Php 在Symfony中按Id访问项目?
我在Symfony3中有一个控制器。所有数据的返回方式如下:Php 在Symfony中按Id访问项目?,php,symfony,Php,Symfony,我在Symfony3中有一个控制器。所有数据的返回方式如下: [ { "id": 13, "name": "testing", "entry_date": { "date": "2017-12-20 15:23:59.000000", "timezone_type": 3, "timezone": "Europe/London" }, "last_update": { "date": "2
[
{
"id": 13,
"name": "testing",
"entry_date": {
"date": "2017-12-20 15:23:59.000000",
"timezone_type": 3,
"timezone": "Europe/London"
},
"last_update": {
"date": "2017-12-20 15:23:59.000000",
"timezone_type": 3,
"timezone": "Europe/London"
}
},
{
"id": 30,
"name": "testing2",
"entry_date": {
"date": "2017-12-20 22:02:37.000000",
"timezone_type": 3,
"timezone": "Europe/London"
},
"last_update": {
"date": "2017-12-20 22:02:37.000000",
"timezone_type": 3,
"timezone": "Europe/London"
}
}
]
/**
* @Method("GET")
* @Route("/item/{item_id}")
* @View()
* @ApiDoc(
* resource = true,
* description = "Get an item record",
* section = "Spark DTR",
* )
*/
public function getItem($item_id)
{
$em = $this->getDoctrine()->getManager('app');
$mi_repo = $em->getRepository('AppBundle:Item')->find($item_id);
if(empty($mi_repo)) {
return new JsonResponse("Invalid Item ID", 404);
}
return new JsonResponse($mi_repo, 200);
}
我正在尝试按其id返回单个项目。到目前为止,我的方法如下所示:
[
{
"id": 13,
"name": "testing",
"entry_date": {
"date": "2017-12-20 15:23:59.000000",
"timezone_type": 3,
"timezone": "Europe/London"
},
"last_update": {
"date": "2017-12-20 15:23:59.000000",
"timezone_type": 3,
"timezone": "Europe/London"
}
},
{
"id": 30,
"name": "testing2",
"entry_date": {
"date": "2017-12-20 22:02:37.000000",
"timezone_type": 3,
"timezone": "Europe/London"
},
"last_update": {
"date": "2017-12-20 22:02:37.000000",
"timezone_type": 3,
"timezone": "Europe/London"
}
}
]
/**
* @Method("GET")
* @Route("/item/{item_id}")
* @View()
* @ApiDoc(
* resource = true,
* description = "Get an item record",
* section = "Spark DTR",
* )
*/
public function getItem($item_id)
{
$em = $this->getDoctrine()->getManager('app');
$mi_repo = $em->getRepository('AppBundle:Item')->find($item_id);
if(empty($mi_repo)) {
return new JsonResponse("Invalid Item ID", 404);
}
return new JsonResponse($mi_repo, 200);
}
但是,此方法当前返回“无效项ID”(如果没有项),或者
{}
如果有项目!我想返回项目的内容。感谢您的帮助,
$em->getRepository('AppBundle:Item')->find($item_id);
结果是得到一个对象,JsonResponse()需要一个数组作为参数
这里有两个选项,您可以安装serializer组件()并将对象直接序列化为Json并返回
$jsonContent = $serializer->serialize($mi_repo, 'json');
或者,如果不想设置序列化程序,也可以使用带getArrayResult()的条令查询生成器返回数组而不是对象
$query = $em->createQueryBuilder()
->select('p')
->from('Products', 'p')
->where('p.id= :item_id')
->setParameter('id', $item_id)
->getQuery();
$mi_repo = $query->getArrayResult();
希望这有帮助!试试这个:$mi_repo=$this->getDoctrine()->getRepository('AppBundle:Item')->find($Item_id);