【发布时间】:2013-04-22 07:12:10
【问题描述】:
我已尝试在 PDO 中禁用模拟准备,但无法使其正常工作。其他一切都有效。查询成功。我认为它不起作用的原因是因为它没有转义引号等所以我得到语法错误。
我尝试了两种不同的方法。
$this->dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
和
$insert = $database->$con->prepare($insert, array(PDO::ATTR_EMULATE_PREPARES => false));
我还注意到 getAttribute 不起作用。
通过这样做...
$emul = $database->$con->getAttribute(PDO::ATTR_EMULATE_PREPARES);
var_dump($emul);
...我收到此错误
SQLSTATE[IM001]: Driver does not support this function: driver does not support that attribute
这是我的数据库类,操作发生在其中。 (我可能在测试时留下了一些不必要/愚蠢的代码。)
<?php
class Database
{
public $dbh;
public $dbh1;
public $dbh2;
private static $instance;
public $numResults;
private $result = array(); // Results that are returned from the query
public function __construct()
{
try
{
$this->dbh = new PDO(DB_TYPE.':host='.DB_HOST.';dbname='.DB_NAME.';charset=utf8', DB_USER, DB_PASS);
$this->dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$this->dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$this->dbh1 = new PDO(DB_TYPE1.':host='.DB_HOST1.';dbname='.DB_NAME1.';charset=utf8', DB_USER1, DB_PASS1);
$this->dbh1->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$this->dbh1->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$this->dbh2 = new PDO(DB_TYPE2.':host='.DB_HOST2.';dbname='.DB_NAME2.';charset=utf8', DB_USER2, DB_PASS2);
$this->dbh2->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$this->dbh2->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
catch (PDOException $e)
{
die("Database Error: ". $e->getMessage() . "<br />");
}
}
public static function getInstance()
{
if (!isset(self::$instance))
{
$object = __CLASS__;
self::$instance = new $object;
}
return self::$instance;
}
private function tableExists($table, $con)
{
switch($con)
{
case 'dbh':
$db_name = DB_NAME;
break;
case 'dbh1':
$db_name = DB_NAME1;
break;
case 'dbh2':
$db_name = DB_NAME2;
break;
}
$database = Database::getInstance();
if(is_array($table))
{
for($i = 0; $i < count($table); $i++)
{
$tablesInDb = $database->$con->prepare('SHOW TABLES FROM '.$db_name.' LIKE "'.$table[$i].'"');
$tablesInDb->execute();
$rowCount = $tablesInDb->rowCount();
if($tablesInDb)
{
if($rowCount <> 1)
{
die('Error: Table does not exist'.$table[$i]);
}
}
}
}else
{
$tablesInDb = $database->$con->prepare('SHOW TABLES FROM '.$db_name.' LIKE "'.$table.'"');
$tablesInDb->execute();
$rowCount = $tablesInDb->rowCount();
if($tablesInDb)
{
if($rowCount <> 1)
{
die('Error: Table does not exist'.$table);
}
}
}
return true;
}
public function insert($con, $table, $values, $cols = null)
{
if($this->tableExists($table, $con))
{
$insert = 'INSERT INTO '.$table;
if($cols != null)
{
$cols = implode(',', $cols);
$insert.= '('.$cols.')';
}
for($i = 0; $i < count($values); $i++)
{
if(is_string($values[$i]))
$values[$i] = "'".$values[$i]."'";
}
$values = implode(',', $values);
$insert .= ' VALUES ('.$values.')';
$database = Database::getInstance();
$insert = $database->$con->prepare($insert, array(PDO::ATTR_EMULATE_PREPARES => false));
$insert->execute();
if($insert)
{
return true;
}else
{
return false;
}
}
}
public function getResult()
{
return $this->result;
}
}
?>
【问题讨论】:
-
语法错误? Ask a question about this very problem,而不是基于你建立的假设。
-
嗯,这完全取决于
DB_TYPE是什么......而且并非每种类型都可能有准备好的语句。 -
@Wrikken DB_TYPE 是 mysql。
-
我看到您的代码的某些部分使用了准备好的语句,但我看不到占位符或参数:-?