Thank you, Hamish, that was enlightening and, of course, embarrassing as well. I have two zope versions running (under windows NT). I had copied the registry parameters of the 243 version to the the 251 version and changed the z2.py path, but forgot to change the python.exe path. So my 251 version was running under python 2.1.0 from the 243 install. (Nevertheless SOFTWARE_HOME shows 'E:\zope\251\lib\python') I still have to figure out some firewall settings, but I guess the sendmail tag will behave then as expected. Thanks again. cb ----- Original Message ----- From: "Hamish Lawson" <hbl@st-andrews.ac.uk> To: <zope@zope.org> Sent: Wednesday, November 06, 2002 12:18 PM Subject: [Zope] attribute error 'sendall' in sendmail tag
Cornelis J. de Brabander wrote:
The following code produces a attribute error with value "sendall' in 2.5.1
If your problem is like the one I had, then it's because your _socket module doesn't have a sendall method. To check this out, create and run the following External Method:
def test_socket(): import _socket sock = _socket.socket(_socket.AF_INET, _socket.SOCK_STREAM) return "<p>%s</p>\n<p>%s</p>" % (_socket.__file__, str(dir(sock)))
This will print out the path to where the _socket is being loaded from, followed by the supported attributes for a socket connection.
* If the sendall attribute is missing, check whether your _socket module is indeed being loaded from your Zope's Python installation; if from elsewhere, try fixing up your paths.
* If the location for _socket seems OK then try downloading Python 2.1.3 and copying the _socket.pyd into the appropriate place in your Zope's Python installation.
If, however, the sendall attribute is in fact listed, then I'm afraid I don't know what the problem is!
Hamish Lawson
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