CakePHP prepopulate form with data from a link - php

Assume I'm in my items controller.
Ok say I am in my view action (the url would be something like /items/view/10012?date=2013-09-30) which lists a list of items that belongs to a client on a given date.
I want to link to add a new item. I would use the htmlhelper like so:
echo $this->Html('action'=>'add');
In my add action I have a form which has fields like client_id and item_date.
When I'm in my view action I know these values as I am viewing the items for a specific client on a specific date. I want to pass these variables to my add action so it will prefill those fields on the form.
If I add a query string in my link ('?' => array('client_id'=>$client_id)) it breaks the add action as it will give an error if the request is not POST. If I use a form->postLink I get another error as the add action's POST data must only be used for adding the record, not passing data to prefill the form.
I basically want to make my link on the view page pass those 2 variables to the add action in the controller so I can define some variables to prefill the form. Is there a way to do this?
Here is my add controller code. It may differ in content a bit from my question above as I have tried to simplify the question a bit but the concept should still apply.
public function add(){
if ($this->request->is('post')) {
$this->Holding->create();
if ($this->Holding->save($this->request->data)) {
$this->Session->setFlash(__('Holding has been saved.'), 'default', array('class' => 'alert alert-success'));
return $this->redirect(array('action' => 'index'));
}
$this->Session->setFlash(__('Unable to add your holding.'), 'default', array('class' => 'alert alert-danger'));
}
$this->set('accounts', $this->Holding->Account->find('list'));
$sedol_list = $this->Holding->Sedol->find('all', array(
'fields' => array(
'id', 'sedol_description'
),
'recursive' => 0,
'order' => 'description'
)
);
$this->set('sedols', Hash::combine($sedol_list, '{n}.Sedol.id', '{n}.Sedol.sedol_description') );
}

Why not use proper Cake URL parameters?
echo $this->Html->link('Add Item', array(
'action' => 'add',
$client_id,
$item_date
));
This will give you a much nicer URL like:
http://www.example.com/items/add/10012/2013-09-30
And then in your controller, you modify the function to receive those parameters:
public function add($client_id, $item_date) {
// Prefill the form on this page by manually setting the values
// in the request data array. This is what Cake uses to populate
// the form inputs on your page.
if (empty($this->request->data)) {
$this->request->data['Item']['client_id'] = $client_id;
$this->request->data['Item']['item_date'] = $item_date;
} else {
// In here process the form data normally from when the
// user has submitted it themselves...
}
}

Related

CakePHP: Check for line item in iframe before submitting parent form

The app I'm working on (an order form) allows the user to enter multiple sub-records within an iframe. These sub-records are joined to the main record via a foreign key.
main_records line_items
----------- ----------
id int(11) PK etc. id int(11) PK etc.
main_record_id (FK)
I need the app to check whether at least one line item exists within this iframe before form submission. I would like to take advantage of the $validate functionality within the model, but I'm unsure how to proceed. Here's what I've tried in the Main model:
App::uses('AppModel', 'Model', 'LineItem');
public $hasMany = array(
'LineItem' => array(
'className' => 'LineItem',
'foreignKey' => 'main_record_id',
'dependent' => false
)
);
public $validate = array(
'main_record_id' = array(
'allowEmpty' => false,
'rule' => 'checkForLineItem',
'message' => 'You must enter at least one line item!'
)
);
//Check to make sure there is at least one line item before saving changes/submitting for approval
function checkForLineItem($id) {
$lines = $this->LineItem->find('all', array(
'fields' => array('LineItem.main_record_id'),
'conditions' => array('LineItem.main_record_id'=>$id, 'LineItem.deleted_record'=>0))
);
if(!empty($lines)) {
return true;
} else {
return false;
}
}
I also track whether the line item has been deleted. If it has, then it is not added to $lines.
I know I can accomplish this in the Controller, but as far as I know, that would require the form to post, and the user would lose any changes upon postback (I haven't yet implemented jQuery on this form). Am I on the right track with how to do this? What changes should I make to get this to work?
Your code looks about right, but validation indeed happens in form submit. If you want to check it prior to that you have to do in JavaScript (jquery). E.g. create a controller action that return if there are existing line items for given main record id and call it via AJAX.

Displaying search results on a separate page

I have a SilverStripe site with some code to display a search form. The for allows you to search for something based on 3 things. Problem is, I'm not sure how to get the results to display correctly on a separate page.
My code:
class InstitutionSearchPage extends Page {
}
class InstitutionSearchPage_Controller extends Page_Controller {
private static $allowed_actions = array(
'Search'
);
public function Form() {
$fields = new FieldList(array(
DropdownField::create('DegreeType', 'Degree', QualificationType::get()->filter('ParentID', 0)->map()),
DropdownField::create('Course', 'Course', Qualification::get()->map()),
DropdownField::create('City', 'City', City::get()->map()),
));
$actions = new FieldList(array(
FormAction::create('Search')->setTitle('Find a College')
));
$validator = ZenValidator::create();
$validator->addRequiredFields(array(
'DegreeType' => 'Please select a Degree',
'Course' => 'Please select a course',
'City' => 'Please select a city',
));
$form = new Form($this, 'Search', $fields, $actions, $validator);
$form->setLegend('Criteria')->addExtraClass('form-horizontal')->setAttribute('data-toggle', 'validator');
// Load the form with previously sent data
$form->loadDataFrom($this->request->postVars());
return $form;
}
public function Search() {
return array('Content' => print_r($this->getRequest()->postVars(), true));
}
}
It seems to be displaying results on the same page but gives me a bunch of weird data. For example, I got this when I tested the form: Array ( [DegreeType] => 53 [Course] => 1 [City] => 1 [SecurityID] => 02718d0283e27eeb539eff19616e0b23eadd6b94 [action_Search] => Find a College )
The result is supposed to be an organized list of colleges (along with other data).
I guess that what you are seeing is expected behavior, as the form will post to the Search function, and that one is just returning a print_r of an array with the post vars which will be picked up by the template.
Otherwise, there are a lot of things not corresponding with the Silverstripe default way to handle forms. Please take a look here and change your form accordingly: https://docs.silverstripe.org/en/3.4/tutorials/forms/
For example, give the form the same name as your function (or in your case, change the function name to the form name). Then implement the action function.

