Annoying Sonata Admin Behaviour - doctrine-orm

I have gone through this problem before but cannot remember how I fixed it, I guess it's been a long day.
I have a field that is nullable in my doctrine definition but is rendered as required on the Sonata Admin entry form. I don't want to turn off HTML validation therefore I am looking for a solution to get sonata admin to render the fields correctly.
I have tried to google but I have not been lucky. Any help will be appreciated

Have you tried setting the required option to false in the configureFormFields function of your entities admin class?
This is an example of one of my classes, where I use it to set it explicitly to true:
class MenuAdmin extends Admin
{
protected function configureFormFields(FormMapper $formMapper)
{
$formMapper
->add('name', null, [])
->add('renderSubPages', null, [])
->add(
'pages', 'sonata_type_model', [
'class' => 'ApplicationSonataPageBundle:Page',
'property' => 'name',
'required' => true,
'multiple' => true
]
);
}
}
You could try setting the option in your class like this:
class EntityAdmin extends Admin
{
protected function configureFormFields(FormMapper $formMapper)
{
$formMapper
->add('yourField', null, ['required' => false ]);
}
}

Related

Yii2 Web Service not returning a single row from database

I've made a web service using yii2 basic template I got a table called 'ely_usuario' when I call it with:
http://localhost/basic/web/index.php/ely-usuario/
it works fine and returns me all the rows in ely_usuario table
but when I try to get just one record, for example:
http://localhost/basic/web/index.php/ely-usuario/29
it doesn't work, show me a not found page, I've made the model class using gii
here's my Controller:
<?php
namespace app\controllers;
use yii\rest\ActiveController;
class ElyUsuarioController extends ActiveController
{
public $modelClass = 'app\models\ElyUsuario';
}
My configs:
'urlManager' => [
'enablePrettyUrl' => true,
'enableStrictParsing' => false,
'showScriptName' => false,
'rules' => [
['class' => 'yii\rest\UrlRule', 'controller' => 'ely-usuario'],
],
],
Another weird thing that you might noticed is that 'enableStrictParsing' is false, in the yii2 guide it says to be true but for me it only works with false
Thanks
You need to change the code in your configs.I hope you will get an idea from the following code.It works fine for me!
'urlManager' => [
'enablePrettyUrl' => true,
'showScriptName' => false,
'rules' => [
'<controller:(ely-usuario)>/<action>/<id:\d+>' => '<controller>/<action>',
],
],
And in your controller please check your specific action.It must be coded like as:
public function actionTransactions($id=null){
if($id!=null){
//retrieve single row
}else{
//retrieve multiple rows
}
Also please check this link for reference:
Why RESTfull API request to view return 404 in Yii2?
I hope it helps!

ZF2 / PHPUnit: Mock a Zend/Db/Adapter/Adapter for further usage

The goal: Unit test an InputFilter in Zend Framework 2.
The problem: A mocked DbAdapter is needed.
As I am relatively new to Unit Testing I just got started with mocking classes. After doing a lot of research I'm still unable to find a proper solution for my problem so here we go with my filter to start things off:
class ExampleFilter extends Inputfilter
{
protected $dbAdapter;
public function __construct(AdapterInterface $dbAdapter)
{
$this->dbAdapter = $dbAdapter;
}
public function init()
{
$this->add(
[
'name' => 'example_field',
'required' => true,
'filters' => [
['name' => 'StringTrim'],
['name' => 'StripTags'],
],
'validators' => [
[
'name' => 'Db\NoRecordExists',
'options' => [
'adapter' => $this->dbAdapter,
'table' => 'example_table',
'field' => 'example_field',
],
],
],
]
);
}
}
Without the need of an adapter, testing this filter would be rather easy. My problem is to create the Filter in my TestClass as seen here:
class ExampleFilterTest extends \PHPUnit_Framework_TestCase
{
protected $exampleFilter;
protected $mockDbAdapter;
public function setUp()
{
$this->mockDbAdapter = $this->getMockBuilder('Zend\Db\Adapter')
->disableOriginalConstructor()
->getMock();
$this->exampleFilter = new ExampleFilter($this->mockDbAdapter);
}
}
When creating the filter like this the ExampleFilter class will end up saying that I did provide a wrong class to its constructor. It's receiving a mock object when expecting one of type Zend\Db\Adapter\Adapter.
I could create a real adapter of course but I want to avoid performing actual queries to the database as it is a unit test and this would go far beyond the border of my unit to test.
Can anyone tell me how I can achieve my goal of testing the filter with a mocked DbAdapter?
Well... As I was commenting on gontrollez hint I already smelled my mistake. I had to create a mock of 'Zend/Db/Adapter/AdapterInterface' instead of just '/Zend/Db/Adapter'.
Thanks for bringing me on the right path anyway gontrollez :)

