如何在 cakephp 中保存关联的 joinData 3.x

How to save associated joinData in cakephp 3.x

我在多对多关系中有问题和字段。连接 table fields_problems 有一个名为 fieldvalue 的字段 我正在尝试创建一个表单,将问题记录和多条记录插入 fields_problems .

/src/Model/Table/ProblemsTable.php

class ProblemsTable extends Table
{
    public function initialize(array $config)
    {
    parent::initialize($config);

    $this->table('problems');
    $this->displayField('id');
    $this->primaryKey('id');
    $this->addBehavior('Timestamp');
    $this->belongsToMany('Fields', [
        'foreignKey' => 'problem_id',
        'targetForeignKey' => 'field_id',
        'joinTable' => 'fields_problems'
    ]);
    }
...

/src/Model/Table/FieldsTable.php

class FieldsTable extends Table
{

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

    $this->table('fields');
    $this->displayField('name');
    $this->primaryKey('id');
    $this->addBehavior('Timestamp');
    $this->belongsToMany('Problems', [
        'foreignKey' => 'field_id',
        'targetForeignKey' => 'problem_id',
        'joinTable' => 'fields_problems'
    ]);
    }
...

/src/Model/Table/FieldsProblemsTable.php

class FieldsProblemsTable extends Table
{
    public function initialize(array $config)
    {
    parent::initialize($config);

    $this->table('fields_problems');
    $this->displayField('id');
    $this->primaryKey('id');
    $this->belongsTo('Fields', [
        'foreignKey' => 'field_id',
        'joinType' => 'INNER'
    ]);
    $this->belongsTo('Problems', [
        'foreignKey' => 'problem_id',
        'joinType' => 'INNER'
    ]);
    }
...

而且我想添加一个新问题,link 将其添加到字段中,并将值添加到连接 table 中的 fieldvalue 字段。

所以我有这个 /src/Template/Problems/add.ctp

<div class="problems form large-10 medium-9 columns">
    <?= $this->Form->create($problem) ?>
    <fieldset>
        <legend><?= __('Add Problem') ?></legend>
        <?php
            echo $this->Form->input("Problems.id");
            echo $this->Form->input('Problems.summary');

            echo $this->Form->input('Problems.Fields.0._ids', [
                'type' => 'select',
                'multiple' => false,
                'options' => $fields,
            ]);
            echo $this->Form->input('Problems.Fields.0._joinData.fieldvalue');

            echo $this->Form->input('Problems.Fields.1._ids', [
                'type' => 'select',
                'multiple' => false,
                'options' => $fields,
            ]);
            echo $this->Form->input('Problems.Fields.1._joinData.fieldvalue');
        ?>

    </fieldset>
    <?= $this->Form->button(__('Submit')) ?>
    <?= $this->Form->end() ?>
</div>

而这个 add() 在 /src/Controller/ProblemsController.php

public function add()
    {
        $problem = $this->Problems->newEntity();
        if ($this->request->is('post')) {
            $problem = $this->Problems->patchEntity($problem, $this->request->data, ['associated'=>['Fields._joinData']] );
            //$problem->dirty('fields',true);
            if ($this->Problems->save($problem)) {
                $this->Flash->success(__('The problem has been saved.'));
                return $this->redirect(['action' => 'index']);
            } else {
                $this->Flash->error(__('The problem could not be saved. Please, try again.'));
            }
        }
        $fields = $this->Problems->Fields->find('list', ['limit' => 200]);
        $this->set(compact('problem', 'fields'));
        $this->set('_serialize', ['problem']);
    }

当我填写并提交应用程序表单时,问题记录已保存,但关联并未保存,没有任何内容插入 fields_problems。

我做错了什么导致关联的 joinData 无法保存?

尽管说明书 (http://book.cakephp.org/3.0/en/views/helpers/form.html) 说要使用特殊的 _ids 键,但不要这样做!

将“_ids”更改为 "id" 修复了表单,现在它可以正常将数据保存到可连接对象中。

这是我用

构建应用程序的食谱中的示例
echo $this->Form->input('tags.0.id');

echo $this->Form->input('tags._ids', [
    'type' => 'select',
    'multiple' => true,
    'options' => $tagList,
]);

这是应该的样子

echo $this->Form->input('tags.0.id', [
    'type' => 'select',
    'multiple' => false,
    'options' => $tagList,
]);