in CakePHP, when applying the debug function on a User model :
debug($user);
we get a result that hide the login or password.
'User' => array(
'login' => '*****',
'id' => (int) 2,
'pwd' => 'fjiogjfdlmgjdomngdjm',
'avatar' => null,
'prenom' => 'Fake',
'nom' => 'Admin',
'email' => 'blabla#domain.fr',
'i18n_code_appli' => '',
'numtel' => ''
),
How could we configure what model has that field hidden or not ?
Like you can see in the example below, as password field named 'pwd', it is not hidden, but login is.
And this is my actual case. I would want to toggle this to view 'login' and hide 'pwd' in debug mode.
As agreed in the comments of OP, I'll add my comment as an answer.
Just use var_dump() instead of debug() if you have to check the password hash. I have had the same problem a long while ago and I thought I'd used var_dump then. I guess you aren't keeping the call to debug() in the code on release so it shouldn't do any harm.
Or you could use the CakePHP Debug Kit. It has a list of all the set variables which you can collapse and extend. That way you have a clear overview of all the data in a deep multidimensional array.
Related
I have created a module that adds a field to the user profile. I used field_create_field and field_create_instance to do this. When I disable the module, I want the field to no longer show up on the user profile, but I do not want to destroy it. I want to be able to enable the module and have the field show up and the data to still be there as originally entered. Can anyone tell me how to accomplish this?
Here is the code I used to create the field:
$field = array(
'field_name' => $field_name,
'type' => 'text',
'visibility' => 1,
'category' => 'API',
);
$field = field_create_field($field);
$field_instance = array(
'field_name' => $field_name,
'entity_type' => 'user',
'bundle' => 'user',
'label' => t('API Token'),
'cardinality' => 1,
'translatable' => 0,
'description' => t('By using this API token, you agree to the site Terms and Conditions and to acknowledge that your submission does not include protected health information or personal identifiers.'),
'widget' => array(
'type' => 'text_textfield',
'weight' => 10,
),
'formatter' => array(
'label' => t('field formatter label'),
'format' => 'text_default'
),
'settings' => array(
),
);
When you have created field using drupal entity like user, node etc then on that entity crud operation automatically apply.
As you have used api to "field_create_field" field then it automatically create field using api of entity vise verse its delete field when you uninstalled module.
First tell me when you uninstall your custom module then your custom field deleted from profile.? If yes then it's difficult to handle your use case. If no then in system table of Drupal you get status of your module whether it's disable or enable if status is 0 then used form alter hook of user profile and hide field
I was not able to accomplish exactly what I wanted, but I ended up installing the field extra widgets module and hiding the field completely on the edit form. Then, I used hook menu alter to create a local task tabs and I display the field on that tab.
All,
I currently have slug setup in my app where I generate the following as links:
http://www.domain.com/article/my-first-news-article.
And this is the current route that I use to accomplish that in addition to the view acepting a slug instead of id:
Router::connect('/article/*',array('controller' => 'articles', 'action' => 'view'));
However, I wanted to improve that a little bit more by adding published date to the URL like the following:
http://www.domain.com/article/2012/06/27/my-first-news-article
I have the following code from the CakePHP manual, but it doesnt seem to work:
Router::connect(
'/article/:year/:month/:day/:slug',
array(
'controller' => 'articles',
'action' => 'view'
),
array(
'year' => '[12][0-9]{3}',
'month' => '0[1-9]|1[012]',
'day' => '0[1-9]|[12][0-9]|3[01]'
)
);
For the sake of this, I really dont care what date is passed to the view. I just care about the slug that is passed. Then I will use function view($slug) to find the article and display it. However the URL needs to be http://www.domain.com/2012/06/27/slug-slug-slug
Thank you....
How do the $this->Html->link() calls look like in the places where you want to use the date in the link?
Are you adding the required params there?
$this->Html->link('Article', array('year' => 2012, 'month' => 01, 'day' => 01, 'action' => 'view', 'slug' => $article['Article']['slug']));
For the sake of completeness, and because somebody else might be interested in this as well, I'll show you my solution to a similar problem:
I want to have URLS in the format
http://mydomain.com/blog/2012/06/slug-slug-slug
i.e. /blog/yyyy/mm/slug
For this, I use
// view a post by year and month and title
Router::connect('/blog/:year/:month/:title/*', array(
'controller' => 'posts',
'action' => 'view'
), array(
'year' => '[12][0-9]{3}',
'month' => '0[1-9]|1[012]',
'title' => '[a-z0-9-]+'
));
With the rules below, you can make sure the correct formats are entered. It also prevents users from entering values such as &"% as slugs.
