问题:如何使用我已经知道URL地址的Python在本地保存图像?
我知道Internet上图像的URL。
例如http://www.digimouth.com/news/media/2011/09/google-logo.jpg,其中包含Google的徽标。
现在,如何使用Python下载此图像,而无需在浏览器中实际打开URL并手动保存文件。
回答 0
Python 2
如果您要做的只是将其保存为文件,这是一种更简单的方法:
import urllib
urllib.urlretrieve("http://www.digimouth.com/news/media/2011/09/google-logo.jpg", "local-filename.jpg")
第二个参数是应在其中保存文件的本地路径。
Python 3
正如SergO所建议的,以下代码应与Python 3配合使用。
import urllib.request
urllib.request.urlretrieve("http://www.digimouth.com/news/media/2011/09/google-logo.jpg", "local-filename.jpg")
回答 1
import urllib
resource = urllib.urlopen("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
output = open("file01.jpg","wb")
output.write(resource.read())
output.close()
file01.jpg
将包含您的图像。
回答 2
我编写了一个脚本来执行此操作,并且可以在我的github上找到该脚本供您使用。
我利用BeautifulSoup允许我解析任何网站的图像。如果您要进行大量的网络抓取(或打算使用我的工具),建议您使用sudo pip install BeautifulSoup
。可在此处获得有关BeautifulSoup的信息。
为了方便起见,这是我的代码:
from bs4 import BeautifulSoup
from urllib2 import urlopen
import urllib
# use this image scraper from the location that
#you want to save scraped images to
def make_soup(url):
html = urlopen(url).read()
return BeautifulSoup(html)
def get_images(url):
soup = make_soup(url)
#this makes a list of bs4 element tags
images = [img for img in soup.findAll('img')]
print (str(len(images)) + "images found.")
print 'Downloading images to current working directory.'
#compile our unicode list of image links
image_links = [each.get('src') for each in images]
for each in image_links:
filename=each.split('/')[-1]
urllib.urlretrieve(each, filename)
return image_links
#a standard call looks like this
#get_images('http://www.wookmark.com')
回答 3
这可以通过请求来完成。加载页面并将二进制内容转储到文件中。
import os
import requests
url = 'https://apod.nasa.gov/apod/image/1701/potw1636aN159_HST_2048.jpg'
page = requests.get(url)
f_ext = os.path.splitext(url)[-1]
f_name = 'img{}'.format(f_ext)
with open(f_name, 'wb') as f:
f.write(page.content)
回答 4
Python 3
from urllib.error import HTTPError
from urllib.request import urlretrieve
try:
urlretrieve(image_url, image_local_path)
except FileNotFoundError as err:
print(err) # something wrong with local path
except HTTPError as err:
print(err) # something wrong with url
回答 5
适用于Python 2和Python 3的解决方案
try:
from urllib.request import urlretrieve # Python 3
except ImportError:
from urllib import urlretrieve # Python 2
url = "http://www.digimouth.com/news/media/2011/09/google-logo.jpg"
urlretrieve(url, "local-filename.jpg")
或者,如果的附加要求requests
是可以接受的并且是http(s)URL:
def load_requests(source_url, sink_path):
"""
Load a file from an URL (e.g. http).
Parameters
----------
source_url : str
Where to load the file from.
sink_path : str
Where the loaded file is stored.
"""
import requests
r = requests.get(source_url, stream=True)
if r.status_code == 200:
with open(sink_path, 'wb') as f:
for chunk in r:
f.write(chunk)
回答 6
我在Yup。的脚本上扩展了脚本。我修好了一些东西。现在它将绕过403:禁止的问题。当无法检索图像时,它不会崩溃。它试图避免损坏预览。它获取正确的绝对URL。它给出了更多信息。可以使用命令行中的参数来运行它。
# getem.py
# python2 script to download all images in a given url
# use: python getem.py http://url.where.images.are
from bs4 import BeautifulSoup
import urllib2
import shutil
import requests
from urlparse import urljoin
import sys
import time
def make_soup(url):
req = urllib2.Request(url, headers={'User-Agent' : "Magic Browser"})
html = urllib2.urlopen(req)
return BeautifulSoup(html, 'html.parser')
def get_images(url):
soup = make_soup(url)
images = [img for img in soup.findAll('img')]
print (str(len(images)) + " images found.")
