Question

我目前正在将mySQLi PHP代码转换为PDO以提高安全性，并且在理解如何将下面的代码转换并将其实现到新的PDO PHP中时遇到了一些麻烦

下面的代码应使用最后插入的ID（PDO正在插入的ID），并将列idnum设置为等于NUM（和最后插入的ID）。

如何将其转换并添加到PDO？

if (mysqli_query($conn, $sql)) {
    $last_id = mysqli_insert_id($conn);
    $sql = "UPDATE Equipment SET idnum = CONCAT('NUM', '$last_id') WHERE equipment_id = '$last_id'";
    mysqli_query($conn, $sql);
    echo "New record created successfully. Last inserted ID is: " . last_id;
} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);

我想将代码添加到以下PDO PHP脚本中：

$hostdb = 'localhost';
$namedb = 'dbname';
$userdb = 'userdb';
$passdb = 'passdb';
$charset = 'utf8'; 

if (isset($_POST['name'], $_POST['place'], $_POST['person'] , $_POST['number'] , $_POST['other_name'])) {
    // Connect and create the PDO object
    $options = [
        PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
        PDO::ATTR_EMULATE_PREPARES => false,
    ];
    $conn = new PDO("mysql:host=$hostdb;dbname=$namedb;charset=$charset", $userdb, $passdb, $options);


    $stmt = $conn->prepare( ' INSERT INTO `Table1` (name, place, person, number, other_name, progress)
VALUES (:name,:place,:person,:number,:other_name, "Done") ' );


    $stmt->execute([
        'name' => $_POST['name'],
        'place' => $_POST['place'],
        'person' => $_POST['person'],
        'number' => $_POST['number'],
        'other_name' => $_POST['other_name'],
    ]);

    // Shows the number of affected rows
    echo 'Affected rows : '. $stmt->rowCount();

}

Answer 1

我猜想像这样（如果您希望将它们组合在一起）：

$hostdb = 'localhost';
$namedb = 'dbname';
$userdb = 'userdb';
$passdb = 'passdb';
$charset = 'utf8'; 

if (isset($_POST['name'], $_POST['place'], $_POST['person'] , $_POST['number'] , $_POST['other_name']  )) {
    // Connect and create the PDO object
    $options = [
        PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION,
        PDO::ATTR_EMULATE_PREPARES => false,
    ];
    $conn = new PDO("mysql:host=$hostdb;dbname=$namedb;charset=$charset", $userdb, $passdb, $options);

    try{
        //start a transaction {ACID}
        $conn->beginTransaction();

        $stmt = $conn->prepare('INSERT INTO `Table1` (`name`, `place`, `person`, `number`, `other_name`, `progress`)
    VALUES (:name,:place,:person,:number,:other_name, "Done") ' );

        $stmt->execute([
            'name' => $_POST['name'],
            'place' => $_POST['place'],
            'person' => $_POST['person'],
            'number' => $_POST['number'],
            'other_name' => $_POST['other_name'],
        ]);

        //get the last insert ID
        $last_id = $conn->lastInsertId();

        $stmt = $conn->prepare('UPDATE `Equipment` SET `idnum` = CONCAT("NUM", :last_id_0) WHERE `equipment_id` = :last_id_1');
        //named placeholders must be unique
        $stmt->execute([
            'last_id_0' => $last_id,
            'last_id_1' => $last_id
        ]);
        echo "New record created successfully. Last inserted ID is: " . $last_id;

        //commit the changes
        $conn->commit();
    }catch(PDOException $e){
        //roll back the changes on errors
        $conn->rollback();
        echo $e->getMessage();
    }

    // Shows the number of affected rows this is pointless (for insert 1 row it's always 1 or an error)
    //echo 'Affected rows : '. $stmt->rowCount();

}

事务就像将数据库更改为假装（如果是INNODB表），那么如果查询之一出错，则两者都会失败。

建议在创建相关记录时使用它们，这样您就不会留下孤立的记录或行，而不会只是在关系之间徘徊。

将mysqli转换为PDO时出现问题PHP的MySQL

1 个答案: