在Django中保存文件之前更改文件名 [英] Change filename before save file in Django

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本文介绍了在Django中保存文件之前更改文件名的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

  def get_file_path(在...中,我有下一个代码可以在django admin实例,文件名):
ext = filename.split('。')[ - 1]
filename =%s。%s%(uuid.uuid4(),ext)
return os.path.join('directory /',filename)

class Archivo(models.Model):
archivo = models.FileField(upload_to = get_file_path)

这适用于我,但我想动态传递目录,如下所示:

  def get_file_path(instance,filename,directory_string_var):
ext = filename.split('。')[ - 1]
filename = %s。%s%(uuid.uuid4(),ext)
return os.path.join(directory_string_var,filename)

如果我这样做,我不能将目录参数(变量)传递给archivo字段的upload_to选项中的方法。

解决方案

如果你的目标是上一个引起文件填满给定的目录(这是一个问题,因为依赖于文件系统,一些目录中的一些操作可能很昂贵),upload_to可以包含strftime格式化,将被替换为日期/时间上传

  archivo = models.FileField(upload_to ='path /%Y /%M /%D /')

您可以将参数存储在实例对象中:



$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ %(uuid.uuid4(),ext)
return os.path.join(instance.directory_string_var,filename)

class Archivo(models.Model):
archivo = models.FileField(upload_to = get_file_path)
directory_string_var ='default_directory_string_var'


I have the next code to rename my files when upload in django admin (in models.py)

def get_file_path(instance, filename):
    ext = filename.split('.')[-1]
    filename = "%s.%s" % (uuid.uuid4(), ext)
    return os.path.join('directory/', filename)

class Archivo(models.Model):
    archivo = models.FileField(upload_to = get_file_path)

That works for me, but i want to pass the directory dynamically, something like this:

def get_file_path(instance, filename, directory_string_var):
    ext = filename.split('.')[-1]
    filename = "%s.%s" % (uuid.uuid4(), ext)
    return os.path.join(directory_string_var, filename)

If i do that, i can't pass the directory parameter (variable) to the method in upload_to option of the "archivo" field.

解决方案

If your goal is just preventing the files to fill up the given directory (this is a concern because depending on the filesystem, some operations over a directory with too many entries can be expensive), upload_to can contain strftime formatting, which will be replaced by the date/time of the upload.

archivo = models.FileField(upload_to = 'path/%Y/%M/%D/')

You can store the parameter in the instance object:

def get_file_path(instance, filename):
    ext = filename.split('.')[-1]
    filename = "%s.%s" % (uuid.uuid4(), ext)
    return os.path.join(instance.directory_string_var, filename)

class Archivo(models.Model):
    archivo = models.FileField(upload_to = get_file_path)
    directory_string_var = 'default_directory_string_var'

这篇关于在Django中保存文件之前更改文件名的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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