The /* after title even allows for additional parameters, e.g. for pagination, when your blog entry has several pages (http://mydomain.com/blog/2012/06/slug/page:2).
I also realized that setting the "pass" array is not really necessary (at least in CakePHP 2.0). You can access the parameters directly via $this -> request -> params['year'], etc.
After more research and trials I was able to figure it out.
The following route will work to accept a URL like this:
http://www.mydomain.com/article/2009/06/10/my-first-article-in-mydomain-dot-com.
As a matter of fact, it doesnt actually matter how the date is setup and it will still work. It could also be,
http://www.mydomain.com/article/06/10/2012/my-first-article-in-mydomain-dot-com
Router::connect(
'/article/:year/:month/:day/:slug',
array(
'controller' => 'articles',
'action' => 'view'
),
array(
'pass' => array('slug')
)
);
Just for organization sake, I wanted to use a different table for the authentication component to check, but it doesn't quite work. While I can initially state:
$this->Auth->userModel = "CoreUsers" plus set the loginAction to my proper MVC
works to look at that table just to confirm it's there, but the login doesn't work, it only keeps returning an incorrect password. Something happens in the authentication component; I can't tell what makes it fail. When I rename my table to "Users", it works.
The other part is I'd prefer to actually use the column name of 'email' rather than 'username' since that's really what I'm using anyway.
I am just not having luck finding a complete tutorial and reference sets to do both these successfully with CakePHP 2.x. What is the way forward?
References:
Stack Overflow question How do I use a table other than "Users" for CakePHP's AuthComponent?
Stack Overflow question CakePHP - 'AuthComponent' with a different model name (not 'User')
(I had a look for answers, but I never quite got the whole answer.)
Make sure your database table "core_users" and model "CoreUser" exists.
When you setup component you can put login/logout redirect here.
var $components = array(
"Auth" => array(
'loginRedirect' => array('controller' => 'dashboard', 'action' => 'index'),
'logoutRedirect' => array('controller' => 'core_users', 'action' => 'login')
),
"Session");
Now in beforeFilter medhod you can put the following
function beforeFilter(){
$this->Auth->authenticate = array(
AuthComponent::ALL => array('userModel' => 'CoreUser', 'scope' => array("CoreUser.status" => 1), "fields" => array("username" => "email", "password" => "your_password_field"), 'Form', 'Basic'
);
}
Above example you can ignore status, if you need to pass any other info on the login verification u can use that. Now about you can remove 'Basic' if you only need form validation.
I hope this would work .
First, model names are generally singular. Are you sure you didn't mean CoreUser which would look in the core_users table?
Secondly, you can use the email field instead of username by setting the fields key on your auth setup.
public $components = array(
'Auth' => array(
'authenticate' => array(
'Form' => array(
'userModel' => 'CoreUser',
'fields' => array('username' => 'email')
)
)
)
);
See the book for more information.
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; ?>
I've done a lot of searching but havent been able to solve this yet. I have a user registration form where the user can select a number of store branches as favourites upon registration. I have this relationship set up as a HABTM.
I am able to display the store branches as multiple checkboxes, but I need to display the store name (the store branches belongs to store names) along with the branch name in the label for each checkbox. Something like:
Levi's - Canal Walk
where Canal Walk is the branch name, and Levi's is the store name (coming from a different table)
Can anyone please share some insight on how to do this?
I haven't tested this, but it might work (or help you along the right path):
$branches = $this->Branch->find('list', array(
'fields' => array('Branch.id', 'CONCAT(Store.name, " - ", Branch.name)'),
'joins' => array(
array(
'table' => 'stores',
'alias' => 'Store',
'type' => 'inner',
'foreignKey' => false,
'conditions' => array('Branch.store_id' => 'Store.id')
)
)
));
Set this in your controller, or keep things DRY by placing it in your model in a function such as findFullList()
Hope that helps!
I would do all of this in the view. If your HABTM relationship is setup correctly, a query of the Store model should work something like this:
$stores = $this->Store->findAll();
//Returns
Array([0] => Array(
'Store' => array(),
'Branch' => array(
[0] => array(),
[1] => array()...));
Then pass the $stores variable into the view, and iterate through it with a double nested loop.
foreach($stores as $store){
foreach($store['Branch'] as $branch){
//Code to output checkbox by using $branch['name']
}
}