我正在编写一个Django应用程序，它将获取特定URL的所有图像并将它们保存在数据库中。
但我没有介绍如何在Django中使用ImageField。
settings.朋友
MEDIA_ROOT = os.path.join(PWD, "../downloads/")
# URL that handles the media served from MEDIA_ROOT. Make sure to use a
# trailing slash.
# Examples: "http://example.com/media/", "htp://media.example.com/"
MEDIA_URL = '/downloads/'
models.朋友
class images_data(models.Model):
image_id =models.IntegerField()
source_id = models.IntegerField()
image=models.ImageField(upload_to='images',null=True, blank=True)
text_ind=models.NullBooleanField()
prob=models.FloatField()
download_IMG.朋友
def spider(site):
PWD = os.path.dirname(os.path.realpath(__file__ ))
#site="http://en.wikipedia.org/wiki/Pune"
hdr= {'User-Agent': 'Mozilla/5.0'}
outfolder=os.path.join(PWD, "../downloads")
#outfolder="/home/mayank/Desktop/dreamport/downloads"
print "MAYANK:"+outfolder
req = urllib2.Request(site,headers=hdr)
page = urllib2.urlopen(req)
soup =bs(page)
tag_image=soup.findAll("img")
count=1;
for image in tag_image:
print "Image: %(src)s" % image
filename = image["src"].split("/")[-1]
outpath = os.path.join(outfolder, filename)
urlretrieve('http:'+image["src"], outpath)
im = img(image_id=count,source_id=1,image=outpath,text_ind=None,prob=0)
im.save()
count=count+1
我在一个视图中调用download_imgs.py
if form.is_valid():
url = form.cleaned_data['url']
spider(url)
Django Documentation始终是一个好的起点
class ModelWithImage(models.Model):
image = models.ImageField(
upload_to='images',
)
更新
所以这个脚本有效。
.
import requests
import tempfile
from django.core import files
# List of images to download
image_urls = [
'http://i.thegrindstone.com/wp-content/uploads/2013/01/how-to-get-awesome-back.jpg',
]
for image_url in image_urls:
# Steam the image from the url
request = requests.get(image_url, stream=True)
# Was the request OK?
if request.status_code != requests.codes.ok:
# Nope, error handling, skip file etc etc etc
continue
# Get the filename from the url, used for saving later
file_name = image_url.split('/')[-1]
# Create a temporary file
lf = tempfile.NamedTemporaryFile()
# Read the streamed image in sections
for block in request.iter_content(1024 * 8):
# If no more file then stop
if not block:
break
# Write image block to temporary file
lf.write(block)
# Create the model you want to save the image to
image = Image()
# Save the temporary image to the model#
# This saves the model so be sure that is it valid
image.image.save(file_name, files.File(lf))
一些参考链接：
如果你想保存下载的图像而不先将它们保存到磁盘（不使用save等），那么有一种简单的方法可以做到这一点。
这比下载文件并将其写入磁盘稍快一些，因为它全部在内存中完成。请注意，此示例是为Python 3编写的 - 该过程在Python 2中类似，但略有不同。
NamedTemporaryFile其中from django.core import files
from io import BytesIO
import requests
url = "https://example.com/image.jpg"
resp = requests.get(url)
if resp.status_code != requests.codes.ok:
# Error handling here
fp = BytesIO()
fp.write(resp.content)
file_name = url.split("/")[-1] # There's probably a better way of doing this but this is just a quick example
your_model.image_field.save(file_name, files.File(fp))
是您要保存的模型的实例，your_model是.image_field的名称。
有关更多信息，请参阅ImageField的文档。
作为我认为你问的一个例子：
在forms.py中：
io
在models.py中：
imgfile = forms.ImageField(label = 'Choose your image', help_text = 'The image should be cool.')
因此，将有来自用户的POST请求（当用户完成表单时）。该请求基本上包含数据字典。字典保存提交的文件。要将请求集中在字段中的文件（在我们的示例中为ImageField），您将使用：
imgfile = models.ImageField(upload_to='images/%m/%d')
您可以在构造模型对象时使用它（实例化模型类）：
request.FILES['imgfield']
要以简单的方式保存，您只需使用赋予对象的save（）方法（因为Django非常棒）：
newPic = ImageModel(imgfile = request.FILES['imgfile'])
默认情况下，您的图像将存储在settings.py中为MEDIA_ROOT指定的目录中。
访问模板中的图像：
if form.is_valid():
newPic = Pic(imgfile = request.FILES['imgfile'])
newPic.save()
网址可能很棘手，但这里是一个简单网址模式的基本示例来调用存储的图像：
<img src="{{ MEDIA_URL }}{{ image.imgfile.name }}"></img>
我希望它有所帮助。
尝试这样做而不是分配路径到图像...
urlpatterns += patterns('',
url(r'^media/(?P<path>.*)$', 'django.views.static.serve', {
'document_root': settings.MEDIA_ROOT,
}),
)
像这样使用上面的功能..
import urllib2
from django.core.files.temp import NamedTemporaryFile
def handle_upload_url_file(url):
img_temp = NamedTemporaryFile()
opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:15.0) Gecko/20120427 Firefox/15.0a1')]
img_temp.write(opener.open(url).read())
img_temp.flush()
return img_temp
类似于@ boltfrombluesky上面的回答，你可以在Python 3中做到这一点，没有任何外部依赖，如下所示：
new_image = images_data()
#rest of the data in new_image and then do this.
new_image.image.save(slug_filename,File(handle_upload_url_file(url)))
#here slug_filename is just filename that you want to save the file with.
from os.path import basename
import urllib.request
from urllib.parse import urlparse
import tempfile
from django.core.files.base import File
def handle_upload_url_file(url, obj):
img_temp = tempfile.NamedTemporaryFile(delete=True)
req = urllib.request.Request(
url, data=None,
headers={
'User-Agent': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_3) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/35.0.1916.47 Safari/537.36'
}
)
with urllib.request.urlopen(req) as response:
img_temp.write(response.read())
img_temp.flush()
filename = basename(urlparse(url).path)
result = obj.image.save(filename, File(img_temp))
img_temp.close()
return result