Question

我想使用以下mysql查询：select * from x join y on x.y_id = y.id where x.a= 0

我尝试使用以下教义语法将它放在我的Symfony项目中。

$repositoryX->createQueryBuilder('x')
         ->add('select', 'x')
         ->add('from', 'AppBundle:X x')
         ->leftJoin('AppBundle:Y', 'y', 'WITH', 'x.y_id = y.id')
         ->where('x.a = :a')
         ->setParameter('a', 0)
         ->getQuery()
         ->getResult();

我的问题是我只看到表x的结果，但我也想要表y的结果（因为连接）。当我进行额外选择时，我得到Neither the property "y_id" nor one of the methods "y_id()", "gety_id()"/"isy_id()" or "__call()" exist and have public access in class

＆＃39; *＆＃39;在学说中不起作用。

Answer 1

您只选择x更改：

->add('select', 'x')

到

->add('select', 'x,y')
//The second param might accept and array i dont remember off hand
->add('select', ['x','y'])

或写得更好：

$repository->createQueryBuilder()
    ->select(['x','y'])
    ->from('AppBundle:X','x')
    ->join('x.y','y') // This assumes you have the mapping defined for this property
    ->where('x.a = :a')
    ->setParameter('a', 0)
    ->getQuery()
    ->getResult();

如果使用注释映射，它应该看起来像

//X entity

//...
class X {

    //...
    /**
     * @ORM\OneToOne(targetEntity="Y")
     * @ORM\JoinColumn(name="y_id", referencedColumnName="id")
     */
     protected $y;
}

Symfony：得到两个表的结果

1 个答案: