Question

我正试图绕过PDO事务，一次提交一组相当复杂的MySQL查询。但是，当我运行事务时，它将提交一个查询而不提交另一个查询 - 如果任一查询中存在单个错误，我希望它回滚两个查询而不对这两个表进行更改。

到目前为止：我的connect.php文件：

class DbConnect {
    private $conn;
    function __construct() {        
    }
    /**
     * Establishing database connection
     * @return database connection handler
     */
    function connect() {
        //Where HOST, USER, PASS etc are set
        include_once "./dbconfig.php";

        // Establish the connection
        try {
        $this->conn = new PDO("mysql:host=".HOST.";dbname=".DBNAME, USER, PASS); 
        $this->conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $this->conn->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);

        return $this->conn;
        } catch (PDOException $e) {
            print "Error!: " . $e->getMessage() . "<br/>";
            die();
        }
    }
}

我正在尝试传递同步SQL查询的文件

    public function transaction ($userId, $amount){
        //Creates the PDO connection EDIT: added my DB connection
        $db = new DbConnect();
        $this->conn = $db->connect();

        $con = $this->conn;

            $con->beginTransaction();
            try{
                $sql = "INSERT INTO transactions (id_user, amount) VALUES (?, ?)";
                $trans = $con->prepare($sql);
                $trans->execute([$userId, $amount]);
                //If I purposely create an error here the query above still runs in the database e.g. remove the $amount variable
                $this->updateBalance($userId, $amount);
                $con->commit();
                return true;
            }
            catch (PDOException $e) {
                $con->rollBack();
                throw $e;                    
            }
    }

    private function updateBalance ($userId, $amount){
            $time = time();
            $sql = "UPDATE balance SET balance=balance + ? WHERE user_id = ?";
            $stmt = $this->conn->prepare($sql);
            $stmt->execute([$amount, $userId]);
            $row_count = $stmt->rowCount();
            return $row_count > 0;
        }

以上只是更大更复杂程序的一小部分示例，否则我只是将余额查询放在与事务相同的位置，但是我需要将它保存在单独的函数中。任何想法我怎么能把它变成“全有或全无”提交状态？

Answer 1

首先，您没有检查第二个功能的调用的返回状态

$this->updateBalance($userId, $amount);

那么，即使有错误，您怎么知道错误？

如果你使那个被调用的函数抛出异常而不是返回一个状态，它应该被调用块catch（）块捕获，导致rollback()而不是commit()

像这样的东西

public function transaction ($userId, $amount){
    //Creates the PDO connection
    $con = $this->conn;

    $con->beginTransaction();
    try{
        $sql = "INSERT INTO transactions (id_user, amount) VALUES (?, ?)";
        $trans = $con->prepare($sql);
        $trans->execute([$userId, $amount]);
        // If I purposely create an error here the 
        // query above still runs in the database 
        // e.g. remove the $amount variable
        $this->updateBalance($userId, $amount);
        $con->commit();
        return true;
    }
    catch (PDOException $e) {
        $con->rollBack();
        throw $e;
        return false;
    }
}

/*
* If this function throws an exception 
* rather than returning a status, then it will
* stop execution of the try block and
* be caught by the calling blocks catch() block
*/
private function updateBalance ($userId, $amount){

    $sql = "UPDATE balance SET balance=balance + ? WHERE user_id = ?";
    $stmt = $this->conn->prepare($sql);
    $res = $stmt->execute([$amount, $userId]);
    if ( ! $res ) {
        throw new Exception('It errored');
    }

}

或者，您可以通过在连接到数据库后设置PDO::ERRMODE_EXCEPTION来使所有PDO调用抛出异常。

$dbh = new PDO($dsn, $user, $password);
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

但是，这可能是对PDO错误处理的重大改变，具体取决于您已经生成了多少代码。

嵌套的PDO事务不会回滚所有查询

1 个答案: