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?
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。