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

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。

下载图像

import requests
Picture_request = requests.get(url)

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)

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()

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)