i have the following code to open a URL address, but can you please
tell me how can I check the content type of the url response?
Thank you.
try:
req = Request(url, txdata, txheaders)
handle = urlopen(req)
except IOError, e:
print e
print 'Failed to open %s' % url
return 0;
else:
data = handle.read()
2  2273 
On Apr 15, 2007, at 6:25 PM, John wrote:
>
i have the following code to open a URL address, but can you please
tell me how can I check the content type of the url response?
Thank you.
try:
req = Request(url, txdata, txheaders)
handle = urlopen(req)
except IOError, e:
print e
print 'Failed to open %s' % url
return 0;
else:
data = handle.read()
Not absolutely sure about this, but try handle.headers. get('Content-
Type') before the read.
hth,
Michael
---
Asking a person who he *is* ... is not Pythonic! --Anton Vredegoor
Michael Bentley <mi*****@jedimi ndworks.comwrit es:
On Apr 15, 2007, at 6:25 PM, John wrote:
i have the following code to open a URL address, but can you please
tell me how can I check the content type of the url response?
Thank you.
try:
req = Request(url, txdata, txheaders)
handle = urlopen(req)
except IOError, e:
print e
print 'Failed to open %s' % url
return 0;
else:
data = handle.read()
Not absolutely sure about this, but try handle.headers. get('Content-
Type') before the read.
handle.get_head er("Content-type")
John This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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