Question

PHP代码：

    $xCodesQueryBuilder = $conn->createQueryBuilder();

    $xCodesQueryBuilder->select("l.id","mdsh.xcode","mdso.xcode")
            ->from("location_tree","l")
            ->join("l","location_tree_pos","p","l.id = p.tree_id")
            ->rightJoin("l","hotel","h","h.location_id = l.id")
            ->leftJoin("l","offer_location","ol","l.id=ol.location_id")
            ->leftJoin("ol","mds_offer","mdso","ol.offer_id = mdso.offer_id")
            ->leftJoin("h","mds_hotel","mdsh","h.id = mdsh.hotel_id")
            ->where("p.parent_id IN (:ids)")
            ->andWhere("(mdso.xcode IS NOT NULL OR mdsh.xcode IS NOT NULL)");

    var_dump($xCodesQueryBuilder->getSQL());exit;

结果：

SELECT l.id, mdsh.xcode, mdso.xcode
FROM location_tree l 
INNER JOIN location_tree_pos p ON l.id = p.tree_id 
RIGHT JOIN hotel h ON h.location_id = l.id 
LEFT JOIN offer_location ol ON l.id=ol.location_id 
WHERE (p.parent_id IN (:ids)) 
AND ((mdso.xcode IS NOT NULL OR mdsh.xcode IS NOT NULL))

为什么2个最后加入的任何想法都被省略了？

Answer 1

我刚刚为我工作了。必须修改getSQLForSelect()中的函数QueryBuilder.php。

我打开了一个ticket并向DBAL提交了拉取请求，但与此同时，请随时使用my patched copy。

<强>更新

刚刚意识到解决此问题的另一种方法是始终在任何FROM方法中使用$fromAlias表别名作为第一个参数（join()）。

在您的情况下，您将代码更改为：

$xCodesQueryBuilder = $conn->createQueryBuilder();

$xCodesQueryBuilder->select("l.id","mdsh.xcode","mdso.xcode")
        ->from("location_tree","l")
        ->join("l","location_tree_pos","p","l.id = p.tree_id")
        ->rightJoin("l","hotel","h","h.location_id = l.id")
        ->leftJoin("l","offer_location","ol","l.id=ol.location_id")
        ->leftJoin("l","mds_offer","mdso","ol.offer_id = mdso.offer_id")
        ->leftJoin("l","mds_hotel","mdsh","h.id = mdsh.hotel_id")
        ->where("p.parent_id IN (:ids)")
        ->andWhere("(mdso.xcode IS NOT NULL OR mdsh.xcode IS NOT NULL)");

var_dump($xCodesQueryBuilder->getSQL());exit;

Answer 2

我认为Doctrine 2.3.1支持这一点。

可以将连接转换为from和where，如果需要，可以使用group by。卡住了2-3个小时，最后使用了这个解决方法因为即使升级后也没有帮助我（试图找出原因）

Doctrine DBAL查询构建器省略了一些连接

2 个答案: