In python how to receive and save image in HTTP post request without using requests and other libraries that...
0
I try to receive image from HTTP post request in python. I am using BaseHTTPHolder and do_POST() function. I try to receive image from localhost address http:127.0.0.1:8080/photo while running server on localhost. I tried to use code below, it saves photo but doesn't allow to open because photo is not received fully.
img = urllib2.urlopen('http://127.0.0.1:8080/photo').read()
content_length = int(self.headers.getheader('content-length',0))
file_content = self.rfile.read(content_length)
with open('/Users/kasymhan/Desktop/sprint2/file01.jpg','wb') as s:
s.write(file_content)
EDIT
My do_POST() function
def do_POST(self):
url = 'http://127.0.0.1:8080/photo/file02.jpg'
request_headers = self.headers
content_length = request_headers.getheaders('content-length')
length = int(content_length[0]) if content_length else 0
file_content = self.rfile.read(length)
img = urllib2.urlopen(url).read()
with open('/Users/kasymhan/Desktop/sprint2/image.jpg','wb') as s:
s.write(img)
python http post urllib2 urllib
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
add a comment |
0
I try to receive image from HTTP post request in python. I am using BaseHTTPHolder and do_POST() function. I try to receive image from localhost address http:127.0.0.1:8080/photo while running server on localhost. I tried to use code below, it saves photo but doesn't allow to open because photo is not received fully.
img = urllib2.urlopen('http://127.0.0.1:8080/photo').read()
content_length = int(self.headers.getheader('content-length',0))
file_content = self.rfile.read(content_length)
with open('/Users/kasymhan/Desktop/sprint2/file01.jpg','wb') as s:
s.write(file_content)
EDIT
My do_POST() function
def do_POST(self):
url = 'http://127.0.0.1:8080/photo/file02.jpg'
request_headers = self.headers
content_length = request_headers.getheaders('content-length')
length = int(content_length[0]) if content_length else 0
file_content = self.rfile.read(length)
img = urllib2.urlopen(url).read()
with open('/Users/kasymhan/Desktop/sprint2/image.jpg','wb') as s:
s.write(img)
python http post urllib2 urllib
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
Please include a larger chunk of your code to see how you're using thedo_POST()function.
– Edgar R. Mondragón
Nov 13 '18 at 19:59
add a comment |
0
0
0
I try to receive image from HTTP post request in python. I am using BaseHTTPHolder and do_POST() function. I try to receive image from localhost address http:127.0.0.1:8080/photo while running server on localhost. I tried to use code below, it saves photo but doesn't allow to open because photo is not received fully.
img = urllib2.urlopen('http://127.0.0.1:8080/photo').read()
content_length = int(self.headers.getheader('content-length',0))
file_content = self.rfile.read(content_length)
with open('/Users/kasymhan/Desktop/sprint2/file01.jpg','wb') as s:
s.write(file_content)
EDIT
My do_POST() function
def do_POST(self):
url = 'http://127.0.0.1:8080/photo/file02.jpg'
request_headers = self.headers
content_length = request_headers.getheaders('content-length')
length = int(content_length[0]) if content_length else 0
file_content = self.rfile.read(length)
img = urllib2.urlopen(url).read()
with open('/Users/kasymhan/Desktop/sprint2/image.jpg','wb') as s:
s.write(img)
python http post urllib2 urllib
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
I try to receive image from HTTP post request in python. I am using BaseHTTPHolder and do_POST() function. I try to receive image from localhost address http:127.0.0.1:8080/photo while running server on localhost. I tried to use code below, it saves photo but doesn't allow to open because photo is not received fully.
img = urllib2.urlopen('http://127.0.0.1:8080/photo').read()
content_length = int(self.headers.getheader('content-length',0))
file_content = self.rfile.read(content_length)
with open('/Users/kasymhan/Desktop/sprint2/file01.jpg','wb') as s:
s.write(file_content)
EDIT
My do_POST() function
def do_POST(self):
url = 'http://127.0.0.1:8080/photo/file02.jpg'
request_headers = self.headers
content_length = request_headers.getheaders('content-length')
length = int(content_length[0]) if content_length else 0
file_content = self.rfile.read(length)
img = urllib2.urlopen(url).read()
with open('/Users/kasymhan/Desktop/sprint2/image.jpg','wb') as s:
s.write(img)
python http post urllib2 urllib
python http post urllib2 urllib
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
edited Nov 13 '18 at 20:03
Kassymkhan Bekbolatov
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
asked Nov 13 '18 at 0:15
Kassymkhan BekbolatovKassymkhan Bekbolatov
13
13
Please include a larger chunk of your code to see how you're using thedo_POST()function.
– Edgar R. Mondragón
Nov 13 '18 at 19:59
add a comment |
Please include a larger chunk of your code to see how you're using thedo_POST()function.
– Edgar R. Mondragón
Nov 13 '18 at 19:59
Please include a larger chunk of your code to see how you're using the
do_POST() function.– Edgar R. Mondragón
Nov 13 '18 at 19:59
Please include a larger chunk of your code to see how you're using the
do_POST() function.– Edgar R. Mondragón
Nov 13 '18 at 19:59
add a comment |
1 Answer
1
active
oldest
votes
0
Why not directly write the contents of the image to the output file:
import urllib2
url = "http://farm5.static.flickr.com/4105/5063092779_17b2133825_b.jpg"
img = urllib2.urlopen(url).read()
with open('./image.jpg','wb') as s:
s.write(img)
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
Why not directly write the contents of the image to the output file:
import urllib2
url = "http://farm5.static.flickr.com/4105/5063092779_17b2133825_b.jpg"
img = urllib2.urlopen(url).read()
with open('./image.jpg','wb') as s:
s.write(img)
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
add a comment |
0
Why not directly write the contents of the image to the output file:
import urllib2
url = "http://farm5.static.flickr.com/4105/5063092779_17b2133825_b.jpg"
img = urllib2.urlopen(url).read()
with open('./image.jpg','wb') as s:
s.write(img)
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
add a comment |
0
0
0
Why not directly write the contents of the image to the output file:
import urllib2
url = "http://farm5.static.flickr.com/4105/5063092779_17b2133825_b.jpg"
img = urllib2.urlopen(url).read()
with open('./image.jpg','wb') as s:
s.write(img)
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
Why not directly write the contents of the image to the output file:
import urllib2
url = "http://farm5.static.flickr.com/4105/5063092779_17b2133825_b.jpg"
img = urllib2.urlopen(url).read()
with open('./image.jpg','wb') as s:
s.write(img)
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
answered Nov 13 '18 at 18:30
Edgar R. MondragónEdgar R. Mondragón
1,5212719
1,5212719
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
add a comment |
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
I am sending image from computer to server that is running on 127.0.0.1. I am not sure how to specify file in '127.0.0.1:8080/photo' this url.
– Kassymkhan Bekbolatov
Nov 13 '18 at 19:55
add a comment |
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Please include a larger chunk of your code to see how you're using the
do_POST()function.– Edgar R. Mondragón
Nov 13 '18 at 19:59