想要使用自定義上傳圖片功能pythondjano上傳圖片

這是我的自定義圖片上傳功能

def upload_image(file, dir_name, filename):
    try:
        target_path = '/static/images/'   dir_name   '/'   filename
        path = storage.save(target_path, file)
        return storage.url(path)
    except Exception as e:
        print(e)

這是我的模型

class MenuOptions(models.Model):
    name = models.CharField(max_length=500, null=False)
    description = models.CharField(max_length=500, null=True)
    image_url = models.ImageField(upload_to=upload_image())

    def __str__(self):
        return f'{self.name}'

我想使用我的 upload_image 函式上傳影像，如您所見，它采用 3 個引數檔案、dir_name 和 file_name。如何在我的 model.ImageField() 中傳遞這些引數另外，我想將 image_url 存盤到我的資料庫中，如 upload_image 函式回傳的那樣，它將檔案存盤在 DB 中還是 URL 中？

uj5u.com熱心網友回復：

首先，您應該傳遞upload_image，而不是呼叫upload_image()：

image_url = models.ImageField(upload_to=upload_image)

第二件事，使用傳遞的值upload_to自動呼叫。因此不需要Storage.save()呼叫內部。應該只回傳應該保存檔案的路徑。upload_image()upload_image

函式upload_to應該有 2 個引數，instance并且filename.

參考

uj5u.com熱心網友回復：

這是我存盤檔案的解決方案。

您只需實作以下功能并在模型中使用它。

import os
import uuid
from django.utils.deconstruct import deconstructible


@deconstructible
class MediaFileNameHash(object):
    def __init__(self, base_dir, path):
        self.address = base_dir   "/"   path
        self.path = path
        if not os.path.exists(self.address):
            os.makedirs(self.address, exist_ok=True)

    def __call__(self, _, filename):
        # @note It's up to the validators to check if it's the correct file type in name or if one even exist.
        filename = os.path.splitext(filename)
        return self.path   '/'   filename[0]   "@"   str(uuid.uuid4())   filename[1]

例如：

class MenuOptions(models.Model):
    name = models.CharField(max_length=500, null=False)
    description = models.CharField(max_length=500, null=True)
    image_url = models.ImageField(
         upload_to=MediaFileNameHash("/media/", "files"),
         verbose_name=_("Address")
    )

uj5u.com熱心網友回復：

另外要傳遞函式名稱，您首先需要在模型中撰寫一個函式來處理這樣的路徑！

# In you models.py 
# will automatically save your image to your custom view 

def get_upload_image_filepath(self, filename):
    return 'Images/'   str(self.pk)   '/default.png'

class MenuOptions(models.Model):
    name = models.CharField(max_length=500, null=False)
    description = models.CharField(max_length=500, null=True)
    image_url = models.ImageField(upload_to=get_upload_image_filepath)

    def __str__(self):
        return f'{self.name}'

好吧，我只是要指導你如何處理這種情況，你可以重構你的代碼，因為它會在同樣的意義上作業。這是我關于如何撰寫自定義視圖的解決方案：

# this would be at the top of your view 
import base64
import os

TEMP_UPLOAD_IMAGE_NAME = "default.png"


def save_temp_upload_image_from_base64String(imageString, user):
    INCORRECT_PADDING_EXCEPTION = "Incorrect padding"
    try:
        if not os.path.exists(settings.TEMP):
            os.mkdir(settings.TEMP)
        if not os.path.exists(settings.TEMP   "/"   str(user.pk)):
            os.mkdir(settings.TEMP   "/"   str(user.pk))
        url = os.path.join(settings.TEMP   "/"   str(user.pk),TEMP_UPLOAD_IMAGE_NAME)
        storage = FileSystemStorage(location=url)
        image = base64.b64decode(imageString)
        with storage.open('', 'wb ') as destination:
            destination.write(image)
            destination.close()
        return url
    except Exception as e:
        print("exception: "   str(e))
        # workaround for an issue I found
        if str(e) == INCORRECT_PADDING_EXCEPTION:
            imageString  = "=" * ((4 - len(imageString) % 4) % 4)
            return save_temp_upload_image_from_base64String(imageString, user)
    return None

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