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这些是我的表：

    Table Gift:
        -id 
        -price
        ... 



    Table Couple:
         -id
         -name 
        ... 



    table registry:  //provide a many-many relation between gifts and couples 
         -id 
         -coupleId 
         -giftId 



    table purchase:
         -amount
         -registryId


我已经编写了一个 sql 查询来获取特定情侣的所有礼物信息

    $qb = $this->createQueryBuilder('g') //gift
    ->from('\BBB\GiftBundle\Entity\Registry', 'reg')
    ->select('g.id , g.price')
    ->where('reg.gift = g.id')
    ->andWhere('reg.couple = :coupleID')
    ->orderBy('reg.id','DESC')
    ->setParameter('coupleID', $coupleID);


或者

     SELECT g.id , g.price,
    FROM gift g, registry reg       
    WHERE reg.gift_id = g.id AND reg.couple_id = 1



我还想获得已购买的礼物（如果有）
EX 的总金额。SUM(purchase.amount) as totalContribute

我试过了：

    $qb = $this->createQueryBuilder('g')
    ->from('\BBB\GiftBundle\Entity\Purchase', 'p')
    ->from('\BBB\GiftBundle\Entity\Registry', 'reg')
    ->select('g.id ,  g.price')
    ->addSelect('SUM(p.amount) as totalContribute')
    ->leftJoin('p','pp', 'ON','reg.id = pp.registry')
    ->where('reg.gift = g.id')
    ->andWhere('reg.couple = :coupleID')
    ->orderBy('reg.id','DESC')
    ->setParameter('coupleID', $coupleID);

但它给了我以下错误：

    [Semantical Error] line 0, col 145 near 'pp ON reg.id': Error: Identification Variable p used in join path expression but was not defined before.
4

1 回答 1

2

首先，您应该在连接后的 SQL 语句中定义连接条件，而不是在 WHERE 子句中。原因是它真的没有效率。所以查询应该是这样的：

 SELECT g.id , g.price,
 FROM gift g JOIN registry reg ON reg.gift_id = g.id
 WHERE reg.couple_id = 1

但是关于您的 Doctrine 查询，您会收到错误，因为您以错误的方式定义连接。您的查询应该更像：

 $qb = $this->createQueryBuilder('g') // You don't have put "from" beacuse I assume you put this into GiftRepository and then Doctrine knows that should be \BBB\GiftBundle\Entity\Gift
    ->select('g.id ,  g.price')
    ->addSelect('SUM(p.amount) as totalContribute')
    ->join('g.purchase','p')          // pay attention for this line: you specify relation basing on entity property - I assume that is "$purchase" for this example        
    ->leftJoin('p.registry', 'reg')   // here you join with \BBB\GiftBundle\Entity\Purchase
    ->where('reg.couple = :coupleID')
    ->orderBy('reg.id','DESC')
    ->setParameter('coupleID', $coupleID);

请将此视为伪代码 - 我没有检查它是否有效，但它应该更像这样。

还有一件事 - 如果您的存储库方法返回 X 实体的对象，您应该将此方法放入 XRepository。

于 2012-10-15T07:42:33.877 回答