<?= '<li>'
. Html::beginForm(['/site/logout'], 'post')
. Html::submitButton(
'Exit(admin)',
['class' => 'btn btn-warning']
)
. Html::endForm()
. '</li>'?>
This code is working, but I want to change submit button to html::a. When I try to do it, it's not working.
It is not recommend to use a link as submit button, anyhow if you want, here is code
<?php echo Html::a('Exit(admin)',$url=false,['class'=>'btn btn-warning','onclick'=>'submitForm()']); ?>
in javascript
function submitForm(){
document.getElementById('FormId').submit;
}
Related
I have a problem, i'm trying to render 2 forms (login and register) on one layout scrpt (header.phtml), every time i submit on one of the forms both actions for the controller are getting fired and i'm unsure how to fix it.
The forms are getting rendered fine within the layout, however when you click 'Login' or 'Register' on the forms the code fires in both the 'login' and 'register actions.
the header layout script snippet:-
<div class="left">
<h1>Already a member? <br>Then Login!</h1>
<?php
echo $this->action('panlogin', 'user');
?>
</div>
<div class="left right">
<h1>Not a member yet? <br>Get Registered!</h1>
<?php
echo $this->action('panregister', 'user');
?>
</div>
the action scripts (phtmls)
panregister.phtml
<div id="pan-register">
<?php
$this->registerForm->setAction($this->url);
echo $this->registerForm;
?>
</div>
panlogin.phtml
<div id="pan-login">
<?php
$this->loginForm->setAction($this->url);
?>
</div>
the user controller actions:-
class Ajfit_UserController extends Zend_Controller_Action
{
protected $_loginForm;
protected $_registerForm;
public function init()
{
$this->_loginForm = new Ajfit_Form_User_Login(array(
'action' => '/user/login',
'method' => 'post',
));
$this->_registerForm = new \Ajfit\Form\User\Registration(array(
'action' => '/user/register',
'method' => 'post'
));
}
//REGISTER ACTIONS
public function panregisterAction(){
$this->registerAction();
}
public function registerAction(){
$request = $this->_request;
if ($this->_request->isPost()){
$formData = $this->_request->getPost();
}
$this->view->registerForm = $this->_registerForm;
}
//LOGIN ACTIONS
public function panloginAction(){
$this->loginAction();
}
public function loginAction(){
$request = $this->_request;
if(!$auth->hasIdentity()){
if ($this->_request->isPost()){
$formData = $this->_request->getPost();
}
}
$this->view->loginForm = $this->_loginForm;
}
}
Please can someone with a little more knowlegde with the action('act','cont'); ?> code with in a layout script help me out with this problem.
Thanks
Andrew
While David is correct where best practices are concerned, I have on occasion just added another if() statement. Kinda like this:
if ($this->getRequest()->isPost()) {
if ($this->getRequest()->getPost('submit') == 'OK') {
just make sure your submit label is unique.
Eventually I'll get around to refactoring all those actions I built early in the learning process, for now though, they work.
Now to be nosy :)
I noticed: $formData = $this->_request->getPost(); while this works, if you put any filters on your forms retrieving the data in this manner bypasses your filters. To retrieve filtered values use $formData = $this->getValues();
from the ZF manual
The Request Object
GET and POST Data
Be cautious when accessing data from the request object as it is not filtered in any way. The router and
dispatcher validate and filter data for use with their tasks, but
leave the data untouched in the request object.
From Zend_Form Quickstart
Assuming your validations have passed, you can now fetch the filtered
values:
$values = $form->getValues();
Don't render the actions in your layout. Just render the forms:
<div class="left">
<h1>Already a member? <br>Then Login!</h1>
<?php
echo new \Ajfit\Form\User\Login(array(
'action' => '/user/login',
'method' => 'post'
));
?>
</div>
<div class="left right">
<h1>Not a member yet? <br>Get Registered!</h1>
<?php
echo new \Ajfit\Form\User\Registration(array(
'action' => '/user/register',
'method' => 'post'
));
?>
</div>
Then, whichever form gets used will post to its own action.
In a symfony 1.4 view I'm attempting to pass some html/javascript in the "attributes" parameter of the sfFormField::renderRow function:
<?php echo $form['ownership_status_id']->renderRow(array('onFocus' => 'displayHelp("<p>help text</p>");'), 'Own/Rent')?>
Unfortuantely the when the page gets rendered, all of the javascript/html output is escaped:
<select name="address[ownership_status_id]" onFocus="displayHelp("("<p>help text</p>");" id="address_ownership_status_id">
I'm not clear on how to prevent this content from being escaped, can someone help?