Yii2:redirect to specific action from submit button

I am not able to redirect a custom form to specific action.
What I am trying is
<?= Html::submitButton( 'delete-selected' ,['class' => 'btn btn-primary']) ?>
here delete-selected is my custom action in controller appointment.
I have also tried like this:
public function actionDeleteForm()
{
return $this->render('delete');
return $this->redirect(['delete-selected']);
}
public function actionDeleteSelected()
{
Appointment::deleteAll(['doctor_name' =>4]);
return $this->redirect(['index']);
}
What I am trying to do is actually delete some records using the form. The form name is delete having a select drop-down field.
I want to post the data to action deleteselected and use the $_POST variable in the delete query.
How can I do this?
Thanks.
Any submit button that you put on your form will submit to the url specified in the action parameter of the form. If you haven't specified one, then Yii will use the current controller/action of the form. If you want to override this behavior, then you will need to specify an action for the form. e.g.
$form = ActiveForm::begin([
'action' => 'appointment/delete-selected'
]);
in actionDeleteForm you have
return $this->render('delete');
before
return $this->redirect(['delete-selected'])
this second instruction will never be executed because you have already made a return to the function and then control has already been returned to the caller
This is somewhat a note for Joe Miller's answer. If you are supposed to override the form's action with an action of a controller, make sure you make the value of 'action' as an array:
$form = ActiveForm::begin([
'action' => ['appointment/delete-selected']
]);
It will treat the action as a route to action delete-selected in controller appointment.

laravel using data in view that has been sent from controller

I always send data from view like this ->with(array('message'=>'there is an error ')), and it works.
I want to allow the customer to edit some information, so when he/she clicks on the the edit like, this function in a controller is being executed:
public function edit($id)
{
$waitingTimes = WaitingTimes::find($id);
return View::make('waitingtimes.edit')->with(array(
'verticalMenu' => 'none',
'verticalMenuTab' => 'none',
'data' => $waitingTimes
));
}
So later in the view, I should be able to say this:
$data->startTime, $data->endTime, $data->id, $data->restaurant_id
but every time I do that, I got $data->startTime printed on the browser, but I should have got the value of the startTime attribute.
This is my view:
<div class="oneInfo">
{{ Form::text('startTime', '', array('class' => 'time ui-timepicker-input', 'id' => 'startTime', 'autocomplete' => 'off'))}}
<span class="errorMessage">
<?php
echo $errors->first('startTime');
?>
</span>
</div>
The view has an input text and I want that input text to be filled with the data that has been sent from the controller.
how could I do that please?
The Form::text() method's second parameter allows you to pass it the value to be assigned to the input element. Your form input declarations are currently setting the value of the inputs to an empty string.
The best way to handle this would be to replace your empty string with $value=null.
{{ Form::text('startTime', $value=null, array('class' => 'time ui-timepicker-input', 'id' => 'startTime', 'autocomplete' => 'off'))}}
This will automatically replace the value with your models data or the data input by the user (should validation fail and you redirect back to the form).
From the looks of things, you could also make things a bit easier for yourself by using form model binding to bind the WaitingTimes model to your form.

How to pass id from "input select" to its action?

Hello guys let me explain:
$this->view->form = new Application_Form_Kursdaty();
$url = $this->view->url(
array(
'action' => 'details',
'controller' => $this->getRequest()->getControllerName(),
'module' => $this->getRequest()->getModuleName(),
'id' => #####SELECTFIELD[selected]######
),
'default',
true
);
$this->view->form->setAction($url);
I basically want my select field to redirect to page which i choose.
I dont really know what to do, please help!
If your form is instance of Zend_Form and populated with sent data then you can use $form->getValue($nameOfField) or $form->getElement($elementName)->getValue()
Here's your problem!
The value of the select element is not set until after the submit button is pressed. So you don't have access to the value until after the post.
So a way to do what you want to do in PHP would be to use 2 actions. The first action takes the post and performs the validation then you could take the data and forward it to the action of your choice depending on the value of that select.
public function formAction() {
$form = new Form();
$form->setAction('/url');
if ($this->_request->isPost()){
if ($form->isValid($this->_request->getPost()){
$select = $form->getValue('select');
switch($select) {
case ('value1') :
$this->_forward('action');
break;
case ('value2') :
$this->_forward('action');
break;
default :
$this->_forward('action');
}
}
} else {
$this->view->form = $form
}
This might be the kind of work flow you would look at for the first action. You'll need to work out the details but a _forward should be doable in this instance as it should perform the redirect without losing the current request object.

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