Question

我将文件名作为实例获取，文件很容易保存到数据库中，没有任何问题。但我需要检查文件扩展名并将文件保存为单独的

我期待这样的输出单独的文件扩展名并相应保存

phots
 image1.jpg
 image2.jpg
text
 text1.txt
 text2.txt

这是我的模特课

class Album(models.Model):
    file_upload = models.FileField(upload_to=content_file_name)
    name_content = models.CharField(max_length=100)

功能就在这里

def content_file_name(instance, filename):
    upload_dir = os.path.join('uploads', 'resource')
    return os.path.join(upload_dir, filename)

Answer 1

使用splitext获取扩展程序：

def content_file_name(instance, filename):
    name, ext = os.path.splitext(filename)
    ext = ext.lstrip(".")                          # Remove leading dot
    if ext in ("jpg", "jpeg", "png"):
        subdir = "phots"
    elif ext in ("txt", "asc"):
        subdir = "text"
    else:
        subdir = extension
    upload_dir = os.path.join('uploads', 'resource', subdir)
    return os.path.join(upload_dir, filename)

Answer 2

您可以使用.endswith功能并检查

    if filename.endswith("jpg") or silename.endswith("jpeg"):
         subdir = "phots"
    elif filename.endswith("txt") or filename.endswith("asc"):
         subdir = "docs"

Answer 3

我更喜欢地图而不是if / else语句，因为它更具扩展性。

dir_map = {".jpg"  : "photos",
           ".jpeg" : "photos",
           ".png"  : "photos",
           ".txt"  : "text",
           ".asc"  : "text"
          }
_, ext = os.path.splitext(filename)
subdir = dir_map.get(ext, "other")
return os.path.join("uploads", "resource", dir_map.get(ext, "other"), filename)

python单独的文件扩展名并相应保存到文件夹中

3 个答案: