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
从请求中获取一个类似文件的对象，并将其复制到文件中。这也将避免将整个内容一次性读入内存。
import shutil
import requests
url = 'http://example.com/img.png'
response = requests.get(url, stream=True)
with open('img.png', 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
del response
其他回答
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')
从请求中获取一个类似文件的对象，并将其复制到文件中。这也将避免将整个内容一次性读入内存。
import shutil
import requests
url = 'http://example.com/img.png'
response = requests.get(url, stream=True)
with open('img.png', 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
del response
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)
这可能比使用请求更容易。这是我唯一一次建议不要使用请求来做HTTP的事情。
使用urllib的两个内衬:
>>> import urllib
>>> urllib.request.urlretrieve("http://www.example.com/songs/mp3.mp3", "mp3.mp3")
还有一个叫做wget的Python模块，它非常容易使用。在这里找到。
这说明了设计的简单性:
>>> import wget
>>> url = 'http://www.futurecrew.com/skaven/song_files/mp3/razorback.mp3'
>>> filename = wget.download(url)
100% [................................................] 3841532 / 3841532>
>> filename
'razorback.mp3'
享受。
编辑:您还可以添加一个out参数来指定路径。
>>> out_filepath = <output_filepath>
>>> filename = wget.download(url, out=out_filepath)
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。
