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
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)
其他回答
这里有一个更友好的答案，仍然使用流媒体。
只需定义这些函数并调用getImage()。默认情况下，它将使用与url相同的文件名并写入当前目录，但两者都可以更改。
import requests
from StringIO import StringIO
from PIL import Image
def createFilename(url, name, folder):
dotSplit = url.split('.')
if name == None:
# use the same as the url
slashSplit = dotSplit[-2].split('/')
name = slashSplit[-1]
ext = dotSplit[-1]
file = '{}{}.{}'.format(folder, name, ext)
return file
def getImage(url, name=None, folder='./'):
file = createFilename(url, name, folder)
with open(file, 'wb') as f:
r = requests.get(url, stream=True)
for block in r.iter_content(1024):
if not block:
break
f.write(block)
def getImageFast(url, name=None, folder='./'):
file = createFilename(url, name, folder)
r = requests.get(url)
i = Image.open(StringIO(r.content))
i.save(file)
if __name__ == '__main__':
# Uses Less Memory
getImage('http://www.example.com/image.jpg')
# Faster
getImageFast('http://www.example.com/image.jpg')
getImage()的请求内容基于这里的答案，getImageFast()的请求内容基于上面的答案。
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。
下载图像
import requests
Picture_request = requests.get(url)
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')
这是谷歌搜索如何下载带有请求的二进制文件时出现的第一个响应。如果你需要下载带有请求的任意文件，你可以使用:
import requests
url = 'https://s3.amazonaws.com/lab-data-collections/GoogleNews-vectors-negative300.bin.gz'
open('GoogleNews-vectors-negative300.bin.gz', 'wb').write(requests.get(url, allow_redirects=True).content)
