Question

功能：

def custom_path(path):
    new_path = os.path.abspath(path)
    def get_path():
        return new_path
    return get_path

问题出在custom_path中，当在此get_image_path函数中调用该函数时，它会创建一个具有绝对路径的目录，例如<function custom_path at 0x7f432d99e158>/somedir/finalName.*ext，但我希望它以字符串表示形式返回。我的代码：

def get_image_path(instance, file):
    '''
    saves file to some location
    '''
    file_path   = custom_path*
    new_file    = random.randint(1, 45360789120)
    name, ext   = get_file_ext(file)
    ranstr      = random_string_generator(size=4)
    final_      = f'{ranstr}_{new_file}{ext}'
    return f'{file_path}/{new_file}/{final_}'

发件人：

from ... import custom_path, get_image_path

custom_path('posts')

class Post(models.Model):
    ...
    ...
    image   = models.FileField(upload_to=get_image_path, null=True,
              blank=True, verbose_name='article image (optional)')
    ...
    ...

否则，我必须将此功能修改为...

def get_image_path(instance, file):
    ...
    ...
    return f'posts*/{new_file}/{final_}' # and so on for *different models..

我也尝试过类似的事情...

def get_image_path(instance, file):
    ...
    ...
    return f'{new_file}/{final_}'

base = os.path.dirname('posts/')
upload_path = base + os.path.join(str(get_image_path))

class Post(models.Model):
    ...
    ...
    image   = models.FileField(upload_to=upload_path, null=True,
              blank=True, verbose_name='article image (optional)')
    ...
    ...

输出：''/newDir/newFile.*ext，如何摆脱它？ ️️

Answer 1

函数AVPlayerItem返回内部函数PlayingState.swift，因此它的工作方式如下：

custom_path

或更短的代码：

get_path

这是我的考试：

foo = custom_path()  # foo is a function
path = foo()  # path is str now

在您的情况下，您必须在f字符串中添加大括号：

path = custom_path()()

django-卡在路径生成问题中

1 个答案: