以下插入查询在PDO::ATTR_EMULATE_PREPARES
为真时工作正常,但在设置为false时会产生SQLSTATE[HY093]: Invalid parameter number
。这是预期的吗?当查询使用相同的值两次时,在数组中包含具有相同值的单独索引总是好的做法吗?例如,$sql='INSERT INTO t1(id,v1) VALUES(:id,:v1) ON DUPLICATE KEY UPDATE v1=:v1_dup'; ... $stmt->execute(array('id'=>1,'v1'=>123, 'v1_dup'=>123));
不知道它是否重要,但我使用PHP的本地mysqlnd驱动程序用于MySQL。
<?php
class db
{
private static $instance = NULL;
private function __construct() {} //Make private
private function __clone(){} //Make private
public static function db() //Get instance of DB
{
if (!self::$instance)
{
try{self::$instance = new PDO("mysql:host=localhost;dbname=myDB;charset=utf8",'myUser','myPW',array(PDO::ATTR_EMULATE_PREPARES=>false,PDO::ATTR_ERRMODE=>PDO::ERRMODE_EXCEPTION,PDO::ATTR_DEFAULT_FETCH_MODE=>PDO::FETCH_ASSOC));}
//try{self::$instance = new PDO("mysql:host=localhost;dbname=myDB;charset=utf8",'myUser','myPW',array( PDO::ATTR_ERRMODE=>PDO::ERRMODE_EXCEPTION,PDO::ATTR_DEFAULT_FETCH_MODE=>PDO::FETCH_ASSOC));}
catch(PDOException $e){die(library::sql_error($e));}
}
return self::$instance;
}
}
try{
$sql = 'CREATE TEMPORARY TABLE t1(id INT UNSIGNED NOT NULL AUTO_INCREMENT, v1 INT NOT NULL,PRIMARY KEY (id) )';
db::db()->exec($sql);
$sql='INSERT INTO t1(id,v1) VALUES(:id,:v1) ON DUPLICATE KEY UPDATE v1=:v1';
$stmt = db::db()->prepare($sql);
$stmt->execute(array('id'=>1,'v1'=>123));
$stmt->execute(array('id'=>1,'v1'=>321));
$stmt=db::db()->query('SELECT * FROM t1');
$rs=$stmt->fetchAll(PDO::FETCH_ASSOC);
echo('<pre>'.print_r($rs,1).'</pre>');
}
catch(PDOException $e){die('<pre>'.print_r($e,1).'</pre>');}
?>
答案 0 :(得分:2)
是
由于“此功能可能导致意外行为”这样的原因,为本机预处理语句关闭了多个绑定。
使用其他占位符绑定它
$sql='INSERT INTO t1(id,v1) VALUES(:id,:v1) ON DUPLICATE KEY UPDATE v1=:v1d';
$stmt = db::db()->prepare($sql);
$stmt->execute(array('id'=>1,'v1'=>123,'v1d'=>123));
$stmt->execute(array('id'=>1,'v1'=>321,'v1d'=>321));
或
你可以使用mysql-way:
$sql='INSERT INTO t1(id,v1) VALUES(:id,:v1) ON DUPLICATE KEY UPDATE v1=values(v1)';