如何在python中找到文件名

Question

我有个问题。我对此有点困惑。我有来自其中的图像和视频的 URL。我想从该链接中获取名称并检查它是图像还是视频并对其执行操作。喜欢if filename.endwith(.jpg or .png):   print("it's an image")else filename.endwith(.mp4, .avi):   print("it's a video) 文件名是一个列表。所有数据都存储在其中：        ls = ["C:/Users/OB/Desktop/DevoMech Project/6.JPG" , "C:/Users/OB/Desktop/DevoMech Project/7.JPG"]    if ls.endwith(.JPG or .JPEG):        item1 = QtWidgets.QListWidgetItem(QIcon(ls), "item1")        self.dlistWidget.addItem(item1)        #item2 = QtWidgets.QListWidgetItem(QIcon("C:/Users/OB/Desktop/DevoMech Project/7.JPEG"), "item2")        self.dlistWidget.addItem(item2)而不是第一项，而是显示实际名称。

Answer 1

如果可能的话，我总是更喜欢使用 os 内置的路径并捕获不受支持的类型

import os

video_types = ('.mp4', '.avi', '.jpeg')

image_types = ('.png', '.jpg')

filenames = ["/test/1.jpg","/test/2.avi","/test/unknown.xml","/test/noextention"]

for filename in filenames:

    print(filename)

    if os.path.splitext(filename)[1] in video_types:

        print("Its a Video")

    elif os.path.splitext(filename)[1] in image_types:

        print("Its an Image")

    else:

        print("No Idea")

/test/1.jpg

Its an Image

/test/2.avi

Its a Video

/test/unknown.xml

No Idea

/test/noextention

No Idea

Answer 2

from urlparse import urlparse

from os.path import splitext

url = "sample/test/image.png"

image_formats = [".png", ".jpeg"]

video_formats = [".mp4", ".mp3", ".avi"]

def get_ext(url):

    """Return the filename extension from url, or ''."""

    parsed = urlparse(url)

    root, ext = splitext(parsed.path)

    return ext  

if get_ext(url) in image_formats:

   print("it's an image")

elif get_ext(url) in video_formats:

    print("it's a video")

else:

    print("some differnet format")

它适用于任何类型的网址

www.example.com/image.jpg

https://www.example.com/page.html?foo=1&bar=2#fragment

https://www.example.com/resource