具有与传入参数的值一样多的结果的 SQL 准则查询

Question

假设我给一个函数一个数组：$arrValues = ['19', '4', '4', '18', '19']到一个函数：$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？那可能吗 ？谢谢。

Answer 1

使用 Doctrine 实现这一点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;