I've got a template X and I want this to become the default style. Meaning I don't want to do this:
echo $this->Form->input('phone',array('class'=>'form-control'));?>
I want to just do this:
echo $this->Form->input('phone');
Do I edit the FormHelper, create a new helper, or do I rewrite the styles in cake's style.css with the those from template X?
The best and cleanest way is to override the FormHelper. So you create a new helper, and then in the respective controllers you can add:
public $helpers = array('Form' => array('className' => 'newFormHelper'));
Look here for more references: http://blog.nlware.com/2012/02/07/cakephp-2-0-how-to-extend-the-formhelper/
You can specify inputDefaults in your create method.
$this->Form->create('Model', array(
'inputDefaults' => array(
'class' => 'form-control'
)
);
There are more options you could use if you wish, it sounds like this is for bootstrap so you might want to checkout BoostCake
https://github.com/slywalker/cakephp-plugin-boost_cake
Related
I am working on CakePHP and I have a URL http://admin.example.com/Pages .
How can I create http://admin.example.com/Pages.html ? Is there any solution or component to solve this issue?
According to CakeBook , You can define extensions you want to parse in routes
E.g. Write the code below in app/Config/routes.php
Router::parseExtensions('html');
This will allow you to send .html extenstion in routes(url)
Not create a link
E.g:
$this->Html->link('Link title', array(
'controller' => 'pages',
'action' => 'index',
'ext' => 'html'
));
When you click that link you will get url in browser something like this
http://admin.example.com/pages/index.html
Do you really need this? CakePHP is automatically render view files from View folder
You can create (if its PagesController and index methond) index.ctp in app/View/Pages folder and it will be automatically render.
You can use file_get_contents function to read files. Something like this:
// in controller
function index() {
$content = file_get_contents('path/to/file.html');
echo $content;
die();
}
how can i access method $this->someModel->find('all') when im in different model form example:
class DevicesController extends AppController {
public function add(){
$departments = $this->Department->find('all', array(
'fields' => array('id', 'mac')
));
$this->set(compact('departments'));
.....
.....
}
right now there is error because $this doesn't "see" Department
what i need to do to make it happen.
If the models are in fact related (and properly set up so in the models), it would be
$this->Device->Department->...
otherwise use loadModel() as documented.
If models are not related, try this way:
$this->loadModel('Department');
$departments = $this->Department->find('all', array(
'fields' => array('id', 'mac')
));
$this->set(compact('departments'));
u can't just use loadModel() because it is controller's method.
if your models aren't related u can do this like that
$this->ModelName = ClassRegistry::init('ModelName');
$this->ModelName->find...
if they are related just use
$this->ModelName->find...
Try this code for loading different models on a controller.
public $uses = array('ModelName1', 'ModelName2')
$this->ModelName1->find()........
I am making blog and url route is like this-
Router::connect('/blog/c/:catid/*',
array('controller' => 'blogarticles', 'action' => 'index'));
it works well with url as- /blog/c/3/other-articles
but when i use paginator in view as
echo $this->Paginator->numbers();
it generates url as- /blogarticles/index/other-articles/page:2
What changes should in make in paginator to generate proper url.
Please suggest possible solution , Thanks in advance
This should solve your problem:
$this->Paginator->options(
array(
'controller' => 'blog',
'action' => 'c',
$catid,
$title
)
);
The trick is to pass blog as the controller and c as the action, and all other variables (NOT LIMITED TO $catid and $title) as additional parameters, sequentially!
NOTE: I assumed here that you have "set" $catid and $title from your Controller, to the current "category id" and "title" respecting. I also assumed that your URLs are always in the format: /blog/c/:catid/:title
You may also want to view my answer to a similar question: https://stackoverflow.com/a/25097693/2862423
What you want is to set the options for the PaginatorHelper for that view:
<?php $this->Paginator->options(array('url' => '/blog/c/3/other-articles')); ?>
CakePHP Book Section on the PaginatorHelper Options Function: http://book.cakephp.org/2.0/en/core-libraries/helpers/paginator.html#modifying-the-options-paginatorhelper-uses
I'm working with CakePHP and trying to understand the best ways to make my application consistent and logical.
Now I'm trying to working with Model data validation and handling validation errors in the view, I have a doubt on how should I do if I like to insert some link inside the returned error, for example for a forgotten password.
Is it good to use (if it's possibile) HtmlHelper inside the Model to return consistent links inside my application, or should I think about another way?
<?php
App::import('Helper', 'Html');
class User extends AppModel {
var $name = 'User';
var $validate = array (
'email' => array (
'checkEmail' => array (
'rule' => array('email', true),
'message' => 'Email not valid message.'
),
'checkUnique' => array (
'rule' => 'isUnique',
'message' => 'This email is allready in the db, if you forgot the password, '.(string)$this->Html->link('click here', array('controller' => 'users', 'action' => 'password-recover')).'.'
)
)
// the rest of the code...
This doesn't work because it seems I can't chain the message string with HTML string.
Does exist e smartest way to do that, or should I simply insert the html string without the HtmlHelper?
If you really want HTML in your validation messages CakePHP provides a way to do this, no breaking Cake, no writing a lot of code.
In your $validation just use whatever HTML you would like to have presented to the user.