Try to unescape the $form variable as so:
sfOutputEscaperGetterDecorator::unescape($form);
Then call renderRow():
<?php echo $form['ownership_status_id']->renderRow(array('onFocus' => 'displayHelp("<p>help text</p>");'), 'Own/Rent'); ?>
I had to use this:
<?php echo sfOutputEscaperGetterDecorator::unescape($form['ownership_status_id']->renderRow(array('onFocus' => 'displayHelp("<p>help text</p>");'), 'Own/Rent')); ?>
I have this view:
<?php echo form_open(); ?>
<?php echo form_input('username', '', ''); ?>
<?php echo form_submit('submit', 'Submit'); ?>
<?php echo form_close(); ?>
<?php echo validation_errors(); ?>
and this controller method:
function test() {
$username = $this->input->post('username');
$this->form_validation->set_rules($username, 'Username', 'required|min_length[4]');
if ($this->form_validation->run() !== FALSE) {
echo 'Tada!';
}
$this->load->view('test');
}
but when I leave the username field blank, nothing happens. However, if I type in something in it will tell me the field is required. I've downloaded and given CI a fresh install almost ten times now, trying to load with and without different helpers etc. It's becoming really frustrating, please help.
Try using this:
$this->form_validation->set_rules('username', 'Username', 'required|min_length[4]');
The problem lies in the first parameter of set_rules, which is the field name.
The $username you're passing is basically setting the field name to validate as whatever the user puts in the input field. If you were to type 'username' into the input box, you'd see that your form validates.
Change the line
$this->form_validation->set_rules($username, 'Username', 'required|min_length[4]');
to
$this->form_validation->set_rules('username', 'Username', 'required|min_length[4]');
Maybe is the problem of this line
<?php echo form_open(); ?>
If you leave it blank it basically send back to the controller itself and calling the construct and index function only. In this case your function dealing with form processing is "test()"
try this
<?php echo form_open('yourControllerName/test'); ?> //test is the function dealing with
if it is not working try on this
<?php echo form_open('test'); ?>
Im trying to render a row field in a template with some extra styles, like this:
<?php echo $form['email']->renderRow(array('class' => 'text')) ?>
<?php echo $form['email']->renderError() ?>
The problem occurs when my form doesnt validate on this field... then it displays the error message 2 times!, i.e the renderRow renders one errorMsg string, and the renderError does it again... How can i stop renderRow from displaying the error message?
If I just do this, then it works:
<?php echo $form['email'] ?>
But in that case I cant style the field as I want....
thanks!
I am pretty sure this is also valid for 1.2. Instead of using renderRow, use something like this:
<?php echo $form['FormElementName']->renderLabel() ?> //display form element label
<?php echo $form['FormElementName']->renderError() ?> //display form element error (if exist)
<?php echo $form['FormElementName']->render(array('class' => 'text')); ?> //display form element
renderRow does them all at once.
EDIT From comments (Flask) - added ->render(array('class' => 'text'));
I'm writing my first CakePHP application and am just writing the second part of a password reset form where a user has received an email containing a link to the site and when they click it they're asked to enter and confirm a new password.
The url of the page is like this:
/users/reset_password_confirm/23f9a5d7d1a2c952c01afacbefaba41a26062b17
The view is like:
<?php echo $form->create('User', array('action' => 'reset_password_confirm')); ?>
<?php
echo $form->input('password', array('label' => 'Password'));
echo $form->input('confirm_password', array('type' => 'password', 'label' => 'Confirm password'));
echo $form->hidden('static_hash');
?>
<?php echo $form->end('Reset password'); ?>
However this produces a form like:
<form id="UserResetPasswordConfirmForm" method="post" action="/users/reset_password_confirm/8">
The problem is the user id (8 in this case) is being added to the form action. It's not really a problem here, but when I want to pass through the hash to my controller:
function reset_password_confirm($static_hash=null) {
// function body
}
$static_hash is now populated with 8 rather than the hash from the URL.
I know I could sort this out by creating the form tag myself rather than using $form->create but is there a more cakey way of doing this?
$form->create('User', array('action' => '…', 'id' => false));
Just explicitly set params you don't want passed to null or false. This is unfortunately a case where Cake tries to be a little too intelligent for its own good. ;o)
You could probably also do something like this to POST to the same URL again:
$form->create('User', $this->here);
How about passing it as a parameter instead of form data :
<?php
echo $form->create('User', array('action' => 'reset_password_confirm', $static_hash));
echo $form->input('password', array('label' => 'Password'));
echo $form->input('confirm_password', array('type' => 'password', 'label' => 'Confirm password'));
echo $form->end('Reset password');
?>
and in the controller :
function reset_password_confirm($static_hash = null) {
// Check if form is submitted
if (!empty($this->data)) {
// if it submitted then do your logic
} else {
$this->set('static_hash', $static_hash); // Else, pass the hash to the view, so it can be passed again when form is submitted
}
}
Hope this help :)