展名时没有错误:

>>> "filename".split(".")[-1]
'filename'

但你必须小心:

>>> "png".split(".")[-1]
'png'    # But file doesn't have an extension

也不会在Unix系统中处理隐藏文件:

>>> ".bashrc".split(".")[-1]
'bashrc'    # But this is not an extension

对于一般用途,首选os.path.splitext

只需加入所有pathlib后缀。

>>> x = 'file/path/archive.tar.gz'
>>> y = 'file/path/text.txt'
>>> ''.join(pathlib.Path(x).suffixes)
'.tar.gz'
>>> ''.join(pathlib.Path(y).suffixes)
'.txt'

试试看:

files = ['file.jpeg','file.tar.gz','file.png','file.foo.bar','file.etc']
pen_ext = ['foo', 'tar', 'bar', 'etc']

for file in files: #1
    if (file.split(".")[-2] in pen_ext): #2
        ext =  file.split(".")[-2]+"."+file.split(".")[-1]#3
    else:
        ext = file.split(".")[-1] #4
    print (ext) #5

获取列表中的所有文件名拆分文件名并检查倒数第二个扩展名,它是否在penext列表中?如果是,则使用最后一个扩展名连接它,并将其设置为文件的扩展名如果没有,则只将最后一个扩展名作为文件的扩展名然后检查一下