On 7/11/06, John Zimmerman [gmail] <[EMAIL PROTECTED]
> wrote:
I just created a form in /app/elements/login.thtml with a username box and a password box with a "login" button.
I conditionally check to see if the user is logged in within my element to display the current name of the logged in user with a logout button.if($this->controller->Session->check('LoggedInUser')) {
//display current username and logout button. logout button posts to /users/logout
} else {
//display login form. login form posts to /users/login
}
I can render the login element on any page using $this->renderElement('login'); in my view or page layout.
Works well for me. I can capture the page they were on using a hidden element in the form and redirect back to that page after login, or I can get that page from the cake controller.
The above is just a general description and rough code for what I did. I didn't copy and paste so you will most likely have to tweak it to make it worth anything.
I do have this working quite well on a couple of sites I am developing so post back if you can't get it and I will put code into the CakeBin with some instructions.On 7/11/06, John David Anderson (_psychic_) < [EMAIL PROTECTED]> wrote:
On Jul 11, 2006, at 6:44 PM, codecowboy wrote:
>
> Hi,
>
> I have followed the instuctions in chapter 21 of the manual to
> create a
> user authentication system. I have that system completely working.
>
> I want to create a login element that functions the same way. This
> way, I can place the login element in the left column of my layout.
>
> The problem that I am running into is that cake requires the view
> to be
> related to an action. I do not know how to tell cake that the login
> element is the view for the users/login action.
I'd just leave it as an action and use $this->requestAction() to pull
it into places you want it displayed.
<?php echo $this->requestAction('/users/login', array('return')); ?>
-- John
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