小编Mas*_*iro的帖子

我有自定义表单,它是各种实体的组合,使用以下代码对最终用户有意义:

$form = $this->container->get('form.factory')->createNamedBuilder(null, 'form')
    ->add('country', 'entity', array(
            'class'         => 'ACME\MyBundle\Entity\Country',
            'query_builder' => function(EntityRepository $er) {
                return $er->createQueryBuilder('c')->orderBy('c.en_name', 'ASC');
            },
            'label'         => '* Country',
            'required'      => true
        ),
    )

即使在查阅http://symfony.com/doc/current/reference/forms/types/entity.html#reference-forms-entity-choices文档时,代码似乎也很好,但我一直收到以下错误:

The options "class", "query_builder" do not exist. Known options are: "action",
"attr", "auto_initialize", "block_name", "by_reference", "cascade_validation", 
"choice_list", "choices", "compound", "constraints", "csrf_field_name", 
"csrf_message", "csrf_protection", "csrf_provider", "csrf_token_id", 
"csrf_token_manager", "data", "data_class", "disabled", "empty_data", "empty_value", 
"error_bubbling", "error_mapping", "expanded", "extra_fields_message", "inherit_data", 
"intention", "invalid_message", "invalid_message_parameters", "js_validation", 
"label", "label_attr", "label_render", "mapped", "max_length", "method", 
"multiple", "pattern", …

我有以下两个类具有ManyToMany关系并保存在一个名为的连接表中countries_involved.

class CountriesInvolved
{
/**
 * @var integer
 */
private $id;

/**
 * @var string
 */
private $description;

/**
 * @var \DateTime
 */
private $createdAt;

/**
 * @var \DateTime
 */
private $updatedAt;

/**
 * @var \ACME\SouthBundle\Entity\Country
 */
private $country;

/**
 * @var \Doctrine\Common\Collections\Collection
 */
private $involvement;
}

和

class Involvement
{
/**
 * @var integer
 */
private $id;

/**
 * @var string
 */
private $name;

/**
 * @var string
 */
private $description;
}

该关系在YML中定义如下

manyToMany:
    involvement: …

我已经发布了这3次,似乎看不到帖子不知道我做错了什么.

我创建了我的批处理操作,我的管理类如下:

namespace ACME\MyBundle\Admin;

use Sonata\AdminBundle\Admin\Admin;
use Sonata\AdminBundle\Show\ShowMapper;
use Sonata\AdminBundle\Datagrid\ListMapper;
use Sonata\AdminBundle\Datagrid\DatagridMapper;
use Sonata\AdminBundle\Validator\ErrorElement;
use Sonata\AdminBundle\Route\RouteCollection;

class JournalistProfileAdmin extends Admin
{
  ...........
  ...........
    public function getBatchActions()
{

    $lists = $this->getModelManager()->createQuery('ACME\MyBundle\Entity\ContactList', 'c')->execute();
    $listsArray = array();

    foreach ($lists as $list)
    {
        $listsArray[$list->getId()] = $list->getName();
    }
    $actions = parent::getBatchActions();

    $actions['addToGroup'] = array(
            'label' => $this->trans('action_add_to_group', array(), 'SonataAdminBundle'),
            'ask_confirmation' => true,
            'secondary' => $listsArray,
        );

    return $actions;
}
}

然后扩展CRUDController,如下面的文件:

namespace ACME\MyBundle\Controller;

use Sonata\AdminBundle\Controller\CRUDController as Controller;
use Sonata\DoctrineORMAdminBundle\Datagrid\ProxyQuery as ProxyQueryInterface;
use Symfony\Component\HttpFoundation\RedirectResponse;

class JournalistProfileAdminController …

谁能明确地告诉我如何将我的symfony2应用程序连接到heroku上的ClearDB？

我输入了heroku上的连接字符串

$ heroku config

但是当我把结果放在我的parameters.yml上时,我得到一个SQLSTATE [HY000] [2002]连接拒绝错误.

提前致谢

我试图在网上搜索这个问题的一个明确的解决方案,但那里的新手确实没有具体的解决方案.

我有一个条目有很多EntryListing.在我的EntryAdmin listMapper中,我可以轻松地按语句列出条目

->add('listings')

它只返回EntryListing __toString()函数中定义的列表.

有没有办法通过覆盖getExportFields()函数导出数据时实现相同的方法,如下所示:

public function getExportFields()
{
    return array('name','tel','email','deviceType','postedOn','createdAt','facilitator','listings');
}

非常感谢您的帮助

I have a JObject item that looks like this:

{

        "part":
         {
             "values": ["engine","body","door"]
             "isDelivered": "true"
        },
        {
        "manufacturer":
         {
             "values": ["Mercedes"]
             "isDelivered": "false" 
         }
}

How can I get the values as a single string in C#?

我在我的管理员中包含了一个相关模型,如下所示

->add('parameters', 'sonata_type_collection', array(
        'type_options' => array(
            // Prevents the "Delete" option from being displayed
            'delete' => false,
             )
    ), array(
        'edit' => 'inline',
        'inline' => 'table',
        'sortable' => 'position',
    ))

包含子管理员,我可以添加新行.但是,当我尝试添加第二个孩子时,我收到以下错误:

PropertyAccessor requires a graph of objects or arrays to operate on, but it found type "NULL" while trying to traverse path "parameters[0]" at property "0". 

我无法解释发生了什么,我正在使用symfony 2.7.3并使用Admin bundle的dev-master分支.我不知道这是否已被报道,我已经尝试检查问题列表但是没有看到它

我正在为仪表板创建一个自定义块,我想在其中显示持久存储在数据库中的信息.如何在块服务中获取容器或doctrine实体管理器的实例？

尝试谷歌搜索很多但到目前为止没有任何实质性的

我已经清除了我的symfony2应用程序,现在无法运行该应用程序,因为无法从命令行为实体生成代理,而默认情况下实际生成了一些代理.

我试过运行下面的命令,这通常在早期版本的symfony/doctrine中做了一些技巧:

php bin/console doctrine:ensure-production-settings --no-debug --env=prod

但这一次我只收到以下回复:

Query Cache uses a non-persistent cache driver, Doctrine\Common\Cache\ArrayCache.

有谁知道如何解决这个问题？

当我的时候,我的阵列看起来像这样 print_r(delegates[$i])

Array
(
    ['firstName'] => Brady
    ['surname'] => Manning
    ['jobTitle'] => CEO
    ['memberNumber'] => 123456
    ['dieteryRequirements'] => halal
)

然后,当我尝试获取firstName时,echo delegates[$i]['firstName']我得到以下错误

Notice: Undefined index: firstName 

小学生的问题,但我真的很感激这里的一些帮助.