如何在cakephp 3.x中保存关联的joinData
我在多对多的关系中有问题和领域。联接表fields\u problems有一个名为fieldvalue的字段。我正在尝试创建一个表单,该表单将在fields\u problems中插入一条问题记录和多条记录 /src/Model/Table/ProblemsTable.php如何在cakephp 3.x中保存关联的joinData,cakephp,orm,many-to-many,associations,cakephp-3.0,Cakephp,Orm,Many To Many,Associations,Cakephp 3.0,我在多对多的关系中有问题和领域。联接表fields\u problems有一个名为fieldvalue的字段。我正在尝试创建一个表单,该表单将在fields\u problems中插入一条问题记录和多条记录 /src/Model/Table/ProblemsTable.php class ProblemsTable extends Table { public function initialize(array $config) { parent::initialize(
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'
]);
}
...
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'
]);
}
...
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'
]);
}
...
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']);
}
/src/Model/Table/FieldsTable.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'
]);
}
...
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'
]);
}
...
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'
]);
}
...
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']);
}
/src/Model/Table/FieldsProblemsTable.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'
]);
}
...
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'
]);
}
...
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'
]);
}
...
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']);
}
我想添加一个新问题,将其链接到字段,并向联接表中的fieldvalue字段添加值
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>
当我填写并提交应用程序表单时,问题记录已保存,但关联未保存,没有任何内容插入到问题字段中
我做错了什么,阻止保存关联的joinData?尽管食谱()说要使用特殊的_id键,但不要
将“_id”更改为“id”修复了表单,现在它可以正确地将数据保存到jointable中
下面是我构建应用程序时使用的食谱中的示例
echo $this->Form->input('tags.0.id');
echo $this->Form->input('tags._ids', [
'type' => 'select',
'multiple' => true,
'options' => $tagList,
]);
这就是它应该是什么样子
echo $this->Form->input('tags.0.id', [
'type' => 'select',
'multiple' => false,
'options' => $tagList,
]);