[Zope] create Login/Logout functionality

[Andreas Pakulat]

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On 05.Mai 2003 - 12:29:12, Paavo Parkkinen wrote:
> On Mon, 05.05.2003 at 11:19 +0200, Andreas Pakulat wrote:
> > On 03.Mai 2003 - 17:38:10, Troy Farrell wrote:
> > > Look in the source for 'manage_zmi_logout' in=20
> > > zopedir/lib/python/App/Management.py
> >=20
> > That doesn't help me much, I see that I can call manage_zmi_logout to
> > logout the user, but how do I get him to log in?
>=20
> I believe manage_zmi_logout opens the login pop-up window.

Yeah, that's right, but either using the function or copying some of its
content, doesn't give me a proper login-message. I see such a message,
but whatever I type, I get a "could not authenticate"! I also don't have
enough knowledge about the Status and Header-Info that is set in the
function, so if somebody could point me to a tutorial or sth. else that
explains what the different status and headers in an HTTP-Response are,
it would help me much.

> > I don't think that I can do it with such a message box, as far as I can
> > see, this can only be done, by setting the permissions on the file. But
> > this doesn't work for me, as anybody should be able to view a certain
> > page, but only logged in users should see a link that points to an
> > edit-form
> >=20
> > So the question is, which method to call if I have a username and a
> > password, I didn't find anything in the UserFolder API. I will try to
> > look further on the source of the API, but maybe someone can directly
> > point me to the right spot.
>=20
> We had a similar problem, and we used the CookieCrumbler product. It
> doesn't open a pop-up window to login, but you can use a custom login
> form.

So you used cookies to authenticate users? That is not possible within
my project, I need to authenticate against the Zope-Userdatabase.

It could theoretically be done, if I can get a User/SimpleUser Object
=66rom having a username, but it seems that I cannot get this. The
functions of the UserFolder Object are all restricted to UserManagers,
and an Unauthorized User isn't a UserManager.

It could theoretically be done, if I can get a User/SimpleUser Object
=66rom having a username, but it seems that I cannot get this. The
functions of the UserFolder Object are all restricted to UserManagers,
and an Unauthorized User isn't a UserManager.

Andreas

--
Fine day to work off excess energy.  Steal something heavy.


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