PHP/Phalcon-自动嵌套对象
假设我有三张桌子:PHP/Phalcon-自动嵌套对象,php,phalcon,Php,Phalcon,假设我有三张桌子: CREATE TABLE divisions { idDivision INT NOT NULL AUTO_INCREMENT PRIMARY KEY, name VARCHAR (40) NOT NULL } CREATE TABLE clubs { idClub INT NOT NULL AUTO_INCREMENT PRIMARY KEY, idDivision INT NOT NULL, name VARCHAR(40) NOT NULL }
CREATE TABLE divisions {
idDivision INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
name VARCHAR (40) NOT NULL
}
CREATE TABLE clubs {
idClub INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
idDivision INT NOT NULL,
name VARCHAR(40) NOT NULL
}
CREATE TABLE footballers (
idFootballer INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
idClub INT NOT NULL,
name VARCHAR (40) NOT NULL
)
我有一些可爱的Phalcon模型来代表这些
现在,我想做的是:
$divisions = new Divisions();
print json_encode($divisions::findFirst(), JSON_NUMERIC_CHECK);
返回一个JSON对象,如下所示:
{
idDivision: 1,
name: "Welsh Premier League",
clubs: [
{
idClub: 1,
idDivision: 1,
name: "Airbus UK",
players: [
{
idPlayer: 1,
idClub: 1,
name: "Alf Jones"
},
...
]
},
..
]
}
对于Phalcon模型,有没有一种简单的方法可以做到这一点 要自动获取嵌套模型,请使用此递归函数
$getRelations = function($model, $namespace, $alias = null, $instances = null) use (&$getRelations)
{
$modelsManager = $model->getModelsManager();
$relations = $modelsManager->getRelations($namespace);
if (is_null($instances)) {
$response = $model->toArray();
}
if (count($relations)) {
// loop relations
foreach ($relations as $i => $relation) {
$options = $relation->getOptions();
// get alias
if (isset($options['alias'])) {
$subAlias = $options['alias'];
$modelName = $relation->getReferencedModel();
$subModel = new $modelName();
// get from model
if (is_null($alias) && count($model->{$subAlias})) {
$response[$subAlias] = $getRelations(
$subModel, $modelName, $subAlias, $model->{$subAlias}
);
// get from object instance
} elseif (count($instances)) {
foreach ($instances as $k => $instance) {
$response[$k] = $instance->toArray();
$response[$k][$subAlias] = $getRelations(
$subModel, $modelName, $subAlias, $instance->{$subAlias}
);
}
}
}
}
} else {
$response = $instances->toArray();
}
return $response;
};
你可以这样称呼它:
$model = new Division::findFirst($divisionId);
$namespace = 'AppName\Models\Division';
$data = $getRelations($model, $namespace);
$this->response->setJsonContent($data);
确保为每个嵌套模型定义一个别名,如下所示:
class Division extends \Phalcon\Mvc\Model
{
public function initialize()
{
$this->hasMany('id', 'AppName\Models\Club', 'division_id', array(
'alias' => 'clubs'
));
}
}
更新
改为使用下面的代码(将其放在基本模型中)。这段新代码将允许您获取新(空)模型上的关系
从控制器调用也更容易
$model = new Division::findFirst($divisionId);
$data = $model->toArray(null, 1);
$this->response->setJsonContent($data);
在Phalcon中,可以在模型本身内部定义模型之间的关系。 例如: 最后:
$results = FirstModel::find();
我还没有测试过它(所以没有被接受),但是谢谢-我认为这将非常非常有用嗯,我们有这种代码,但速度很慢。它为每个对象生成多个查询。因此,如果我想要一个具有嵌套关系的对象列表,它会变得非常缓慢(300个对象的速度超过1s)@maddanio是的,我记得我很久以前就向Phalcon团队提到过这一点。显然,关系引用只在按对象的基础上工作。
class FirstModel extends \Phalcon\Mvc\Model
{
public function initialization()
{
$this->hasMany('field', SecondModel::class, 'referenceField', [options]);
}
}
class SecondModel extends \Phalcon\Mvc\Model
{
public function initialization()
{
$this->hasMany('field', ThirdModel::class, 'referenceField', [options]);
}
}
class ThirdModel extends \Phalcon\Mvc\Model
{
// ... some code
}
$results = FirstModel::find();