I have a news controller with a 'view' action for each news item. In the 'view' action of each news item I would like to include another view, the 'add' action of the comments controller.
Basically, I need a form on each news item's page to add comments.
I have the two views but I can't manage to link them. I tried with elements, but it seems it doesn't render the view for the add comment view.
What should I do?
Thank you very much!
EDIT
Code in element:
<?php echo $this->Form->create('Comment', array('class' => 'big', 'url' => array('controller' => 'comments', 'action' => 'add', $news_id)));?>
<?php
echo $this->Form->input('comment', array('label' => false));
?>
Form->end(__('Submit', true));?>
Code in view:
<?php echo $this->element('add_comment', array('news_id' => $news['News']['id'])); ?>
move the code from your view into an element and then call the element all over the place.
Related
I'm trying to get a link in cakephp to point to a specific place on another page but what I imagined would work doesn't.
I'm using
<a name="Telstra"></a>
Telstra
Can anyone tell me the correct way?
You need to use the built-in link function of CakePHP. try to use this code.
<?php echo $this->Html->link('NameOfLink', array('controller' => 'ControllerName', 'action' => 'FunctionName/#Telstra')); ?>
See this url from cakephp docs:
http://book.cakephp.org/2.0/en/core-libraries/helpers/html.html#HtmlHelper::url
For creating link only
echo $this->Html->url(array(
"controller" => "posts",
"action" => "search",
"#" => "first"
));
and in your case for creating the link with anchor tag,
echo $this->Html->link('Telestra',array(
"controller" => "sponsors",
"action" => "index",
"#" => "Telstra"
));
If you want to load on a specific part of the page you must use ID on the element of your target rather than Name.
Example:
//Your target element on a page
<a name="Telstra"></a>
//URL that will redirect you to your target element.
Telstra
Or might as well code it the cakePhp style. :)
//URL
<?php echo $this->Html->link('Telstra', array('controller' => 'YourController', 'action' => 'YourFUnction', '#TargetElement'));
//Target Element
<a name="Telstra"></a>
Good Luck!
you need put a route in the folde of configuration /app/config/routes.php:
Router::connect('/index', array('controller' => 'yourController', 'action' => 'index'));
or
Router::connect('/index/*', array('controller' => 'yourController', 'action' => 'index'));
I think the parameter you're meaning to put is 'id' and not 'name' and I think you can just put #idName instead of the whole link.
For example:
<a id="Telstra"></a>
Telstra
For more information you can check here: http://www.w3schools.com/html/html_links.asp
I am a Cakephp beginner
I am creating breadcrumbs from my website, I am not sure what is the difference between using HTML helper and Breadcrumb helper, Html helper seems easier to use, but it seems like I have to add each crumb to each page manually, please correct me if i am wrong.
When I was trying to use the Html helper, I put
<?php echo $this->Html->getCrumbs(' > ', array( 'text' => 'Customers', 'url' => array('controller' => 'customers', 'action' => 'index'), 'escape' => false)); ?>
in my index.ctp
then, i put
<?php $this->Html->addCrumb('Add customer', 'Customers/add'); ?>
in add.ctp
the breadcrumb "Customer" appears on the index.ctp, but when i go to the add.ctp page, no breadcrumb is shown.
I tried putting
echo $this->Html->getCrumbs(' > ', 'Home');
in default.ctp, then the "Add customer" appears after the Home crumb
How can I make it so that, on the add.ctp , the breadcrumb shows like this:
Customers > Add customer
You should take a better look at the documentation in the cookbook, usually it's all there.
Steps are:
Create a place for cakePHP to display the breadcrubs in your layout template, not index.ctp using something like echo $this->Html->getCrumbs(' > ', 'Home');. You should find your default layout in View/Layout/default.ctp.
Add a trail in each page with something like $this->Html->addCrumb('Add customer', 'Customers/add'); for View/Customers/add.ctp
This is all done by the HTML helper, there isn't any different (official) breadcrumb helper.
