Quick questions
Is there Way to Append a textarea using Cakephp
view code:
<?php echo $ajax->link(
$qnote['Qnote']['subject'],
array(
'controller' => 'qnotes',
'action' => 'view', $qnote['Qnote']['id']
),
array( 'update' => 'Textarea_id')
);
?>
controller Code:
function view($id = null) {
$this->Qnote->id = $id;
$this->set('qnote', $this->Qnote->read());
}
the above code Pulls the information But Replaces the entire Text in the textarea.
Is there a way I can just append the textarea with out removing the existing text in the textarea
if possible can somebody point me in the right direction please.
You can try saving the result of your AJAX request to a hidden field and then having it execute and on page javascript function to simply slap the values from the hidden field to the visible text area.
The AJAX helper lets your specify callback functions, so something like this should work:
<?php echo $ajax->link(
$qnote['Qnote']['subject'],
array(
'controller' => 'qnotes',
'action' => 'view', $qnote['Qnote']['id']
),
array( 'update' => 'Textarea_id_hidden', "complete" => "concat_fields()" )
);
?>
and then the JavaScript in the View
<script type="text/javascript">
function concat_fields() {
$('#Textarea_id').val( $('#Textarea_id').val() . $('#Textarea_id_hidden').val() );
}
</script>
Note: My JavaScript example above assumes you're using JQuery, changes will need to be made if you're not.
Related
I have a function in my application which the users can upload files to the webserver. Then these uploaded files will appear in another page wherein another type of users can click on the link. Once the link is clicked, a new tab will open and the file will be shown.
But I can't seem to do it. Using the 'target' => '_blank' is not working, or I may have put it on the wrong part of the code.
In my case, when you click on the link, the file will load on the same tab.
Here's my code:
<?php
echo $this->Html->link($staff_uploads['StaffUpload']['title'], array(
'controller' => 'websites',
'action' => 'view',
'target' => '_blank',
$staff_uploads['StaffUpload']['iduploads']
)
);
?>
Thank you in advance!
The correct code is:
<?php
echo $this->Html->link($staff_uploads['StaffUpload']['title'], array(
'controller' => 'websites',
'action' => 'view',
$staff_uploads['StaffUpload']['iduploads']
), array('target' => '_blank')
);
?>
And do read the documentation as burzum has suggested.
Read the documentation.
HTML attribute options go into the 3rd argument of the link() method, not the second which is the URL as string or array.
Problems like this can be simply resolved by using the documentation.
I have a simple form in a a view and I am trying to access the $this=>passedArgs but it is coming back empty.
I am actualy trying to use the cakeDC search plugin which uses the $this=>passedArgs. It must be something simple I have not done to get the results from the form submit.
find view
<?php
echo $this->Form->create('Member', array(
'url' => array_merge(array('action' => 'find'), $this->params['pass'])
));
echo $this->Form->input('name', array('div' => false));
echo $this->Form->submit(__('Search'), array('div' => false));
echo $this->Form->end();
?>
Controller
public function find() {
debug($this->passedArgs);
exit;
}
I have tried $this->request->params
array(
'plugin' => null,
'controller' => 'members',
'action' => 'find',
'named' => array(),
'pass' => array(),
'isAjax' => false
)
I have add method get to the form.
This question has been asked before but their solution of having lower cases in the public $uses = array('order', 'product'); when it should be public $uses = array('Order', 'Product'); did not work.
Cakephp version 2.3.5
Thanks for any help
Update:
I have set my form to method get and this is the url:
http://localhost/loyalty/members/find?name=searchtext
I have removed the plugin and I still do not get anything $this->passedArgs, but I now get data for $this->request->data['name']. Once I put public $components = array('Search.Prg'); I get noting again for $this->request->data['name'].
I have tried again $this->Prg->parsedParams() with the Search plugin and I just get array()
The documentation is pretty clear on that.
You cannot just debug something that has not been set yet.
So including the plugin itself (and its component) is not enough.
From the readme/documenation:
public function find() {
$this->Prg->commonProcess();
$this->paginate['conditions'] = $this->ModelName->parseCriteria($this->passedArgs);
$this->set('...', $this->paginate());
}
Note the commonProcess() call which then only makes passedArgs contain what you need.
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 ---
}
}
$this->render('user_login');
}
Sorry for my english ! Have a nice day ! :D
The create method on your login form is missing the 'url' key for creating the action attribute. I tried to re-create this once I fixed this and could not. Maybe that will fix it?
Hia
I am using the following in a custom module to customise the form on the node/add/baby page for my BABY content type, however i want to make some slight modifications to this page when its the node/nid/edit page for the baby content type. is this possible?
<code>
<?php
function concept_theme() {
return array(
'baby_node_form' => array(
'arguments' => array(
'form' => NULL,
),
'template' => 'templates/baby-node-form',
'render element' => 'form',
),
);
}
?>
</code>
thank u
In template.php
function my_theme_name_theme($existing, $type, $theme, $path)
$hooks['baby_node_form']=array(
'render element'=>'form',
'template' =>'templates/node--baby-edit',
);
return $hooks;
}
In node--baby-edit.tpl.php
Hide fields like this:
<?php hide($form['title']); ?>
Manipulate fields like this:
<?php print render($form['field_image']); ?>
Use this to print all/rest of content:
<?php print drupal_render_children($form); ?>
Clear cache.
You can alter the node creation form by implementing this hook,
something similar to this:
hook_form_alter(&$form, &$form_state, $form_id) {
if($form_id == 'node_baby_form') {
//do modification to form array $form
}
}
Or if your node is defined by hook_node_info (which I believe is not the case), just changing the elements in hook_form()
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; ?>