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Php 如何在codeigniter中使用条令查询生成器?_Php_Codeigniter_Doctrine Orm - Fatal编程技术网

Php 如何在codeigniter中使用条令查询生成器?

Php 如何在codeigniter中使用条令查询生成器?,php,codeigniter,doctrine-orm,Php,Codeigniter,Doctrine Orm,我有一个问题: 我使用的是条令2和代码点火器2。我想知道条令查询生成器是如何工作的?我想在条令查询生成器上使用select查询,但我做不到。我也读过条令文件并尝试过,但没有成功 尝试此操作后,我收到以下错误消息: 致命错误:未捕获异常“条令\ORM\Query\QueryException”,消息为“选择u.username、u.password、u.email FROM Entity\Userss u,在C:\xampp\htdocs\hotell\application\libraries\

我有一个问题: 我使用的是条令2和代码点火器2。我想知道条令查询生成器是如何工作的?我想在条令查询生成器上使用select查询,但我做不到。我也读过条令文件并尝试过,但没有成功

尝试此操作后,我收到以下错误消息:

致命错误:未捕获异常“条令\ORM\Query\QueryException”,消息为“选择u.username、u.password、u.email FROM Entity\Userss u,在C:\xampp\htdocs\hotell\application\libraries\doctor\ORM\Query\QueryException.php:39堆栈跟踪:0 C:\xampp\htdocs\hotell\application\libraries\doctor\ORM\Query\Parser.php429:doctor\ORM\Query\QueryException::dqlError'SELECT u.userna…'1C:\xampp\htdocs\hotell\application\libraries\Doctrine\ORM\Query\Parser.php854:Doctrine\ORM\Query\Parser->语义错误'Class'Userss'…',数组2 C:\xampp\htdocs\hotell\application\libraries\doctor\ORM\Query\Parser.php1529:doctor\ORM\Query\Parser->AbstractSchemaName 3 C:\xampp\htdocs\hotell\application\libraries\doctor\Query\Parser.php1426:doctor\ORM\Query\Parser->RangeVariableDeclaration 4C:\xampp\htdocs\hotell\application\libraries\doctor\ORM\Query\Parser.php1173:doctor\ORM\Query\Parser->IdentificationVariableDeclaration 5 C:\xampp\htdocs\hotell\applic在C:\xampp\htdocs\hotell\application\libraries\doctor\ORM\Query\Query\QueryException.php第49行

我的实体代码:

<?php    
namespace Entity;
use Doctrine\Common\Collections\ArrayCollection;

/**
 * @Entity
 * @Table(name="user")
 */
class Userss
{
    /**
     * @Id
     * @Column(type="integer", nullable=false)
     * @GeneratedValue(strategy="AUTO")
     */
    public $id;

    /**
     * @Column(type="string", length=32, unique=true, nullable=false)
     */
    public $username;

    /**
     * @Column(type="string", length=64, nullable=false)
     */
    public $password;

    /**
     * @Column(type="string", length=255, unique=true, nullable=false)
     */
    public $email;

    /**
     * The @JoinColumn is not necessary in this example. When you do not specify
     * a @JoinColumn annotation, Doctrine will intelligently determine the join
     * column based on the entity class name and primary key.
     *
     * @ManyToOne(targetEntity="Group")
     * @JoinColumn(name="group_id", referencedColumnName="id")
     */

    public function setUsername($Username)
    {
      $this->username = $Username;
    }

    public function getUsername()
    {
      return $this->username;
    }

    // and more...
}
这是我从CI_模型扩展而来的模型中的代码:

<?php

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
require_once APPPATH . 'modules/user/models/entity/user_modell.php';


  class User_model extends CI_Model
  {
    public $em;
    public function __construct() {
      parent::__construct();

      $this->em = $this->doctrine->em;
    }


    function test()
    {
      $repo = $this->em->getRepository('Entity\\Userss', 'u');
      $qb = $repo->createQueryBuilder('u')
                 ->select('u.username, u.password, u.email')
                 ->from('Userss', 'u')
                 ->orderBy('u.username', 'ASC')
                 ->getQuery()->getArrayResult();
      var_dump($qb);
      die;
    }
  }
试着改变

->getQuery()->getArrayResult(); 
进入


它对我有效。

你看到了吗:?是的,我阅读并应用了此文档,但我得到了相同的错误:它有效!我评论这个:名称空间实体;但我不知道它的原因。。。
->getQuery()->getResult();