我正在开发一个扩展,它显示sys_file_reference中引用的所有文件。还可以根据sys_file_metadata中的新自定义属性对每个文件进行分类和过滤。
获取和过滤所有引用文件作为TYPO3\CMS\Core\Resource\File
模型的最佳方法是什么?
在我的帮助下,TYPO3\CMS\Core\Resource\FileRepository
我能够获得File
包括元数据的模型。但是存储库总是期望 auid
检索一个File
. 现在我用我的存储库获取所有引用的文件并将它们传递给findFileReferenceByUid()
方法。
class DocumentRepository extends \TYPO3\CMS\Extbase\Persistence\Repository {
/**
* Disables pid constraint
*
* @return void
*/
public function initializeObject() {
$querySettings = $this->objectManager->create('TYPO3\\CMS\\Extbase\\Persistence\\Generic\\Typo3QuerySettings');
$querySettings->setRespectStoragePage(FALSE);
$this->setDefaultQuerySettings($querySettings);
}
/**
* Finds all referenced documents returning them as File modules
*
* @return void
*/
public function findAllReferenced() {
$fileRepository = \TYPO3\CMS\Core\Utility\GeneralUtility::makeInstance('TYPO3\\CMS\\Core\\Resource\\FileRepository');
$query = $this->createQuery();
$documents = $query->execute();
$references = array();
foreach ($documents as $document) {
$references[] = $fileRepository->findFileReferenceByUid($document->getUid());
}
return $references;
}
}
有没有更简单的方法,我该如何实现过滤器?
更新
现在我绕过 extbase 并直接使用数据库连接。TYPO3 在内部经常这样做。
/**
* Finds all referenced documents returning them as File models
*
* @return array
*/
public function findAllReferenced() {
$rows = $this->getDatabaseConnection()->exec_SELECTgetRows(
'sys_file AS sf
LEFT JOIN tt_content AS ttc ON ttc.header_link=CONCAT("file:", sf.uid)
LEFT JOIN sys_file_reference AS ref ON sf.uid=ref.uid_local
LEFT JOIN sys_file_metadata AS meta ON sf.uid=meta.file',
'((ttc.uid IS NOT NULL AND ttc.hidden = 0 AND ttc.deleted = 0) OR (ref.uid IS NOT NULL AND ref.hidden = 0 AND ref.deleted = 0))',
'sf.uid'
);
$documents = array();
if (!empty($rows)) {
foreach ($rows as $row) {
$documents[] = $this->createDomainObject($row);
}
}
return $documents;
}