如何在Symfony2中进行LIKE数据库查询

Acy*_*yra 12 doctrine dql query-builder symfony doctrine-orm

这应该很简单,但我找不到一个有效的例子.这是一个控制器方法,它抛出错误"无效的参数号:绑定变量的数量与令牌的数量不匹配".我成功发布了"searchterm"变量,但无法使查询生效.缺什么?谢谢!

 public function searchAction()
{
    $request = $this->getRequest();

    $searchterm = $request->get('searchterm');

    $em = $this->getDoctrine()->getEntityManager();

    $query = $em->createQuery("SELECT n FROM AcmeNodeBundle:Node n WHERE n.title LIKE '% :searchterm %'")
             ->setParameter('searchterm', $searchterm);

    $entities = $query->getResult();

    return array('entities' => $entities);

}
Run Code Online (Sandbox Code Playgroud)

小智 28

我的Symfony2项目的工作示例:

$qb = $this->createQueryBuilder('u');
$qb->where(
         $qb->expr()->like('u.username', ':user')
     )
     ->setParameter('user','%Andre%')
     ->getQuery()
     ->getResult();
Run Code Online (Sandbox Code Playgroud)


Tjo*_*rie 9

您应该转储创建的查询以便于调试.

我只能建议你也试试querybuilder:

$qb = $em->createQueryBuilder();
$result = $qb->select('n')->from('Acme\NodeBundle\Entity\Node', 'n')
  ->where( $qb->expr()->like('n.title', $qb->expr()->literal('%' . $searchterm . '%')) )
  ->getQuery()
  ->getResult();
Run Code Online (Sandbox Code Playgroud)

DOC


Ima*_*iev 5

我认为此选项也有帮助:

$qb = $this->createQueryBuilder('u');
$qb->where('u.username like :user')
     ->setParameter('user','%hereIsYourName%')
     ->getQuery()
     ->getResult();
Run Code Online (Sandbox Code Playgroud)