Question

我是初学的symfony开发者。我的表格有很大问题我想将数据保存到我的表名称广告活动，如下所示：

Campaigns:
id, user_id, recipients, message, sender, type, created

和下一个

我在收件人系统的基础上有2个表

contacts_groups: 
id, user_id, name, description, created

和

contacts:
id, user_id, group_id, email, city, created.

现在让我们假设我的表有以下记录：

    contacts_groups:
        id, user_id, name, description, created
        1, 1, Test Group, Newest recipients, 2016-20-02 14:24:00
        2, 1, Awesome Group, Pro players, 2016-10-02 11:22:41
    contacts:
        id, user_id, group_id, email, city, created
        1, 1 , 1, exmple@email.com, New York, 2016-12-12 12:12:12
        2, 1 , 1, test@emailer.com, New York, 2016-12-12 12:12:12
        3, 1 , 2, proplayer@games.com, New York, 2016-12-12 12:12:12

现在我想用2个表记录做一个表格例如：

TextField  => text for sender name row
ChoiceType => select with options with contacts_groups records
ChoiceType => select with options with contacts where group_id == choosed contacts_groups record above.

我有那段代码：

class CampaignsType extends AbstractType{  


public function buildForm(FormBuilderInterface $builder, array $options) {
    $Campaigns = $options['data'];
    $user      = $Campaigns->tmpusr; // variable transferred from the controller ( logged user id )

    $builder
          ->add('sender', TextType::class, array(
             'trim'     => true,
             'label'    => 'Odbiorca',
             'label_attr' => array('class' => 'col-sm-2 control-label'),
             'attr' => array('class' => 'form-control', 'maxlength' => '11')
                ))
         ->add('recipients', EntityType::class, array(
            'label' => 'Grupa odbiorców',
            'label_attr' => array('class' => 'col-sm-2 control-label'),
            'attr' => array('class' => 'form-control'),
            'class' => 'MyAppPanelBundle:ContactsGroups',
            'query_builder' => function (\Doctrine\ORM\EntityRepository $er) use ($user) {
                return $er->createQueryBuilder('u')
                    ->select('u')
                    ->add('where', 'u.user_id =  ' . $user)
                    ->orderBy('u.id', 'DESC');

            },
            'choice_value' => 'id',
            'choice_label' => 'name',
                 ));     
}

有人知道如何使用数据库中的联系人添加下一个字段，其中group_id将与所选组的ID相同吗？

谢谢大家。

Answer 1

是Ajax：

在您的CampaignsType中添加此内容：

    ->add('contact', ChoiceType::class, array(
        'required' => true,
        'placeholder' => 'choose contact'
    ))

在您的观点中：

                <div class="form-group">
                    {{ form_label(form.contact, label|default(null), { 'label_attr': { 'class': 'font-secondary' } }) }}
                    {{ form_widget(form.contact, {'attr' : {'class' : 'form-control'}} )}}
                    {{ form_errors(form.contact) }}
                </div>

：

`

/**
 * @Route("/select/contact", name="_select_contact")
 */
public function contactsAction(Request $request) {
    $group_id = $request->request->get('group_id');
    $em = $this->getDoctrine()->getManager();
    $contacts = $em->getRepository('AppBundle:Contact')->getContacts($group_id);
    return new JsonResponse($contacts);
}

：

         $("#campaigns_group").change(function () {
            var data = {
                group_id: $(this).val()
            };
           //$('form').spin();
            $.ajax({
                type: 'post',
                url: Routing.generate('_select_contact'),
                data: data
            }).done(function (response) {
                var $contacts_selector = $('#campaigns_contact'),
                    contacts = response.contact;
                $contacts_selector.html('<option>choise contact</option>');
                for (var i = 0, total = contacts.length; i < total; i++) {
                    $contacts_selector.append('<option value="' + contacts[i].id + '">' + contacts[i].email + '</option>');
                }

            }).fail(function () {
                alert("error");
            }).always(function () {
                //$('form').find('.spinner').hide()
            });
        });

只需在您的联系人存储库中创建getContacts（）并调整javascript即可满足您的需求。

希望这有帮助！

具有多个表选择类型

1 个答案: