我有三张桌子:
1 lab_categories(列包括id
, category
)
class LabCategoriesTable extends Table
{
public function initialize(array $config)
{
parent::initialize($config);
$this->table('lab_categories');
$this->displayField('id');
$this->primaryKey('id');
$this->hasMany('LabTests', [
'foreignKey' => 'lab_category_id'
]);
$this->belongsToMany('Laboratories', [
'foreignKey' => 'lab_category_id',
'targetForeignKey' => 'laboratory_id',
'joinTable' => 'laboratories_lab_categories'
]);
}}
2 个实验室(列包括id
, name
)
class LaboratoriesTable extends Table
{
public function initialize(array $config)
{
parent::initialize($config);
$this->table('laboratories');
$this->displayField('name');
$this->primaryKey('id');
$this->hasMany('LabRefValues', [
'foreignKey' => 'laboratory_id'
]);
$this->belongsToMany('LabCategories', [
'foreignKey' => 'laboratory_id',
'targetForeignKey' => 'lab_category_id',
'joinTable' => 'laboratories_lab_categories'
]);
}}
3 个实验室_实验室_类别(列包括id
, laboratory_id
, lab_category_id
)
class LaboratoriesLabCategoriesTable extends Table
{
public function initialize(array $config)
{
$this->table('laboratories_lab_categories');
$this->displayField('id');
$this->primaryKey('id');
$this->belongsTo('Laboratories', [
'foreignKey' => 'laboratory_id',
'joinType' => 'INNER'
]);
$this->belongsTo('LabCategories', [
'foreignKey' => 'lab_category_id',
'joinType' => 'INNER'
]);
}}
我希望能够选择lab_categories.name
与指定的关联,laboratory.id
以便能够将其添加到我创建的 ajax 调用中。我想生成类似于此查询将生成的结果;选择 lab_categories.id,category From lab_categories 加入laboratory_lab_categories wherelaboratory_lab_categories.laboratory_id=$id AND Laboratory_lab_categories.lab_category_id=lab_categories.id
我正在使用cakephp 3.1
,桌子已经烤好了。我已经搜索了相关问题,但不适用于我。