當我將「PDO :: commit()」方法放入「try」塊時,它不起作用,即使SQL代碼是確定的。在「嘗試」塊中PHP/PDO :: commit不起作用
$conn->dbh->beginTransaction();
$stmt = $conn->dbh->prepare('some SQL-code');
$stmt->bindValue(...);
try
{
$stmt->execute();
$conn->dbh->commit();
}
catch (Exception $e)
{
$dbh->rollBack();
echo $e->getMessage();
}
但是,如果我把它放在「catch」塊之後,一切正常。
...
try
{
$stmt->execute();
}
catch (Exception $e)
{
$dbh->rollBack();
echo $e->getMessage();
}
$conn->dbh->commit();
是,這是一個預期behaviuor爲什麼PDO ::提交()裏面的「嘗試」塊不起作用?
只是想澄清一下,因爲我花了整整兩個小時才找到解決方案,並且我不確定該解決方案是否正確。
下面的代碼:
Class Connection
{
public $dbh;
private static $instance;
private function __construct()
{
$config = parse_ini_file('config.ini');
$dsn = $config['db.dbms'] . ':host=' . $config['db.host'] .
';dbname=' . $config['db.dbname'] .
';port=' . $config['db.port'] .
';connect_timeout=15';
$this->dbh = new PDO($dsn, $config['db.user'], $config['db.password'], array(PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8"));
$this->dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
}
public static function getInstance()
{
if (!isset(self::$instance))
{
$object = __CLASS__;
self::$instance = new $object;
}
return self::$instance;
}
}
Class NestedSet
{
public function insertAsLastChildOf ($parentnode)
{
$fields = '';
$params = '';
$conn = Connection::getInstance();
$conn->dbh->beginTransaction();
foreach ($this->_modelfields as $field => $type)
{
$fields .= '`'.$field.'`, ';
$params .= ':'.$field.', ';
}
$stmt = $conn->dbh->prepare('UPDATE `' . $this->_tablename . '`
SET `rgt` = `rgt` + 2
WHERE `rgt` >= :parentnodergt;
UPDATE `' . $this->_tablename . '`
SET `lft` = `lft` + 2
WHERE `lft` > :parentnodergt;
INSERT INTO `' . $this->_tablename . '`
(' . $fields . '
`lft`,
`rgt`,
`level`)
SELECT
' . $params . '
`rgt` - 2,
`rgt` - 1,
`level` + 1
FROM `' . $this->_tablename . '`
WHERE `id` = :parentnodeid;');
foreach ($this->_modelfields as $field => $type)
{
$pdoparam = (stripos($type, 'int') === 0) ? PDO::PARAM_INT : PDO::PARAM_STR;
$stmt->bindValue(':'.$field.'', $this->$field, $pdoparam);
}
$stmt->bindValue(':parentnodergt', $parentnode->rgt, PDO::PARAM_INT);
$stmt->bindValue(':parentnodeid', $parentnode->id, PDO::PARAM_INT);
try
{
$stmt->execute();
$conn->dbh->commit();
}
catch (PDOException $e)
{
$conn->dbh->rollBack();
echo $e->getMessage() . '<br/> file - ' . __FILE__ . '<br/> line - ' . __LINE__ . '<br/>';
}
unset($stmt);
}
}
我試圖重寫代碼,使我有準備每一個語句的查詢。在這種情況下,一切工作正常 - 如果沒有錯誤,我得到代碼執行,並得到一個異常,否則回滾。
下面的代碼
$stmt_1 = $conn->dbh->prepare('UPDATE `' . $this->_tablename . '`
SET `rgt` = `rgt` + 2
WHERE `rgt` >= :parentnodergt;');
$stmt_1->bindValue(':parentnodergt', $parentnode->rgt, PDO::PARAM_INT);
$stmt_2 = $conn->dbh->prepare('UPDATE `' . $this->_tablename . '`
SET `lft` = `lft` + 2
WHERE `lft` > :parentnodergt;');
$stmt_2->bindValue(':parentnodergt', $parentnode->rgt, PDO::PARAM_INT);
$stmt_3 = $conn->dbh->prepare('INSERT INTO `' . $this->_tablename . '`
(' . $fields . '
`lft`,
`rgt`,
`level`)
SELECT
' . $params . '
`rgt` - 2,
`rgt` - 1,
`level` + 1
FROM `' . $this->_tablename . '`
WHERE `id` = :parentnodeid;');
foreach ($this->_modelfields as $field => $type)
{
$pdoparam = (stripos($type, 'int') === 0) ? PDO::PARAM_INT : PDO::PARAM_STR;
$stmt_3->bindValue(':'.$field.'', $this->$field, $pdoparam);
}
$stmt_3->bindValue(':parentnodeid', $parentnode->id, PDO::PARAM_INT);
try
{
$stmt_1->execute();
$stmt_2->execute();
$stmt_3->execute();
$conn->dbh->commit();
}
catch (PDOException $e)
{
$conn->dbh->rollBack();
echo $e->getMessage() . '<br/> file - ' . __FILE__ . '<br/> line - ' . __LINE__ . '<br/>';
}
unset($stmt);
這是否意味着PDO ::準備()不應該與多個查詢中使用?
我在問,因爲我在PDO手冊中沒有找到任何有關這方面的內容。
您是否爲您的PDO實例啓用了異常拋出? '$ dbh-> setAttribute(PDO :: ATTR_ERRMODE,PDO :: ERRMODE_EXCEPTION);' – hakre 2012-03-06 17:45:00
是的,我把它放在Connection類(Singleton)的__construct()中:'$ this-> dbh = new PDO(。 ..); $ this-> dbh-> setAttribute(PDO :: ATTR_ERRMODE,PDO :: ERRMODE_EXCEPTION);'如果我在SQL代碼中有錯誤,則會拋出異常。 – Placido 2012-03-06 17:49:01
試圖挑起異常,看看是否真的有效。 – hakre 2012-03-06 17:50:23