print 'Downloading images to current working directory.'
image_links = [each.get('src') for each in images]
for each in image_links:
try:
filename = each.strip().split('/')[-1].strip()
src = urljoin(url, each)
print 'Getting: ' + filename
response = requests.get(src, stream=True)
# delay to avoid corrupted previews
time.sleep(1)
with open(filename, 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
except:
print ' An error occured. Continuing.'
print 'Done.'
if __name__ == '__main__':
url = sys.argv[1]
get_images(url)
回答 7
使用请求库
import requests
import shutil,os
headers = {
'user-agent': 'Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/78.0.3904.108 Safari/537.36'
}
currentDir = os.getcwd()
path = os.path.join(currentDir,'Images')#saving images to Images folder
def ImageDl(url):
attempts = 0
while attempts < 5:#retry 5 times
try:
filename = url.split('/')[-1]
r = requests.get(url,headers=headers,stream=True,timeout=5)
if r.status_code == 200:
with open(os.path.join(path,filename),'wb') as f:
r.raw.decode_content = True
shutil.copyfileobj(r.raw,f)
print(filename)
break
except Exception as e:
attempts+=1
print(e)
ImageDl(url)
回答 8
这是很简短的答案。
import urllib
urllib.urlretrieve("http://photogallery.sandesh.com/Picture.aspx?AlubumId=422040", "Abc.jpg")
回答 9
Python 3版本
我为Python 3调整了@madprops的代码
# getem.py
# python2 script to download all images in a given url
# use: python getem.py http://url.where.images.are
from bs4 import BeautifulSoup
import urllib.request
import shutil
import requests
from urllib.parse import urljoin
import sys
import time
def make_soup(url):
req = urllib.request.Request(url, headers={'User-Agent' : "Magic Browser"})
html = urllib.request.urlopen(req)
return BeautifulSoup(html, 'html.parser')
def get_images(url):
soup = make_soup(url)
images = [img for img in soup.findAll('img')]
print (str(len(images)) + " images found.")
print('Downloading images to current working directory.')
image_links = [each.get('src') for each in images]
for each in image_links:
try:
filename = each.strip().split('/')[-1].strip()
src = urljoin(url, each)
print('Getting: ' + filename)
response = requests.get(src, stream=True)
# delay to avoid corrupted previews
time.sleep(1)
with open(filename, 'wb') as out_file:
shutil.copyfileobj(response.raw, out_file)
except:
print(' An error occured. Continuing.')
print('Done.')
if __name__ == '__main__':
get_images('http://www.wookmark.com')
回答 10
使用Requests对于Python 3来说有些新鲜:
代码中的注释。准备使用功能。
import requests
from os import path
def get_image(image_url):
"""
Get image based on url.
:return: Image name if everything OK, False otherwise
"""
image_name = path.split(image_url)[1]
try:
image = requests.get(image_url)
except OSError: # Little too wide, but work OK, no additional imports needed. Catch all conection problems
return False
if image.status_code == 200: # we could have retrieved error page
base_dir = path.join(path.dirname(path.realpath(__file__)), "images") # Use your own path or "" to use current working directory. Folder must exist.
with open(path.join(base_dir, image_name), "wb") as f:
f.write(image.content)
return image_name
get_image("https://apod.nasddfda.gov/apod/image/2003/S106_Mishra_1947.jpg")
回答 11
较晚的答案,但是python>=3.6
您可以使用dload,即:
import dload
dload.save("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
如果您需要使用图像bytes
,请使用:
img_bytes = dload.bytes("http://www.digimouth.com/news/media/2011/09/google-logo.jpg")
使用安装 pip3 install dload
回答 12
img_data=requests.get('https://apod.nasa.gov/apod/image/1701/potw1636aN159_HST_2048.jpg')
with open(str('file_name.jpg', 'wb') as handler:
handler.write(img_data)