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我有一个具有 2 个属性的实体，name并且photo. 该name属性是从数据库中读取的，但我必须photo用一些其他信息填充该属性。

我已经按照文档中的编写自定义规范化器教程进行操作，并制作了自定义规范化器：

<?php

namespace App\Serializer;

use App\Entity\Style;
use Symfony\Component\Serializer\Normalizer\DenormalizerInterface;
use Symfony\Component\Serializer\Normalizer\NormalizerInterface;
use Vich\UploaderBundle\Templating\Helper\UploaderHelper;

final class StyleNormalizer implements NormalizerInterface, DenormalizerInterface
{
    private $normalizer;

    private $uploaderHelper;

    public function __construct(NormalizerInterface $normalizer, UploaderHelper $uploaderHelper)
    {
        if (!$normalizer instanceof DenormalizerInterface) {
            throw new \InvalidArgumentException('The normalizer must implement the DenormalizerInterface');
        }

        $this->normalizer = $normalizer;
        $this->uploaderHelper = $uploaderHelper;
    }

    public function denormalize($data, $class, $format = null, array $context = [])
    {
        return $this->normalizer->denormalize($data, $class, $format, $context);
    }

    public function supportsDenormalization($data, $type, $format = null)
    {
        return $this->normalizer->supportsDenormalization($data, $type, $format);
    }

    public function normalize($object, $format = null, array $context = [])
    {
        if ($object instanceof Style) {
            $object->setPhoto('http://api-platform.com');
        }

        return $this->normalizer->normalize($object, $format, $context);
    }

    public function supportsNormalization($data, $format = null)
    {
        return $this->normalizer->supportsNormalization($data, $format);
    }
}

但是该photo属性没有填写所需的信息。

经过一点调试，我发现该supportsNormalization方法执行了两次（对于每个数据库元素）。如果我打印$data变量，我name第一次得到实体属性，第二次得到photo具有null值的属性。我从来没有得到整个Style实体。然后该supportsNormalitzation方法总是返回false。

如何获取完整Style实体并修改其属性？

谢谢！

4

1 回答 1

1

尝试将此添加到您的supportsNormalization方法中：

public function supportsNormalization($data, $format = null)
{
    return
        $this->normalizer->supportsNormalization($data, $format)
        && is_object($data) && $data instanceof Style::class
        ;
}
于 2018-03-26T12:30:05.200 回答