以编程方式将图像保存到Django ImageField

我有一些代码可以从web上获取图像并将其存储在模型中。重要的是:

from django.core.files import File  # you need this somewhere
import urllib


# The following actually resides in a method of my model

result = urllib.urlretrieve(image_url) # image_url is a URL to an image

# self.photo is the ImageField
self.photo.save(
    os.path.basename(self.url),
    File(open(result[0], 'rb'))
    )

self.save()

• 不 存储在upload_to文件夹中,urllib.urlretrieve()将其存储为tempfile,然后将其丢弃。
• save()方法采用文件名(os.path.basename位)和django.core.files.File对象。

编辑:为清晰起见,以下是模型(减去所有必需的导入语句):

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0], 'rb'))
                    )
            self.save()

如果尚未创建模型,则超级简单:

,将图像文件复制到上载路径(假定为= “路径/” 在下面的代码片段中)。

第二 ,使用类似于:

class Layout(models.Model):
    image = models.ImageField('img', upload_to='path/')

layout = Layout()
layout.image = "path/image.png"
layout.save()

在django 1.4中测试并运行,它可能也适用于现有模型。

只是一句话。tvon answer可以工作,但是,如果你在windows上工作,你可能想 open() 文件与 'rb' . 这样地:

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0], 'rb'))
                    )
            self.save()

0x1A 字节

以下是一种工作良好的方法,允许您将文件转换为特定格式(以避免“无法将模式P写入JPEG”错误):

import urllib2
from django.core.files.base import ContentFile
from PIL import Image
from StringIO import StringIO

def download_image(name, image, url):
    input_file = StringIO(urllib2.urlopen(url).read())
    output_file = StringIO()
    img = Image.open(input_file)
    if img.mode != "RGB":
        img = img.convert("RGB")
    img.save(output_file, "JPEG")
    image.save(name+".jpg", ContentFile(output_file.getvalue()), save=False)

下面是一个使用示例:

p = ProfilePhoto(user=user)
download_image(str(user.id), p.image, image_url)
p.save()

希望这有帮助

好的,如果您只需要将现有的图像文件路径与ImageField关联,那么此解决方案可能会很有用:

from django.core.files.base import ContentFile

with open('/path/to/already/existing/file') as f:
  data = f.read()

# obj.image is the ImageField
obj.image.save('imgfilename.jpg', ContentFile(data))

好的,如果认真的话,已经存在的图像文件将不会与ImageField关联,但是该文件的副本将在upload_to dir中创建为'imgfilename.jpg',并将与ImageField关联。

from django.core.files.storage import FileSystemStorage

class CustomStorage(FileSystemStorage):

    def _open(self, name, mode='rb'):
        return File(open(self.path(name), mode))

    def _save(self, name, content):
        # here, you should implement how the file is to be saved
        # like on other machines or something, and return the name of the file.
        # In our case, we just return the name, and disable any kind of save
        return name

    def get_available_name(self, name):
        return name

然后,在我的模型中,对于我的ImageField,我使用了新的自定义存储:

from custom_storage import CustomStorage

custom_store = CustomStorage()

class Image(models.Model):
    thumb = models.ImageField(storage=custom_store, upload_to='/some/path')

from django.core.files import File

with open('path_to_file', 'r') as f:   # use 'rb' mode for python3
    data = File(f)
    model.image.save('filename', data, True)

model_instance.myfile = model_instance.myfile.field.attr_class(model_instance, model_instance.myfile.field, 'my-filename.jpg')

这些答案中有很多都过时了,我花了很多时间在沮丧中(我对Django&web开发一般来说都是新手)。然而,我发现@iambibhas有一个很好的要点: https://gist.github.com/iambibhas/5051911

import requests

from django.core.files import File
from django.core.files.temp import NamedTemporaryFile


def save_image_from_url(model, url):
    r = requests.get(url)

    img_temp = NamedTemporaryFile(delete=True)
    img_temp.write(r.content)
    img_temp.flush()

    model.image.save("image.jpg", File(img_temp), save=True)

这可能不是你想要的答案。但是您可以使用charfield来存储文件的路径,而不是ImageFile。通过这种方式,您可以通过编程将上载的图像关联到字段,而无需重新创建文件。

与Django 3, 对于这样的模型:

class Item(models.Model):
   name = models.CharField(max_length=255, unique=True)
   photo= models.ImageField(upload_to='image_folder/', blank=True)

如果图像已经上传,我们可以直接执行以下操作:

Item.objects.filter(...).update(photo='image_folder/sample_photo.png')

