Question

我正在尝试将以下sql查询传递给Doctrine：

SELECT id, name, count(*) FROM (SELECT r.id, r.name, f.timestamp FROM `food_log` as f inner join recipe as r on f.recipe_id = r.id WHERE f.user_id = 6 and timestamp > "2016-09-01" and recipe_id is not null group by f.timestamp, r.id) a GROUP BY a.id ORDER BY count(*) DESC

它应该返回给我的食谱，其中包含特定用户从所选时间戳使用单个食谱的次数。

现在我正在尝试使用Doctrine 1.2和Symfony 1.4，但是我不知道如何从子查询进行查询，我试图做这样的事情

        $subQuery = Doctrine_Query::create()
        ->select('r.id, r.name, f.timestamp')
        ->from('FoodLog f')
        ->innerJoin('f.Recipe r')
        ->where('f.user_id = ?', $userId)
        ->andWhere('timestamp > ?', "2016-09-01")
        ->andWhere('f.recipe_id is not null')
        ->andWhere('r.is_premium = ?', $premium)
        ->groupBy('f.timestamp, r.id')
        ->getSqlQuery();

    $query = Doctrine_Query::create()
    ->select('id, name, count(*)')
    ->from($subQuery)
    ->groupBy('id')
    ->orderBy('count(*) DESC');

    return $query->fetchArray();

有人知道我错在哪里吗？非常感谢您的回复！

Answer 1

基本上，您无法在此版本的Doctrine中执行嵌套查询。我建议通过doctrine连接器使用原始SQL查询：

$sql = 'SELECT id, name, count(*) FROM (SELECT r.id, r.name, f.timestamp FROM `food_log` as f inner join recipe as r on f.recipe_id = r.id WHERE f.user_id = 6 and timestamp > "2016-09-01" and recipe_id is not null group by f.timestamp, r.id) a GROUP BY a.id ORDER BY count(*) 
DESC';
$conn = Doctrine_Manager::getInstance()->getCurrentConnection();
$result = $conn->execute($sql);

foreach($result as $data){
    // do something
}

由于你没有保湿物品，你应该会发现这很有效。

从子查询进行查询

1 个答案: