使用python urllib2返回url文件的流
我有一段Python代码,它得到了一个正确的url 从Amazon S3获得权限并将其复制到本地目录文件:使用python urllib2返回url文件的流,python,file-management,Python,File Management,我有一段Python代码,它得到了一个正确的url 从Amazon S3获得权限并将其复制到本地目录文件: def fileDownload(self, some_Id, localDir, name, range=None): # [...] something happens here # get the Amazon URL fileUrl = we_get_the_amazon_url_here req = urllib2.urlopen(fileUrl
def fileDownload(self, some_Id, localDir, name, range=None):
# [...] something happens here
# get the Amazon URL
fileUrl = we_get_the_amazon_url_here
req = urllib2.urlopen(fileUrl)
if len(range):
req.headers['Range']='bytes=%s-%s' % (range[0], range[1])
# Do the download
with open(os.path.join(localDir,name), 'wb') as fp:
shutil.copyfileobj(req, fp)
return 1
我不熟悉urllib2,但我想做的是
这个fileDownload
方法进入fileStreaming
方法,以便
可以通过管道将文件的结果内容传送到下游的工具中
你知道怎么做吗?我想你应该阅读一些文档。 但是,您的“req”对象类似于文件,所以您可以使用read()方法获取其内容
def fileDownload(self, some_Id, range=None):
# [...] something happens here
# get the Amazon URL
fileUrl = we_get_the_amazon_url_here
req = urllib2.urlopen(fileUrl)
if len(range):
req.headers['Range']='bytes=%s-%s' % (range[0], range[1])
return req.read()