Question

我有2个表（drugs_and_medications和drug_images），它们与belongsTo链接在一起，如下所示：

public function initialize(array $config)
    {
        parent::initialize($config);

        $this->setTable('drugs_and_medications');
        $this->setDisplayField('id');
        $this->setPrimaryKey('id');

        $this->addBehavior('Timestamp');

        $this->belongsTo('DrugImages', [
            'foreignKey' => 'drug_image_id',
        ]);
    }

在我的Drugs_And_Medications表的编辑功能中，我有一个表格，可让我从Drug_Image表的图片中获取ID。

Drugs_and_medications表的编辑功能：

public function edit($id = null)
    {
        $drugsAndMedication = $this->DrugsAndMedications->get($id, [
            'contain' => ['Prescriptions'],
        ]);
        if ($this->request->is(['patch', 'post', 'put'])) {
            $drugsAndMedication = $this->DrugsAndMedications->patchEntity($drugsAndMedication, $this->request->getData());
            if ($this->DrugsAndMedications->save($drugsAndMedication)) {
                $this->Flash->success(__('The drugs and medication has been saved.'));

                return $this->redirect(['action' => 'index']);
            }
            $this->Flash->error(__('The drugs and medication could not be saved. Please, try again.'));
        }
        $drugImages = $this->DrugsAndMedications->DrugImages->find('list', ['limit' => 200]);
        //$prescriptions = $this->DrugsAndMedications->Prescriptions->find('list', ['limit' => 200]);
        $this->set(compact('drugsAndMedication', 'drugImages'));
    }

这是编辑功能的模板：

<div class="drugsAndMedications form large-9 medium-8 columns content">
    <?= $this->Form->create($drugsAndMedication) ?>
    <fieldset>
        <legend><?= __('Edit Drugs And Medication') ?></legend>
        <?php
            echo $this->Form->control('drug_name');
            echo $this->Form->control('drug_cost');
            echo $this->Form->control('drug_available_date');
            echo $this->Form->control('drug_withdrawn_date');
            echo $this->Form->control('drug_description');
            echo $this->Form->control('drug_image_id', ['option' => $drugImages]);
            //echo $this->Form->control('prescriptions._ids', ['options' => $prescriptions]);
        ?>
    </fieldset>
    <?= $this->Form->button(__('Submit')) ?>
    <?= $this->Form->end() ?>
</div>

这是我的drugs_and_medications和drug_images表：

CREATE TABLE `drugs_and_medications` (
  `id` int(10) NOT NULL,
  `drug_name` varchar(255) COLLATE utf8mb4_general_ci NOT NULL,
  `drug_cost` int(3) NOT NULL,
  `drug_available_date` date NOT NULL,
  `drug_withdrawn_date` date NOT NULL,
  `drug_description` text COLLATE utf8mb4_general_ci,
  `drug_image_id` int(11) DEFAULT NULL,
  `created` date NOT NULL,
  `modified` date NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_general_ci;

ALTER TABLE `drugs_and_medications`
  ADD CONSTRAINT `drugs_and_medications_ibfk_1` FOREIGN KEY (`drug_image_id`) REFERENCES `drug_images` (`id`) ON DELETE RESTRICT ON UPDATE RESTRICT;

CREATE TABLE `drug_images` (
  `id` int(11) NOT NULL,
  `name` varchar(255) COLLATE utf8mb4_general_ci NOT NULL,
  `path` varchar(255) COLLATE utf8mb4_general_ci NOT NULL,
  `created` date NOT NULL,
  `modified` date NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_general_ci;

我应该从drug_images表中获取图像ID，并将其放在drugs_and_medications字段的drug_image_id表中，这样我就可以在索引中显示图片，但是无论何时我尝试获取image_id，即使我可以在列表中看到它，它也返回null，甚至可以单击并选择它。所以我无法在应用程序中显示图像。

如果需要，您总是可以要求我显示更多代码。

Cakephp-无法访问toBelong链接表中的数据

0 个答案: