Django中的動態文件路徑

Question

我試圖在django中生成動態文件路徑。我想打一個文件系統是這樣的：

-- user_12 
    --- photo_1 
    --- photo_2 
--- user_ 13 
    ---- photo_1 

我發現了一個相關的問題：Django Custom image upload field with dynamic path

在這裏，他們說，我們可以改變upload_to路徑並導致https://docs.djangoproject.com/en/stable/topics/files/文檔。在文檔中，有一個例子：

from django.db import models 
from django.core.files.storage import FileSystemStorage 

fs = FileSystemStorage(location='/media/photos') 

class Car(models.Model): 
    ... 
    photo = models.ImageField(storage=fs) 

但是，這仍然不是動態的，我想給車輛ID的圖像名稱，汽車的定義完成之前，我不能指定ID。那麼我如何創建一個帶有車輛ID的路徑？

Answer 1

您可以使用在upload_to參數可贖回，而不是使用自定義存儲。請參閱docs，並注意其中的警告，即在調用該函數時可能尚未設置主鍵（因爲在將對象保存到數據庫之前可能需要處理上載），所以可能無法使用ID。您可能會考慮在模型上使用另一個字段，如slug。 E.g：

import os 
def get_upload_path(instance, filename): 
    return os.path.join(
     "user_%d" % instance.owner.id, "car_%s" % instance.slug, filename) 

則：

photo = models.ImageField(upload_to=get_upload_path) 

Answer 2

https://docs.djangoproject.com/en/stable/ref/models/fields/#django.db.models.FileField.upload_to

def upload_path_handler(instance, filename): 
    return "user_{id}/{file}".format(id=instance.user.id, file=filename) 

class Car(models.Model): 
    ... 
    photo = models.ImageField(upload_to=upload_path_handler, storage=fs) 

有一個在文檔的警告，但它不應該影響你，因爲我們是User ID，而不是Car ID後。

In most cases, this object will not have been saved to the database yet, so if it uses the default AutoField, it might not yet have a value for its primary key field.

Answer 3

This guy有辦法做到動態路徑。這個想法是設置你最喜歡的存儲，並用一個函數定製「upload_to（）」參數。

希望這會有所幫助。

Answer 4

我發現了一個不同的解決方案，它很髒，但它的工作原理。您應該創建一個新的虛擬模型，它與原始模型自動同步。我對此並不感到自豪，但沒有找到其他解決方案。在我的情況下，我想上傳文件，並將其存儲在以模型ID命名的目錄中（因爲我將生成更多文件）。

的model.py

class dummyexperiment(models.Model): 
    def __unicode__(self): 
    return str(self.id) 

class experiment(models.Model): 
    def get_exfile_path(instance, filename): 
    if instance.id == None: 
     iid = instance.dummye.id 
    else: 
     iid = instance.id 
    return os.path.join('experiments', str(iid), filename) 

    exfile = models.FileField(upload_to=get_exfile_path) 

    def save(self, *args, **kwargs): 
    if self.id == None: 
     self.dummye = dummyexperiment() 
     self.dummye.save() 
    super(experiment, self).save(*args, **kwargs) 

我在Python和Django中很新，但它似乎是確定我。

另一種解決方案：

def get_theme_path(instance, filename): 
    id = instance.id 
    if id == None: 
    id = max(map(lambda a:a.id,Theme.objects.all())) + 1 
    return os.path.join('experiments', str(id), filename) 

Answer 5

可以使用lambda函數如下，請注意，如果實例新的，那麼它不會有實例ID，所以用別的東西：

logo = models.ImageField(upload_to=lambda instance, filename: 'directory/images/{0}/{1}'.format(instance.owner.id, filename)) 

Answer 6

上有DjangoSnippets兩種解決方案

1. 兩級節省：https://djangosnippets.org/snippets/1129/
2. 預取的ID（僅PostgreSQL的）：https://djangosnippets.org/snippets/2731/

Answer 7

好非常遲到了，但是這一次對我的作品。這樣才

class Album(models.Model): 
    albumname = models.CharField(max_length=100) 
    audiofile = models.FileField(upload_to=content_file_name) 

Answer 8

def content_file_name(instance, filename): 
    upload_dir = os.path.join('uploads',instance.albumname) 
    if not os.path.exists(upload_dir): 
     os.makedirs(upload_dir) 
    return os.path.join(upload_dir, filename) 

型號彷彿模型實例沒有被保存到數據庫中還沒有主鍵（id）可能無法使用，我寫了移動的文件我的FileField子類保存模型，以及刪除舊文件的存儲子類。

存儲：

class OverwriteFileSystemStorage(FileSystemStorage): 
    def _save(self, name, content): 
     self.delete(name) 
     return super()._save(name, content) 

    def get_available_name(self, name): 
     return name 

    def delete(self, name): 
     super().delete(name) 

     last_dir = os.path.dirname(self.path(name)) 

     while True: 
      try: 
       os.rmdir(last_dir) 
      except OSError as e: 
       if e.errno in {errno.ENOTEMPTY, errno.ENOENT}: 
        break 

       raise e 

      last_dir = os.path.dirname(last_dir) 

的FileField：

def tweak_field_save(cls, field): 
    field_defined_in_this_class = field.name in cls.__dict__ and field.name not in cls.__bases__[0].__dict__ 

    if field_defined_in_this_class: 
     orig_save = cls.save 

     if orig_save and callable(orig_save): 
      assert isinstance(field.storage, OverwriteFileSystemStorage), "Using other storage than '{0}' may cause unexpected behavior.".format(OverwriteFileSystemStorage.__name__) 

      def save(self, *args, **kwargs): 
       if self.pk is None: 
        orig_save(self, *args, **kwargs) 

        field_file = getattr(self, field.name) 

        if field_file: 
         old_path = field_file.path 
         new_filename = field.generate_filename(self, os.path.basename(old_path)) 
         new_path = field.storage.path(new_filename) 
         os.makedirs(os.path.dirname(new_path), exist_ok=True) 
         os.rename(old_path, new_path) 
         setattr(self, field.name, new_filename) 

        # for next save 
        if len(args) > 0: 
         args = tuple(v if k >= 2 else False for k, v in enumerate(args)) 

        kwargs['force_insert'] = False 
        kwargs['force_update'] = False 

       orig_save(self, *args, **kwargs) 

      cls.save = save 


def tweak_field_class(orig_cls): 
    orig_init = orig_cls.__init__ 

    def __init__(self, *args, **kwargs): 
     if 'storage' not in kwargs: 
      kwargs['storage'] = OverwriteFileSystemStorage() 

     if orig_init and callable(orig_init): 
      orig_init(self, *args, **kwargs) 

    orig_cls.__init__ = __init__ 

    orig_contribute_to_class = orig_cls.contribute_to_class 

    def contribute_to_class(self, cls, name): 
     if orig_contribute_to_class and callable(orig_contribute_to_class): 
      orig_contribute_to_class(self, cls, name) 

     tweak_field_save(cls, self) 

    orig_cls.contribute_to_class = contribute_to_class 

    return orig_cls 


def tweak_file_class(orig_cls): 
    """ 
    Overriding FieldFile.save method to remove the old associated file. 
    I'm doing the same thing in OverwriteFileSystemStorage, but it works just when the names match. 
    I probably want to preserve both methods if anyone calls Storage.save. 
    """ 

    orig_save = orig_cls.save 

    def new_save(self, name, content, save=True): 
     self.delete(save=False) 

     if orig_save and callable(orig_save): 
      orig_save(self, name, content, save=save) 

    new_save.__name__ = 'save' 
    orig_cls.save = new_save 

    return orig_cls 


@tweak_file_class 
class OverwriteFieldFile(models.FileField.attr_class): 
    pass 


@tweak_file_class 
class OverwriteImageFieldFile(models.ImageField.attr_class): 
    pass 


@tweak_field_class 
class RenamedFileField(models.FileField): 
    attr_class = OverwriteFieldFile 


@tweak_field_class 
class RenamedImageField(models.ImageField): 
    attr_class = OverwriteImageFieldFile 

和我upload_to可調用看起來像這樣：

def user_image_path(instance, filename): 
    name, ext = 'image', os.path.splitext(filename)[1] 

    if instance.pk is not None: 
     return os.path.join('users', os.path.join(str(instance.pk), name + ext)) 

    return os.path.join('users', '{0}_{1}{2}'.format(uuid1(), name, ext)) 

Answer 9

我的解決方案是不優雅，但它的工作原理：

在t他的模型，使用標準的功能，將需要的ID /包

def directory_path(instance, filename): 
    return 'files/instance_id_{0}/{1}'.format(instance.pk, filename) 

在views.py保存這樣的形式：

f=form.save(commit=False) 
ftemp1=f.filefield 
f.filefield=None 
f.save() 
#And now that we have crated the record we can add it 
f.filefield=ftemp1 
f.save() 

它爲我工作。 注意：我的文件區位於模型中，並且允許使用空值。 Null = True