php Doctrine 中的左连接 ON 条件和其他条件语法
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/19185587/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Left join ON condition AND other condition syntax in Doctrine
提问by juuga
I'm using Doctrine's querybuilder in Symfony2 to create a query to fetch entities.
我在 Symfony2 中使用 Doctrine 的查询构建器来创建一个查询来获取实体。
My current code looks like this:
我当前的代码如下所示：
$repository = $this->getDoctrine()->getRepository('AaaBundle:Application');
$queryBuilder = $repository->createQueryBuilder('a');
$queryBuilder
->addSelect('u')
->addSelect('i')
->orderBy('a.created_date', 'DESC')
->leftJoin('a.created_by', 'u')
->leftJoin('a.installations', 'i')
//->where('i.page = :page')
//->setParameter('page', $found)
;
Now I can use this to get all the pages regardless of them having an installation or not.
But I only want to join them it the $foundpage is available (so that if there is an installation for the app, but it's on another page, the installation wont be joined). If I unquote the where clause, it will show only apps that have an installation for the page. I want all apps with or without installations for the page.
现在我可以使用它来获取所有页面，无论它们是否安装。但是我只想在$found页面可用时加入他们（这样，如果应用程序安装了，但它在另一个页面上，则不会加入安装）。如果我取消引用 where 子句，它将仅显示已安装该页面的应用程序。我想要页面上安装或不安装的所有应用程序。
In SQL I can get this by adding ANDto the join
在 SQL 中，我可以通过添加AND到连接来获得它
LEFT JOIN installations i ON a.id = i.app AND i.page = :page
This way I get the installation info for an app that has an installation on the page, but I get null values on the columns for app's that have installations on other pages or not at all.
通过这种方式，我获得了在页面上安装的应用程序的安装信息，但我在其他页面上安装或根本没有安装的应用程序列上获得了空值。
Is there a way to do this in Doctrine or am I better off just getting all the installations for each application and then checking against the found page with php?
有没有办法在 Doctrine 中做到这一点，或者我最好只获取每个应用程序的所有安装，然后用 php 检查找到的页面？
回答by zapcost
You can try this :
你可以试试这个：
use Doctrine\ORM\Query\Expr;
->leftJoin('a.installations', 'i', Expr\Join::WITH, 'i.page = :page')
->setParameter('page', $page)
See function leftJoin in doc: http://docs.doctrine-project.org/projects/doctrine-orm/en/latest/reference/query-builder.html#high-level-api-methods
请参阅文档中的 leftJoin 函数：http://docs.doctrine-project.org/projects/doctrine-orm/en/latest/reference/query-builder.html#high-level-api-methods
