我试图使用python的请求模块从网络下载并保存一张图像。
下面是我使用的(工作)代码:
img = urllib2.urlopen(settings.STATICMAP_URL.format(**data))
with open(path, 'w') as f:
f.write(img.read())
下面是使用请求的新(无效)代码:
r = requests.get(settings.STATICMAP_URL.format(**data))
if r.status_code == 200:
img = r.raw.read()
with open(path, 'w') as f:
f.write(img)
你能帮我从请求中使用响应的什么属性吗?
当前回答
这是一个非常简单的代码
import requests
response = requests.get("https://i.imgur.com/ExdKOOz.png") ## Making a variable to get image.
file = open("sample_image.png", "wb") ## Creates the file for image
file.write(response.content) ## Saves file content
file.close()
其他回答
这个怎么样，一个快速的解决方案。
import requests
url = "http://craphound.com/images/1006884_2adf8fc7.jpg"
response = requests.get(url)
if response.status_code == 200:
with open("/Users/apple/Desktop/sample.jpg", 'wb') as f:
f.write(response.content)
如导入图像和请求一样简单
from PIL import Image
import requests
img = Image.open(requests.get(url, stream = True).raw)
img.save('img1.jpg')
主要有两种方式:
Using .content (simplest/official) (see Zhenyi Zhang's answer): import io # Note: io.BytesIO is StringIO.StringIO on Python2. import requests r = requests.get('http://lorempixel.com/400/200') r.raise_for_status() with io.BytesIO(r.content) as f: with Image.open(f) as img: img.show() Using .raw (see Martijn Pieters's answer): import requests r = requests.get('http://lorempixel.com/400/200', stream=True) r.raise_for_status() r.raw.decode_content = True # Required to decompress gzip/deflate compressed responses. with PIL.Image.open(r.raw) as img: img.show() r.close() # Safety when stream=True ensure the connection is released.
计时两者无明显差异。
我将发布一个答案，因为我没有足够的代表来发表评论，但使用Blairg23发布的wget，您还可以为路径提供一个out参数。
wget.download(url, out=path)
你可以这样做:
import requests
import random
url = "https://images.pexels.com/photos/1308881/pexels-photo-1308881.jpeg? auto=compress&cs=tinysrgb&dpr=1&w=500"
name=random.randrange(1,1000)
filename=str(name)+".jpg"
response = requests.get(url)
if response.status_code.ok:
with open(filename,'w') as f:
f.write(response.content)
