Php 使用显示PDO异常错误的条令添加新字段时
这是我代码的一部分。当我第一次创建employee字段时 输入…它工作得很好…但我尝试添加第三个字段 它显示此错误。请帮助我查找错误Php 使用显示PDO异常错误的条令添加新字段时,php,sql,zend-framework,doctrine-orm,Php,Sql,Zend Framework,Doctrine Orm,这是我代码的一部分。当我第一次创建employee字段时 输入…它工作得很好…但我尝试添加第三个字段 它显示此错误。请帮助我查找错误 ---------- 条令错误显示 [Doctrine\DBAL\DBALException] An exception occurred while executing 'ALTER TABLE leave ADD days VARC
----------
条令错误显示
[Doctrine\DBAL\DBALException]
An exception occurred while executing 'ALTER TABLE leave ADD days VARCHAR(1
00) NOT NULL':
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error i
n your SQL syntax; check the manual that corresponds to your MySQL server v
ersion for the right syntax to use near 'leave ADD days VARCHAR(100) NOT NU
LL' at line 1
[PDOException]
SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error i
n your SQL syntax; check the manual that corresponds to your MySQL server v
ersion for the right syntax to use near 'leave ADD days VARCHAR(100) NOT NU
LL' at line 1
----------
这是我的Model.Leave.php
<?php
namespace Application\Model;
use Doctrine\ORM\Mapping as ORM;
use Doctrine\Common\Collections\ArrayCollection;
/**
* @ORM\Table(name="leave")
* @ORM\Entity
*/
class Leave extends Entity
{
/**
* @ORM\Column(type="string", length=100)
* @var string
*/
protected $employee;
public function getEmployee()
{
return $this->employee;
}
public function setEmployee($emp)
{
$this->employee = $emp;
}
/**
* @ORM\Column(type="string", length=100)
* @var string
*/
protected $type;
public function getType()
{
return $this->type;
}
public function setType($type)
{
$this->type = $type;
}
/**
* @ORM\Column(type="string", length=100)
* @var string
*/
protected $days;
public function getDays()
{
return $this->days;
}
public function setDays($days)
{
$this->days = $days;
}
}
请参见
你需要引用保留的词,例如
/**
* @ORM\Table(name="`leave`")
* @ORM\Entity
*/
“离开”是一个概念,而不是引用它作为一个标识符。你确定你用的是1.2条令吗?我认为1.2没有注释支持或namespaces@Phil:谢谢,我换了假,现在很好用。