In your view when you create your FormHelper::input($fieldName, array $options) pass the following array to $options:
$options = array('error' => array(
'attributes' => array('escape' => false)
))
See this page to learn more about the $options['error'] ...options.
Alternatively, if you want all inputs with no HTML escaping you can pass $options['inputDefaults'] when you create the form.
this is a difficult topic because
you might need to break MVC
validation is as in your case usually in $validate and cannot contain dynamic stuff
for 1)
you can also use Router::url() with manual HTML
you can use BBcode or pseudo-markup and translate this into real links in the view/element of the flashmessage
for 2)
use __construct() and $this->validate to use dynamic elements if needed
In PHP, properties of a class (such as $validate) have to be initialized with constant values.
<?php
class User extends AppModel {
public $validate = array(
'email' => array(
'checkUnique' => array(
'rule' => array('isUnique'),
'message' => 'This email address has already been claimed, possibly by you. If this is your email address, use the reset password facility to regain access to your account'
),
),
);
public function beforeValidate($options = array()) {
$this->validate['email']['checkUnique']['message'] = String::insert(
$this->validate['email']['checkUnique']['message'],
array('link' => Router::url(array('action' => 'password-recover')))
);
return true;
}
You are making it hard on yourself. The helpers are not accessible in the model and controller. And for good reason: the M and C shouldn't be concerned with the V.
There are ways to do exactly as you want (but involves considerably more code). Since you ask for the smartest way: What's wrong with just echo the reset password link in the view, after the login form? Just echo 'Forgot your password? '.$this->Html->link('Click here', array('controller' => 'users', 'action' => 'password-recover'));
I don't agree on breaking the MVC logic. I also tried all the array('escape' => false) possible ways (in Form->input, in Form->error and even in the model) and none of them worked with me! (cakephp 2.0)
"Anh Pham" answer is the easiest and simplest way. In addition to that, I returned empty error message from model validation ('errorMessage' => false ; doesn't work in cakePhp 2.0).
Because I wanted to pass a variable to the view to build the link there (MVC), in the controller I check if the field is invalidated:
$invlaidFields = array_keys($this->Model->validationErrors();
if ( in_array('myField', $invalidFields) ){
...
}
In the view, I check if the field was invalidated, I then echo my error message giving it class error-message so it looks the same as the rest error messages.
if ($this->Form->('myFields')) { ... echo '<span class="error-message">error message'. $this->Html->link(...).'</span>'; }
Hope it helps somebody out there.
P.S. It's always a good practice to mention what cakePHP version you are using...
To cakephp2 you can use the following:
//model validation
'company' => array('notempty' => array('rule' => array('notempty'),'message' => "select one company o send email to contact",),)
//front
<?php if ($this->Form->isFieldError('Register.company')): ?>
<span class="text-danger"><?php echo $this->Form->error('Register.company', null, array('escape'=>false)); ?></span>
<?php endif; ?>
have custom pagination in my cakephp view. before that i made some custom routing changes.
problem is that links leads to pages like
http://localhost/myapp/foos/view/news/page:2
instead of
http://localhost/myapp/news/page:2
so, part with foos/view/ not have to be part of the link.
tried to change url with several custom options, like
$this->Paginator->options(array('url' => $this->passedArgs));
but no luck, because i always have foos/view/ in url.
can you help me how can i get rid of that foos/view?
thank you very much in advance!
UPDATE: i manage to do "something", but not enough, by adding following lines:
$options = array('url'=> array('controller' => 'news' ) );
$paginator->options($options);
now, my link looks like:
http://localhost/myapp/news/index/page:2
how can i get rid of that "index" in url?
The following line is more about passing various pieces of URL information to the view:
$this->Paginator->options(array('url' => $this->passedArgs));
I think what you want to look into is the helper declaration in your Controller:
var $helpers = (
'SomeHelper',
'AnotherHelper',
'Paginator' => array(
'url' => array('controller'=>'news')
)
);
If you want finer control of a custom route like the one you have then try
'url' => '/news'
I haven't used PaginatorHelper in a while - so I could be egregiously on the wrong track - but I believe that's a good start.
Also, take a look at the Paginator Helper page for where it mentions $options and then take a look at Router::url() as the former page recommends.
I had a case where I am working on a project using CakePHP 2.1 (This thread is tagged as 1.3) with a dynamic admin route to display pages like this:
Router::connect('/admin/main/*', array('controller' => 'adminPages', 'action' => 'display'));
With a query string parameter, that produces a dynamic url like this: http://mydomain.com/adminPages/main/...?page=1
The link route, was incorrect for our needs and found I could alter the url directly by using this:
$this->Paginator->options(array(
'url' => array(
'controller' => 'admin/main/my-display',
)
));
For me it made a link: http://mydomain.com/admin/main/my-display?page=1 - which was the correct url we were looking for. If I used a string, as described above, it appends itself to the url, like: http://mydomain.com/adminPages/main/.../admin/main/my-display?page=1
In view :
<?php
$this->Paginator->options(array('url' => array('controller' => '','action' =>'your-custom-url')));
?>
In routes.php :
<?php
Router::connect('/your-custom-url/*', array('controller' => 'Controller', 'action' => 'function'));
?>