或

my_item = Item.objects.get(id=5)
my_item.photo='image_folder/sample_photo.png'
my_item.save()

model.ImageField.path = os.path.join('/Upload', generated_image_path)

class tweet_photos(models.Model):
upload_path='absolute path'
image=models.ImageField(upload_to=upload_path)
image_url = models.URLField(null=True, blank=True)
def save(self, *args, **kwargs):
    if self.image_url:
        import urllib, os
        from urlparse import urlparse
        file_save_dir = self.upload_path
        filename = urlparse(self.image_url).path.split('/')[-1]
        urllib.urlretrieve(self.image_url, os.path.join(file_save_dir, filename))
        self.image = os.path.join(file_save_dir, filename)
        self.image_url = ''
    super(tweet_photos, self).save()

class Pin(models.Model):
    """Pin Class"""
    image_link = models.CharField(max_length=255, null=True, blank=True)
    image = models.ImageField(upload_to='images/', blank=True)
    title = models.CharField(max_length=255, null=True, blank=True)
    source_name = models.CharField(max_length=255, null=True, blank=True)
    source_link = models.CharField(max_length=255, null=True, blank=True)
    description = models.TextField(null=True, blank=True)
    tags = models.ForeignKey(Tag, blank=True, null=True)

    def __unicode__(self):
        """Unicode class."""
        return unicode(self.image_link)

    def save(self, *args, **kwargs):
        """Store image locally if we have a URL"""
        if self.image_link and not self.image:
            result = urllib.urlretrieve(self.image_link)
            self.image.save(os.path.basename(self.image_link), File(open(result[0], 'r')))
            self.save()
            super(Pin, self).save()

工作 您可以使用文件系统存储来保存映像。 检查下面的示例

def upload_pic(request):
if request.method == 'POST' and request.FILES['photo']:
    photo = request.FILES['photo']
    name = request.FILES['photo'].name
    fs = FileSystemStorage()
##### you can update file saving location too by adding line below #####
    fs.base_location = fs.base_location+'/company_coverphotos'
##################
    filename = fs.save(name, photo)
    uploaded_file_url = fs.url(filename)+'/company_coverphotos'
    Profile.objects.filter(user=request.user).update(photo=photo)

class DemoImage(models.Model):
    title = models.TextField(max_length=255, blank=False)
    image = models.ImageField(blank=False, upload_to="images/DemoImages/")

import requests
import urllib.request
from django.core.files import File
url = "https://path/to/logo.jpg"

# Below 3 lines is to fake as browser agent 
# as many sites block urllib class suspecting to be bots
opener = urllib.request.build_opener()
opener.addheaders = [("User-agent", "Mozilla/5.0")]
urllib.request.install_opener(opener)

# Issue command to actually download and create temp img file in memory        
result = urllib.request.urlretrieve(url)

# DemoImage.objects.create(title="title", image=File(open(result[0], "rb"))) 
# ^^ This erroneously results in creating the file like 
# images/DemoImages/path/to/temp/dir/logo_image_file 
# as opposed to 
# images/DemoImages/logo_image_file

# Solution to get the file in images/DemoImages/
reopen = open(result[0], "rb") # Returns a BufferedReader object of the temp image
django_file = File(reopen)     # Create the file from the BufferedReader object 
demoimg = DemoImage()
demoimg.title = "title"
demoimg.image.save("logo.png", django_file, save=True)

如果配置了cloudinary/S3,这种方法也会触发文件上传

class Avatar(models.Model):
    image_file = models.ImageField(upload_to=user_directory_path_avatar)

至少在Django3.15中,更改图像是相当容易的。

在视图中,处理图像时,可以从以下位置获取图像:

self.request.FILES['avatar']

这是InMemoryUploadedFile类型的实例,只要您的html表单具有enctype集和avatar字段。。。

    <form method="post" class="avatarform" id="avatarform" action="{% url avatar_update_view' %}" enctype="multipart/form-data">
         {% csrf_token %}
         <input id="avatarUpload" class="d-none" type="file" name="avatar">
    </form>

然后,在视图中设置新图像就像下面一样简单(其中profile是self.request.user的配置文件模型)

profile.avatar.image_file.save(self.request.FILES['avatar'].name, self.request.FILES['avatar'])

无需保存profile.avatar,图像_字段已经保存,并且由于“upload_to”回调函数而保存到正确的位置。

你可以用 Django REST framework Requests 库以编程方式将图像保存到Django ImageField

以下是一个例子:

import requests


def upload_image():
    # PATH TO DJANGO REST API
    url = "http://127.0.0.1:8080/api/gallery/"

    # MODEL FIELDS DATA
    data = {'first_name': "Rajiv", 'last_name': "Sharma"}

    #  UPLOAD FILES THROUGH REST API
    photo = open('/path/to/photo', 'rb')
    resume = open('/path/to/resume', 'rb')
    files = {'photo': photo, 'resume': resume}

    request = requests.post(url, data=data, files=files)
    print(request.status_code, request.reason) 

我用uuid将图像保存在django 2 python 3中,因为django就是这样做的:

import uuid   
from django.core.files import File 
import urllib

httpUrl = "https://miimgeurl/image.jpg"
result = urllib.request.urlretrieve(httpUrl)            
mymodel.imagefield.save(os.path.basename(str(uuid.uuid4())+".jpg"),File(open(result[0], 'rb')))
mymodel.save()