Symfony测试：Browser-Kit上传文件$_FILES为空但Request->files有值

问题原因 #

Browser-Kit在模拟请求时，是直接在内存中构建Symfony的Request对象，不会触发PHP原生的HTTP请求处理流程，因此不会自动填充$_FILES、$_POST这类超全局变量。而Request::createFromGlobals()方法完全依赖这些超全局变量生成Request实例，所以在测试场景下调用该方法无法获取到文件。

解决方案 #

方案1：直接用注入的Request初始化自定义请求类（推荐） #

不要在自定义请求类中调用Request::createFromGlobals()，而是通过构造函数注入Symfony的Request对象，直接从$request->files中获取文件：

自定义请求类示例：

use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Component\Validator\Validation;

class CustomUploadRequest
{
    private $file;

    public function __construct(Request $request)
    {
        $this->file = $request->files->get('file');
        $this->validate();
    }

    private function validate()
    {
        $constraint = new Assert\File([
            'maxSize' => '10M',
            'mimeTypes' => ['image/jpeg', 'image/png'],
        ]);

        $validator = Validation::createValidator();
        $violations = $validator->validate($this->file, $constraint);

        if (count($violations) > 0) {
            throw new \InvalidArgumentException((string) $violations);
        }
    }

    public function getFile()
    {
        return $this->file;
    }
}

控制器中直接注入自定义请求类：

use Symfony\Component\HttpFoundation\JsonResponse;
use Symfony\Component\Routing\Annotation\Route;

#[Route('upload', name: 'upload', methods: 'POST')]
public function uploadFile(CustomUploadRequest $customRequest): JsonResponse
{
    $uploadedFile = $customRequest->getFile();
    // 执行文件上传逻辑
    return new JsonResponse(['status' => 'success', 'filename' => $uploadedFile->getClientOriginalName()]);
}

方案2：测试中手动填充$_FILES（不推荐） #

如果代码必须依赖$_FILES超全局变量，可在测试中手动填充，但这不符合Symfony最佳实践，仅作为临时兼容方案：

$uploadedFile = new UploadedFile(
    $uploadFilePath,
    $fileName
);

$this->kernelBrowser->request(
    "POST",
    "/upload",
    [],
    [
        "file" => $uploadedFile
    ]
);

// 手动填充$_FILES数组
$_FILES['file'] = [
    'name' => $uploadedFile->getClientOriginalName(),
    'type' => $uploadedFile->getMimeType(),
    'tmp_name' => $uploadedFile->getPathname(),
    'error' => UPLOAD_ERR_OK,
    'size' => $uploadedFile->getSize()
];

内容的提问来源于stack exchange，提问作者pkol96