How to set the default selected value for a Doctrine ObjectRadio element

I've got a simple user registration form where a user can choose their own user type. The user type maps to a role. This is part of a zf2 application using the doctrine2 module.
The relevant part of the init() method of my user fieldset looks like this:
public function init()
{
// ... other field definitions ...
$roleRadio = new ObjectRadio('role');
$roleRadio->setLabel('What type of user are you?')
->setOptions(
array(
'object_manager' => $this->objectManager,
'target_class' => 'MyUser\Entity\Role',
'property' => 'roleId',
'is_method' => true,
'find_method' => array(
'name' => 'findBy',
'params' => array(
'criteria' => array('userselectable' => true),
'orderBy' => array('displayorder' => 'ASC'),
),
),
)
);
$this->add($roleRadio);
// ... more stuff ...
}
I'm using Doctrine's ObjectRadio class for this element to automatically populate the value options. Is there any way to set the default selected value?
I know I can just do something like this:
$form->get('user')->get('role')->setValue(3);
But I don't want to hard code this and I also don't want to put that kind of logic in my controller.
Any suggestions?
I don't know what do you mean by "I don't want to hard code this", but you can do it as you said in your controller, or you can do it in the form definition by setting attributes as the following:
$roleRadio->setAttributes(array('value' => 3));