I have two form on a single page: login form and register form. When I submit the register form, it validates both: form fields that are in login and registeration. How can I handle it if both form have the same model (user model)
Register form
<?php echo $this->Form->create('User', array('url' => array('controller' => 'users', 'action' => 'add'))); ?>
<?php echo $this->Form->input('username', array('label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('email', array('label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('password', array('label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('confirm_password', array('type' => 'password', 'label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->submit(__('Submit', true), array ('class' => 'reg_button', 'div' => false));
echo $this->Form->end();?>
and Login form is below
<?php echo $this->Form->create('User', array('controller' => 'users', 'action' => 'login'))?>
<?php echo $this->Form->input('User.username',array('label'=>false,'div'=>false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('User.password',array('label'=>false,'div'=>false, 'class' => 'reg_input'));?>
<?php echo $this->Form->submit(__('Log in', true), array ('class' => 'reg_button', 'div' => false)); ?>
<?php echo $this->Form->end();?>
When I submit registration form it validates both forms, I want to validate only the registration form.
How can I handle that?
I've come up with a "solution" (I find the approach dirty, but it works) for a different question (very similar to this). That other question worked with elements and views, though. I'll post the entire solution here to see if it helps someone (though I rather someone else comes with a different approach).
So, first: change the creation names for the two forms.
//for the registration
<?php echo $this->Form->create('Registration',
array('url' => array('controller' => 'users', 'action' => 'add'))); ?>
//for the login
<?php echo $this->Form->create('Login',
array('controller' => 'users', 'action' => 'login'))?>
The forms should work, look and post to the same actions, so no harm done.
Second step: I don't have your action code, so I'm going to explain what needs to be done in general
public function login() {
if ($this->request->is('post')) {
//we need to change the request->data indexes to make everything work
if (isset($this->request->data['Login'] /*that's the name we gave to the form*/)) {
$this->request->data['User'] = $this->request->data['Login'];
unset($this->request->data['Login']); //clean everything up so all work as it is working now
$this->set('formName', 'Login'); //we need to pass a reference to the view for validation display
} //if there's no 'Login' index, we can assume the request came the normal way
//your code that should work normally
}
}
Same thing for the registration (only need to change 'Login' to 'Registration').
Now, the actions should behave normally, since it has no idea we changed the form names on the view (we made sure of that changing the indexes in the action). But, if there are validation errors, the view will check for them in
$this->validationErrors['Model_with_errors']
And that 'Model_with_errors' (in this case 'User') won't be displayed in the respective forms because we've changed the names. So we need to also tweak the view. Oh! I'm assuming these both forms are in a view called index.ctp, for example, but if they are on separate files (if you're using an element or similar) I recommend add the lines of code for all the files
//preferably in the first line of the view/element (index.ctp in this example)
if (!empty($this->validationErrors['User']) && isset($formName)) {
$this->validationErrors[$formName] = $this->validationErrors['User'];
}
With that, we copy the model validation of the User to the fake-named form, and only that one. Note that if you have a third form in that view for the same model, and you use the typical $this->form->create('User'), then the validation errors will show for that one too unless you change the form name for that third one.
Doing that should work and only validate the form with the correct name.
I find this a messy approach because it involves controller-view changes. I think everything should be done by the controller, and the view shouldn't even blink about validation issues... The problem with that is that the render function of Controller.php needs to be replaced... It can be done in the AppController, but for every updgrade of Cakephp, you'll have to be careful of copying the new render function of Controller.php to the one replacing it in AppController. The advantage of that approach, though, is that the "feature" would be available for every form without having to worry about changing the views.
Well, it's just not that maintainable anyway, so better to leave it alone if it's just for this one case... If anyone is interested on how to handle this just in the controller side, though, comment and I'll post it.
You can duplicate your model and change his name and define $useTable as the same table name.
