Question

这就是我得到的：

$queryBuilder = $this->createQueryBuilder('a')->select()->where('a.id = ?0')->setParameters(array($id));
$queryBuilder->leftJoin('Doctrine\Entity\AccommodationUser',    'b');
$queryBuilder->leftJoin('Doctrine\Entity\AccommodationContact', 'c');

这不会工作，我得到：[语法错误]第0行，第97行：错误：预期文字，得到了“加入”＃39;错误信息。但如果我尝试一个一个地做到这一点

$queryBuilder = $this->createQueryBuilder('a')->select()->where('a.id = ?0')->setParameters(array($id));
$queryBuilder->leftJoin('Doctrine\Entity\AccommodationContact', 'c');

或

$queryBuilder = $this->createQueryBuilder('a')->select()->where('a.id = ?0')->setParameters(array($id));
$queryBuilder->leftJoin('Doctrine\Entity\AccommodationUser',    'b');

它确实有用！

Answer 1

$ queryBuilder-＆gt; leftJoin（'a。 nameOfPropertyInAEntity '，'b'）;

Answer 2

Doctrine并没有在join语句中使用类名。如果您想加入另一个表格，则需要使用表格WITH src AS (SELECT DISTINCT i.id, iwt.wearlocid, i.sdesc, i.wearprog, i.isoog, i.minlvl, i.maxlvl FROM item i inner join itemweartable iwt ON i.id = iwt.itemid WHERE iwt.wearlocid >= 1 AND iwt.wearlocid <= 22 AND i.minlvl <= 50) SELECT DISTINCT src.id, src.wearlocid, src.sdesc, src.wearprog, src.isoog, src.minlvl, src.maxlvl, iaft.affectid, iaft.amount, NULL AS affectbyID FROM src left join itemaffecttable iaft ON src.id = iaft.itemid UNION ALL SELECT DISTINCT src.id, src.wearlocid, src.sdesc, src.wearprog, src.isoog, src.minlvl, src.maxlvl, NULL AS affectID, NULL AS amount, iafbt.affectbyid FROM src left join itemaffectedbytable iafbt ON src.id = iafbt.itemid ORDER BY wearlocid, sdesc, id

alias.property

来源：Doctrine docs

我无法与Doctrine 2

2 个答案: