php - 具有与参数中传递的值一样多的结果的 SQL 原则查询

Question

假设我给函数一个数组:

$arrValues = ['19', '4', '4', '18', '19']

一个函数:

$objRequeteDoctrine = $this->getEntityManager()->createQueryBuilder()
    ->select('o')
    ->from('xxxxxx', 'o')
    ->where('o.id IN (:values)')
    ->andwhere('o.actif = 1')
    ->setParameter(':values', $arrValues );

return $objRequeteDoctrine->getQuery()->getResult();

在这里，它将检索 3 个对象(删除重复项)。但是，如果我想检索 5 个重复的对象怎么办？这可能吗？

谢谢。

Answer 1

用 Doctrine 实现这一目标可能不是您问题的答案，但可以解决您的问题。 之后生成完整数组怎么样？

$arrValues = ['19', '4', '4', '18', '19'];

$objRequeteDoctrine = $this->getEntityManager()->createQueryBuilder()
    ->select('o')
    ->from('xxxxxx', 'o')
    ->where('o.id IN (:values)')
    ->andwhere('o.actif = 1')
    ->setParameter(':values', $arrValues );

$result = $objRequeteDoctrine->getQuery()->getResult();

// Get the object ids in a separate array to find their position in the result

$idsOnly = [];
foreach($result as $entity) {
    $idsOnly[] = $entity->getId();
}

// Find the key of the object by id in the first result

$fullResult = [];
foreach ($arrValues as $id) {
    $keyFound = array_search($id, $idsOnly);
    if ($keyFound !== false) {
        $fullResult[] = $result[$keyFound];
    }
}

return $fullResult;