在我的Symfony 3应用程序中,我使用规范化器和序列化程序将结果集格式化为
JSON.默认情况下,所有教条日期时间都以以下格式呈现：
"maintenanceStart":{
"timezone":{"name":"Australia\/Melbourne","location":{"country_code":"AU","latitude":-37.81667,"longitude":144.96666,"comments":"Victoria"}},"offset":36000,"timestamp":1372892400}
我可以通过在规范化器上设置回调,以每个字段为基础输出ISO日期格式,如下所示：
$callback = function ($dateTime) {
return $dateTime instanceof \DateTime ? $dateTime->format ( \DateTime::ISO8601 ) : '';
};
$normalizer->setCallbacks ( array (
'maintenanceStart' => $callback
) );
但我想将此更改应用于所有时间戳,而无需指定每个字段.有一种直截了当的方法吗？
实体定义：
/** * @var \DateTime * * @ORM\Column(name="MAINTENANCE_START",type="datetime",nullable=true) */ private $maintenanceStart;
控制器代码：
public function dataAction(Request $request) {
$doc = $this->get ( 'doctrine' );
$repo = $doc->getRepository ( 'AppBundle:Customer' );
$result = $repo->createQueryBuilder ( 'c' )->getQuery ()->getResult ();
$encoder = new JsonEncoder ();
$normalizer = new GetSetMethodNormalizer ();
$callback = function ($dateTime) {
return $dateTime instanceof \DateTime ? $dateTime->format ( \DateTime::ISO8601 ) : '';
};
$normalizer->setCallbacks ( array (
'maintenanceStart' => $callback
) );
$serializer = new Serializer ( array (
$normalizer,new \AppBundle\DateTimeNormalizer()
),array (
$encoder
) );
$json = $serializer->serialize ( $result,'json' );
return $this->render ( 'default/data.json.twig',[
'data' => $json,'base_dir' => realpath ( $this->getParameter ( 'kernel.root_dir' ) . '/..' )
] );
}
在尝试遵循规定的方法(在这里发布
https://stackoverflow.com/a/33149890/403614)时,我忽略了观察到传递给Serializer的规范化器数组是依赖于顺序的.
原文链接：https://www.f2er.com/php/240269.html因此,我上面的代码的问题和修复是颠倒顺序,以便DateTime Normalizer首先应用：
$normalizer = new GetSetMethodNormalizer ();
$serializer = new Serializer ( array (
new \AppBundle\DateTimeNormalizer(),$normalizer
),array (
$encoder
) );