Example :
class Registration extends AppModel {
public $useTable = 'users';
You define the action in form->create like Nunser for your login form
<?php
echo $this->Form->create('User',array(
'url' => array(
'controller' => 'Users',
'action' => 'login',
'user' => true
),
'inputDefaults' => array(
'div' => false,
'label' => false
),
'novalidate'=>true,
));
?>
and your registration form
<?php
echo $this->Form->create('Registration',array(
'url' => array(
'controller' => 'Users',
'action' => 'validation_registration',
'user' => false
),
'inputDefaults' => array(
'div' => false,
'label' => false
),
'novalidate'=>true,
));
?>
In your controller define a method for registration validation and the most important define the render
public function validation_registration(){
$this->loadModel('Registration');
if($this->request->is('post')){
if($this->Registration->save($this->request->data)){
--- code ---
}else{
--- code ---
}
}
$this->render('user_login');
}
Sorry for my english ! Have a nice day ! :D
The create method on your login form is missing the 'url' key for creating the action attribute. I tried to re-create this once I fixed this and could not. Maybe that will fix it?
I'm new with CakePHP and I need your assistance. I need to display a specific widget which is in the form of a div on a specific page i.e. my homepage and disable on the rest of the pages. Essentially I have been able to specifically display specific divs based on log in status as indicated below:
<?php if (!$this->Session->read('Auth.User.id')): ?>
<div class="register link right <?php if ($active == 'register') echo 'active'; ?>"><?php echo $html->link('Register', array('controller' => 'users', 'action' => 'register')); ?></div>
<div class="login link right <?php if ($active == 'login') echo 'active'; ?>"><?php echo $html->link('Login', array('controller' => 'users', 'action' => 'login')); ?></div>
<?php else: ?>
<div class="logout link right"><?php echo $html->link('Logout', array('controller' => 'users', 'action' => 'logout')); ?></div>
<div class="myaccount link right <?php if ($active == 'myaccount') echo 'active'; ?>"><?php echo $html->link('My account', array('controller' => 'account', 'action' => 'summary')); ?></div>
<?php endif; ?>
I was asking for any help with regards to displaying a specific div based on the selection of my homepage.
The pseudocode below indicates my the line of thinking I'm taking to solve this issue:
<?php if (the selected page is homepage or default.ctp)?>
// set the display property for the desired div to none
<?php else: ?>
// do not set the display property for the desired div to none
<?php endif; ?>
In cakephp you cannot use directly $this->Session->read('Auth.User.id') in your view is bettere to do this into your controller:
$this->set('authUser', $this->Auth->user());
and into your view
if (!$authUser)
{
//not logged
}
else{
//logged
}
And if you wanna check which is the page you can try something like that
echo Router::url($this->last, true);
Is what you want?
In your controller you can define something like:
$this->set('pageName', $pageName);
Then you can do in your view:
$class='';
if($pageName=='homepage') {
$class=' hide';
}
echo $this->Html->div($class, 'your content here');
Also think about why you need a structure of this in your view. Maybe you can just not supply the content if it is not needed to the view? So you make the decision in your controller. That makes most of the time more clean views and the smallest amount of data needed in your view.
I have this in my view
<?=$this->Form->create('Company')?>
<?=$this->Form->input('Company.company_category_id')?>
<?=$this->Form->input('Company.county')?>
<?=$this->Form->input('Company.name')?>
// Here i intend to insert all model fields in order to export them
<?=$this->Form->input('ExportField.company_category_id', array('label' => 'Categorie', 'type' => 'checkbox', 'options' => null))?>
// ...
<?=$this->Form->end('Submit')?>
My problem is that the helper is "autoMagically" consider that ExportField.{field} as being the form's main model field (Company in this case).
I can use a workaround to resolve this, but I want to know if I can force it somehow maintaining this approach.
Thank's!
You are declaring model in:
<?=$this->Form->create('Company')?>
As cake doc says, All parameters are optional. Try with:
<?=$this->Form->create()?>
You can use the following:
<?php echo $this->Form->create(null, array('controller' => 'controller_name', 'action' => 'action_name')?>
<?php echo $this->Form->input('Company.company_category_id')?>
<?php echo $this->Form->input('Company.county')?>
<?php echo $this->Form->input('Company.name')?>
// Here i intend to insert all model fields in order to export them
<?php echo $this->Form->input('ExportField.company_category_id', array('label' => 'Category', 'type' => 'checkbox'))?>
// ...
<?php echo $this->Form->end('Submit')?>
If you would use ModelName as null as a first argument in $this->Form->create() method, then you can easily achieve the same you needed.