Custom Prestashop Admin Module

I am developing a module for prestashop (basically, it's a very custom import of data and the only thing I need is to have a form and process data). I have created controller class derived from the ModuleAdminController but the problem is where should I put the tpl file containing the look of my custom form?
I realize that I can put tpl file to the templates but I want to keep all files within my module folder, is it possible (probably somewhere like "/views/templates/admin")?
This is the most easy method to create a basic admin controller / action in Prestashop 1.6
Create basic configuration :
./config.xml
<?xml version="1.0" encoding="UTF-8" ?>
<module>
<name>foo</name>
<displayName><![CDATA[Foo]]></displayName>
<version><![CDATA[2.1.3]]></version>
<description><![CDATA[Bar.]]></description>
<author><![CDATA[your-name]]></author>
<tab><![CDATA[administration]]></tab>
<is_configurable>0</is_configurable>
<need_instance>0</need_instance>
<limited_countries></limited_countries>
</module>
./foo.php
if (!defined('_PS_VERSION_'))
exit;
class BarcodeEasyPrint extends Module
{
public function __construct()
{
$this->name = 'foo';
$this->tab = 'administration';
$this->version = '1.0.0';
$this->author = 'your-name-here';
$this->need_instance = 0;
$this->bootstrap = true;
parent::__construct();
$this->displayName = $this->l('Foo');
$this->description = $this->l('Bar.');
if ((int)Tools::getValue('p'))
$this->page = (int)Tools::getValue('p');
}
}
You need to create the controller with base functions :
./controllers/admin/AdminFooController.php
class AdminFooController extends ModuleAdminController {
public function __construct() {
$this->bootstrap = true;
parent::__construct();
}
public function createTemplate($tpl_name) {
if (file_exists($this->getTemplatePath() . $tpl_name) && $this->viewAccess())
return $this->context->smarty->createTemplate($this- >getTemplatePath() . $tpl_name, $this->context->smarty);
return parent::createTemplate($tpl_name);
}
public function initContent(){
parent::initContent();
$tpl = $this->createTemplate('content.tpl')->fetch();
/* DO STUFF HERE */
$posts = array();
$this->context->smarty->assign('posts', $posts);
}
}
You can use boostrap directly in the template file :
./views/templates/admin/content.tpl
<div class="row">
<div class="col-md-6">
</div>
<div class="col-md-6">
</div>
</div>
If it is an admin module only, then you will have no need to create any views. Because Prestashop provides a nice structure for admin section which is easy to use and we dont need to use any views or .tpl files. For admin section, normally three types of views or .tpl files are required, one for data display in grid, second for form and third for displaying a single record.
Prestashop already created .tpl files for them which you can find in "admin_folder/themes/default/templates". In our controllers for admin, for form and for data grid, we just create arrays and PS handles to view the form and data grid according to the arrays we created.
So if you need a custom form at admin, then create a public function renderForm and create the form array in it, like below:
$this->fields_form = array(
'legend' => array(
'title' => $this->l('Video'),
'image' => '../img/admin/tab-genders.gif'
),
'input' => array(
array(
'type' => 'text',
'label' => $this->l('Video Title:'),
'name' => 'title',
'lang' => true,
'size' => 70,
'hint' => $this->l('Invalid characters:').' 0-9!<>,;?=+()##"�{}_$%:',
'required' => true
),
array(
'type' => 'textarea',
'label' => $this->l('Video Code'),
'name' => 'video_code',
'rows' => 5,
'cols' => 70,
'desc' => $this->l('Place the embed code for the video')
),
),
'submit' => array(
'title' => $this->l('Save'),
'class' => 'button'
)
);
return parent::renderForm();
} /* End of render member */
For other fields, checkout other prestashop admin controllers and you will see that how easily we can create forms in PS using that simple definitions in the arrays and we dont need to create .tpl files.
For front end, we can use the new modules MVC structure, where our module folder have sub folders for controllers (controllers/front, controllers/admin) , views and models .
Hope this will help you.
Thank you
You need to use the helper form, here is the documentation for it, it is really easy to use ;) .
http://doc.prestashop.com/display/PS15/HelperForm
You also can find more information about how and where to use helper form, look for the function getContent() and displayForm().
http://doc.prestashop.com/display/PS15/Creating+a+PrestaShop+module
unfortunately any document not exist to point directly to solve this question but hear i have some URLs really useful and you should combine theme and get your realize:
http://presthemes.com/prestashop-news/modules-classes-and-controller-override-by-julien-breux-4.html
http://doc.prestashop.com/display/PS15/Diving+into+PrestaShop+Core+development
http://doc.prestashop.com/display/PS15/New+Developers+Features+In+PrestaShop+1.5
http://blog.belvg.com/how-to-implement-a-controller.html
best regards
extending the answer from #altafhussain create a folder views/templates/admin in your module and place your customview.tpl
Than append the free text block as below.
$this->fields_form = array(
'legend' => array(
'title' => $this->l('Legend')
),
'input' => array(
array(
'type' => 'free',
'label' => 'Whatever label text',
'desc' => $this->display(__FILE__,'views/templates/admin/customview.tpl'),
'name' => 'FREE_TEXT',
'required' => false
)
),
'submit' => array(
'title' => $this->l('Save'),
'class' => 'button'
)
);
return parent::renderForm();
}

Username in URL using Regex & Route

I have a method in my Users Controller called view, which should display a specified (by URL) user:
public function view($username = null) {
$this->User->username = $username;
if (!$this->User->exists()) {
throw new NotFoundException('Няма такъв потребител!');
}
if (!$username) {
$this->Session->setFlash('Няма такъв потребител!');
$this->redirect(array('action' => 'index'));
}
$this->set('user', $this->User->read());
}
And in the route config:
Router::connect('/:username', array('controller' => 'users', 'action'=> 'view'), array('username' => '^([a-z0-9])+$'));
But when I try: www.example.com/Username it returns a fatal error: Missing controller.
I tried also this:
Router::connect('/users/:username', array('controller' => 'users', 'action'=> 'view'), array('pass' => array('username'), 'username' => '^([a-z0-9])+$'));
Unfortunately for this sort of setup using /:username is too simple, it will pretty much override every single other route. To do this effectively ( + the proper CakePHP way) you need to setup a custom route, here is mine which pretty much achieves the same thing. Just replace "product" with "user" for most cases, read through it though, to make sure you understand what's going on..
Look at my routes config as well if you still can't work it out.
If you want to route /user/dunhamzz to a profile you would set it up like this:
Router::connect('/user/:username',
array('controller' => 'users', 'action' => 'view'),
array('pass' => array('username')
);
Then your view action simply gets the username as the first argument:
public function view($username) {
}