Reading headers from urllib.request.urlopen
up vote
0
down vote
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I've searched for and found many answers, unfortunately all Python2-related, which looks something like this:
r = urllib.urlopen(url)
headers = r.info()
print(headers.getheader('Content-Disposition'))
However this doesn't seem to work with Python3. There is no .getheader() method. All the header data is inside r.info()._headers as a list of tuples. The underscore may suggest that this isn't to be accessed directly, or there's a more "proper" way of reading the headers... if so, what is the proper way of reading the headers?
python-3.x http-headers urllib
asked Nov 9 at 19:54
dtgq
1,15121639
add a comment |
up vote
0
down vote
favorite
I've searched for and found many answers, unfortunately all Python2-related, which looks something like this:
r = urllib.urlopen(url)
headers = r.info()
print(headers.getheader('Content-Disposition'))
However this doesn't seem to work with Python3. There is no .getheader() method. All the header data is inside r.info()._headers as a list of tuples. The underscore may suggest that this isn't to be accessed directly, or there's a more "proper" way of reading the headers... if so, what is the proper way of reading the headers?
python-3.x http-headers urllib
asked Nov 9 at 19:54
dtgq
1,15121639
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've searched for and found many answers, unfortunately all Python2-related, which looks something like this:
r = urllib.urlopen(url)
headers = r.info()
print(headers.getheader('Content-Disposition'))
However this doesn't seem to work with Python3. There is no .getheader() method. All the header data is inside r.info()._headers as a list of tuples. The underscore may suggest that this isn't to be accessed directly, or there's a more "proper" way of reading the headers... if so, what is the proper way of reading the headers?
python-3.x http-headers urllib
asked Nov 9 at 19:54
dtgq
1,15121639
I've searched for and found many answers, unfortunately all Python2-related, which looks something like this:
r = urllib.urlopen(url)
headers = r.info()
print(headers.getheader('Content-Disposition'))
However this doesn't seem to work with Python3. There is no .getheader() method. All the header data is inside r.info()._headers as a list of tuples. The underscore may suggest that this isn't to be accessed directly, or there's a more "proper" way of reading the headers... if so, what is the proper way of reading the headers?
python-3.x http-headers urllib
python-3.x http-headers urllib
asked Nov 9 at 19:54
dtgq
1,15121639
asked Nov 9 at 19:54
dtgq
1,15121639
edited Nov 9 at 20:00
asked Nov 9 at 19:54
dtgq
1,15121639
asked Nov 9 at 19:54
dtgq
1,15121639
asked Nov 9 at 19:54
dtgq
1,15121639
1,15121639
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
1
down vote
accepted
If url uses http or https scheme r is of type http.client.HTTPResponse. You can get headers that way:
import urllib.request
r = urllib.request.urlopen(url)
print(r.getheaders())
print(r.getheader('Content-Disposition'))
You can use print(dir(r)) to list attributes of r.
answered Nov 9 at 20:46
Strigo
964
add a comment |
up vote
0
down vote
r.info() returns a HTTPMessage object which is implemented using the email.message.Message class. From the documentation it looks like headers.get('Content-Disposition') is the method you want.
answered Nov 9 at 20:38
wilkben
415
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If url uses http or https scheme r is of type http.client.HTTPResponse. You can get headers that way:
import urllib.request
r = urllib.request.urlopen(url)
print(r.getheaders())
print(r.getheader('Content-Disposition'))
You can use print(dir(r)) to list attributes of r.
answered Nov 9 at 20:46
Strigo
964
add a comment |
up vote
1
down vote
accepted
If url uses http or https scheme r is of type http.client.HTTPResponse. You can get headers that way:
import urllib.request
r = urllib.request.urlopen(url)
print(r.getheaders())
print(r.getheader('Content-Disposition'))
You can use print(dir(r)) to list attributes of r.
answered Nov 9 at 20:46
Strigo
964
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If url uses http or https scheme r is of type http.client.HTTPResponse. You can get headers that way:
import urllib.request
r = urllib.request.urlopen(url)
print(r.getheaders())
print(r.getheader('Content-Disposition'))
You can use print(dir(r)) to list attributes of r.
answered Nov 9 at 20:46
Strigo
964
If url uses http or https scheme r is of type http.client.HTTPResponse. You can get headers that way:
import urllib.request
r = urllib.request.urlopen(url)
print(r.getheaders())
print(r.getheader('Content-Disposition'))
You can use print(dir(r)) to list attributes of r.
answered Nov 9 at 20:46
Strigo
964
edited Nov 9 at 20:59
answered Nov 9 at 20:46
Strigo
964
answered Nov 9 at 20:46
Strigo
964
answered Nov 9 at 20:46
Strigo
964
964
add a comment |
add a comment |
up vote
0
down vote
r.info() returns a HTTPMessage object which is implemented using the email.message.Message class. From the documentation it looks like headers.get('Content-Disposition') is the method you want.
answered Nov 9 at 20:38
wilkben
415
add a comment |
up vote
0
down vote
r.info() returns a HTTPMessage object which is implemented using the email.message.Message class. From the documentation it looks like headers.get('Content-Disposition') is the method you want.
answered Nov 9 at 20:38
wilkben
415
add a comment |
up vote
0
down vote
up vote
0
down vote
r.info() returns a HTTPMessage object which is implemented using the email.message.Message class. From the documentation it looks like headers.get('Content-Disposition') is the method you want.
answered Nov 9 at 20:38
wilkben
415
r.info() returns a HTTPMessage object which is implemented using the email.message.Message class. From the documentation it looks like headers.get('Content-Disposition') is the method you want.
answered Nov 9 at 20:38
wilkben
415
answered Nov 9 at 20:38
wilkben
415
answered Nov 9 at 20:38
wilkben
415
answered Nov 9 at 20:38
wilkben
415
415
add a comment |
add a comment |
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