Question

如何查找以指定字符开头的记录？

例如：只有用户名以A

开头的用户

用户类

    <?php

    namespace DemoBundle\Entity;

    use FOS\UserBundle\Model\User as BaseUser;
    use Doctrine\ORM\Mapping as ORM;

    /**
     * @ORM\Entity
     * @ORM\Table(name="fos_user")
     */
    class User extends BaseUser
    {

        /**
         * @ORM\Id
         * @ORM\Column(type="integer")
         * @ORM\GeneratedValue(strategy="AUTO")
         */
        protected $id;

        /**
         * @ORM\Column(type="string", length=50)
         */
        protected $name;

        /**
         * @ORM\Column(type="string", length=50)
         */
        protected $lastname;

        public function __construct()
        {
            parent::__construct();
            // your own logic
        }
  //getters and setters
    }

和解决方案@Kwido的控制器操作

public function listAction($letter)
    {
        $queryBuilder = $this->getEntityManager()->createQueryBuilder();
        $queryBuilder
            ->select(['user'])
            ->from(User::class, 'user')
            ->andWhere($queryBuilder->expr()->like('user.lastname', ':lastname'))
            ->setParameter('lastname', $queryBuilder->expr()->literal('A%'));

        $result = $queryBuilder->getQuery()->getResult();

        die(var_dump($result));
}

这就是我所拥有的一切。这给了我错误

尝试调用名为“getEntityManager”的未定义方法 class DemoBundle \ Controller \ DefaultController“

Answer 1

您好@Damian将您的代码更改为。

public function listAction($letter)
{
    $result = $this->getDoctrine()
                     ->getManager()
                     ->createQueryBuilder()
                     ->select('u')
                     ->from(User::class, 'u')
                     ->where('u.lastname LIKE :lastname')
                     ->setParameter('lastname', 'A%')
                     ->getQuery()
                     ->getResult();

    dump($result);

    // return response bellow
}

加载页面后，单击Symfony探查器栏并转到数据库部分。查看生成的查询并尝试直接在数据库上运行它。

Answer 2

使用Doctrine的querybuilder Like expr：

// Example - $qb->expr()->like('u.firstname', $qb->expr()->literal('Gui%'))
public function like($x, $y); // Returns Expr\Comparison instance

要在存储库类中使用Like表达式：

// getEntityManager() = $this->getDoctrine()->getManager();
$queryBuilder = $this->getEntityManager()->createQueryBuilder();
$queryBuilder
    ->select(['user'])
    ->from(User::class, 'user')
    ->andWhere($queryBuilder->expr()->like('user.userName', ':userName'))
    ->setParameter('userName', $queryBuilder->expr()->literal('A%'));

$result = $queryBuilder->getQuery()->getResult();

Symfony2如何查找以指定字符开头的记录？

2 个答案: