import urllib
response = urllib.urlopen('http://stackoverflow.com/documentation/')
Using urllib.urlopen()
will return a response object, which can be handled similar to a file.
print response.code
# Prints: 200
The response.code
represents the http return value. 200 is OK, 404 is NotFound, etc.
print response.read()
'<!DOCTYPE html>\r\n<html>\r\n<head>\r\n\r\n<title>Documentation - Stack. etc'
response.read()
and response.readlines()
can be used to read the actual html file returned from the request. These methods operate similarly to file.read*
import urllib.request
print(urllib.request.urlopen("http://stackoverflow.com/documentation/"))
# Prints: <http.client.HTTPResponse at 0x7f37a97e3b00>
response = urllib.request.urlopen("http://stackoverflow.com/documentation/")
print(response.code)
# Prints: 200
print(response.read())
# Prints: b'<!DOCTYPE html>\r\n<html>\r\n<head>\r\n\r\n<title>Documentation - Stack Overflow</title>
The module has been updated for Python 3.x, but use cases remain basically the same. urllib.request.urlopen
will return a similar file-like object.