how to check the 'content/type' using urlopen

[John]

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()

[Michael Bentley]

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

[John J. Lee]

Michael Bentley <mi*****@jedimindworks.comwrites:

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_header("Content-type")
John