mysql - 在执行简单的 LEFT JOIN 时，如何创建高效的 DQL 语句以匹配高效的 SQL？

Question

我可以制作一个简单的 SQL，当存在具有请求的 id 和与 FS 匹配的相应 model 的项目时返回 1 -%，否则为 0。

但是当我尝试将其编写为 DQL 时，我以各种惊人的方式失败了。请参阅下面的 EXPLAIN 结果。

问题:如何编写高效的 DQL？

SQL(高效)

select count(*) 
from item 
left join product on item.product_id = product.id 
where item.id=2222 and product.model like "FS-%";

使用解释:

+----+-------------+---------+-------+--------------------+---------+---------+-------+------+-------+
| id | select_type | table   | type  | possible_keys      | key     | key_len | ref   | rows | Extra |
+----+-------------+---------+-------+--------------------+---------+---------+-------+------+-------+
|  1 | SIMPLE      | item    | const | PRIMARY,product_id | PRIMARY | 4       | const |    1 |       |
|  1 | SIMPLE      | product | const | PRIMARY            | PRIMARY | 4       | const |    1 |       |
+----+-------------+---------+-------+--------------------+---------+---------+-------+------+-------+

DQL(效率不高)

 $this->getEntityManager()
        ->createQueryBuilder()
        ->select('count(i)')
        ->from(Item::class, 'i')
        ->leftJoin(Product::class, 'p')
        ->where('i.id = :id')
        ->andWhere('p.model like :model')
        ->setParameter('id', 2222)
        ->setParameter('model', 'FS-%')
        ->getQuery()->getSingleResult();

结果 SQL:

SELECT * FROM item i0_ LEFT JOIN product p1_
        ON (i0_.id = 409264 AND p1_.model LIKE 'FS-%');

使用解释:

+----+-------------+-------+------+---------------+------+---------+------+--------+-------+
| id | select_type | table | type | possible_keys | key  | key_len | ref  | rows   | Extra |
+----+-------------+-------+------+---------------+------+---------+------+--------+-------+
|  1 | SIMPLE      | i0_   | ALL  | NULL          | NULL | NULL    | NULL | 276000 |       |
|  1 | SIMPLE      | p1_   | ALL  | NULL          | NULL | NULL    | NULL |    564 |       |
+----+-------------+-------+------+---------------+------+---------+------+--------+-------+
2 rows in set (0.00 sec)            

注意:我使用了https://stackoverflow.com/a/25887143/2883328帮我写DQL。

Answer 1

在你的情况下我会尝试这个查询:

$this->getEntityManager()
        ->createQueryBuilder()
        ->select('count(i)')
        ->from(Item::class, 'i')
        ->leftJoin(Product::class, 'p', 'WITH', 'i.product = p.id')
        ->where('i.id = :id')
        ->andWhere('p.model like :model')
        ->setParameter('id', 2222)
        ->setParameter('model', 'FS-%')
        ->getQuery()->getSingleScalarResult();

将 i.product 中的产品更改为您的属性名称