CakePHP 3.4如何在一个属于并且有很多关系的
在cakePHP 3.4上,我有3个表,其中一个表属于,并且有很多关系:配料、产品和IngCreditsProducts:CakePHP 3.4如何在一个属于并且有很多关系的,cakephp,save,associations,has-and-belongs-to-many,cakephp-3.x,Cakephp,Save,Associations,Has And Belongs To Many,Cakephp 3.x,在cakePHP 3.4上,我有3个表,其中一个表属于,并且有很多关系:配料、产品和IngCreditsProducts: class IngredientsTable extends Table { public function initialize(array $config) { // Use through option because it looks like you // have additional data on your
class IngredientsTable extends Table
{
public function initialize(array $config)
{
// Use through option because it looks like you
// have additional data on your IngredientsProducts table
$this->belongsToMany('Products', [
'through' => 'IngredientsProducts',
]);
}
}
class ProductsTable extends Table
{
public function initialize(array $config)
{
$this->belongsToMany('Ingredients', [
'through' => 'IngredientsProducts',
]);
}
}
class IngredientsProductsTable extends Table
{
public function initialize(array $config)
{
$this->belongsTo('Ingredients');
$this->belongsTo('Products');
}
}
我想完成的是,当我插入一个新产品时,我还想在IngreditsProductsjoiner表中插入与该产品相关的每个成分的成分id和字段数量
我一直在阅读烹饪书,看到在joiner表上保存传统数据(在我的例子中,字段数量如下所述)时,必须使用_joinData属性,因此我的add视图如下所示:
<?php
$this->Form->create($product);
// fields of the Products table
echo $this->Form->control('name',array('class' => 'form-control'));
echo $this->Form->control('retail_price');
echo $this->Form->control('best_before');
echo $this->Form->control('comments');
// ingredient_id and qty fields of the ingredients_products table
echo $this->Form->control('ingredients.0._joinData.ingredient_id',['options' => $ingredients);
echo $this->Form->control('ingredients.0._joinData.qty');
//and repetition of these last two as ingredients.1._joinData to ingredients.N._joinData
$this->Form->button(__('Save'));
$this->Form->end();
?>
$product = $this->Products->newEntity();
if ($this->request->is('post')) {
$product = $this->Products->patchEntity($product, $this->request->getData(),[
'associated' => ['Ingredients._joinData']
]);
if ($this->Products->save($product)) {
$this->Flash->success(__('The product has been saved.'));
return $this->redirect(['action' => 'index']);
}
$this->Flash->error(__('The product could not be saved. Please, try again.'));
}
$this->set(compact('product'));
$this->set('_serialize', ['product']);
但是,当我提交时,它不会保存任何内容。调试将显示正在发布的以下内容:
object(App\Model\Entity\Product) {
'name' => 'salada',
'retail_price' => (float) 23,
'best_before' => (int) 234,
'comments' => 'wer',
'directions' => 'werwewerer',
'ingredients' => [
(int) 0 => object(App\Model\Entity\Ingredient) {
'_joinData' => object(Cake\ORM\Entity) {
'ingredient_id' => (int) 4,
'qty' => (float) 100,
'[new]' => true,
'[accessible]' => [
'*' => true
],
'[dirty]' => [
'ingredient_id' => true,
'qty' => true
],
'[original]' => [],
'[virtual]' => [],
'[errors]' => [],
'[invalid]' => [],
'[repository]' => 'IngredientsProducts'
},
'[new]' => true,
'[accessible]' => [
'*' => true,
'id' => false
],
'[dirty]' => [
'_joinData' => true
],
'[original]' => [
'_joinData' => [
'ingredient_id' => '4',
'qty' => '100'
]
],
'[virtual]' => [],
'[errors]' => [
'name' => [
'_required' => 'This field is required'
]
],
'[invalid]' => [],
'[repository]' => 'Ingredients'
},
(int) 1 => object(App\Model\Entity\Ingredient) {
'_joinData' => object(Cake\ORM\Entity) {
'ingredient_id' => (int) 5,
'qty' => (float) 200,
'[new]' => true,
'[accessible]' => [
'*' => true
],
'[dirty]' => [
'ingredient_id' => true,
'qty' => true
],
'[original]' => [],
'[virtual]' => [],
'[errors]' => [],
'[invalid]' => [],
'[repository]' => 'IngredientsProducts'
},
'[new]' => true,
'[accessible]' => [
'*' => true,
'id' => false
],
'[dirty]' => [
'_joinData' => true
],
'[original]' => [
'_joinData' => [
'ingredient_id' => '5',
'qty' => '200'
]
],
'[virtual]' => [],
'[errors]' => [
'name' => [
'_required' => 'This field is required'
]
],
'[invalid]' => [],
'[repository]' => 'Ingredients'
}
],
'[new]' => true,
'[accessible]' => [
'*' => true,
'id' => false
],
'[dirty]' => [
'name' => true,
'retail_price' => true,
'best_before' => true,
'comments' => true,
'directions' => true,
'ingredients' => true
],
'[original]' => [],
'[virtual]' => [],
'[errors]' => [],
'[invalid]' => [],
'[repository]' => 'Products'
}
有人知道怎么做吗
以下是mySQL上的表结构:
TABLE `Ingredients` (
`id` int(11) NOT NULL,
`name` varchar(255) NOT NULL,
`category_id` int(11) NOT NULL,
`measure_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
Table products
TABLE `Products` (
`id` int(11) NOT NULL,
`name` varchar(255) NOT NULL,
`retail_price` float NOT NULL,
`best_before` int(11) NOT NULL,
`comments` text NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
TABLE `ingredients_products` (
`ingredient_id` int(11) NOT NULL,
`product_id` int(11) NOT NULL,
`qty` double NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
任何帮助或指导都将不胜感激!
谢谢如果保存失败,请检查错误
信息
当保存实体无效时,请检查错误
信息:
'[errors]' => [
'name' => [
'_required' => 'This field is required'
]
],
您的验证规则根据需要定义成分.name
字段,但它不在表单中,因此保存失败
提供目标关联上的主键
仔细看看文档中保存连接数据的示例,数据结构与您的略有不同
从您对配料id
的使用情况来看,我怀疑您想将产品与现有配料联系起来,将配料id
放在\u joinData
中,但这不是有效的方法。编辑/链接现有记录要求记录的主键出现在特殊的\u id
键中,或出现在目标关联(成分
)的主键字段中
因此,在您希望存储其他联接数据的情况下,必须使用配料的id
属性,这样封送员就知道需要加载现有记录:
echo $this->Form->control('ingredients.0.id', [
// defining the type is required, as the input type
// guessing only recognizes `_ids` fields, or fields
// with `_id` appended as possible select types
'type' => 'select'
'options' => $ingredients
]);
echo $this->Form->control('ingredients.0._joinData.qty');
另见