PDO在try / catch [duplicate]中未给出错误响应

Question

此问题已经在这里有了答案：

• Try/Catch block in PHP not catching Exception9个答案

我有一些PDO代码，即使该行未成功插入数据库，也不会出现错误...

try {
    $options = [
        \PDO::ATTR_ERRMODE => \PDO::ERRMODE_EXCEPTION,
        \PDO::ATTR_DEFAULT_FETCH_MODE => \PDO::FETCH_ASSOC,
        \PDO::ATTR_EMULATE_PREPARES => false,
    ];
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);

    // prepare sql and bind parameters
    $stmt = $conn->prepare("INSERT INTO reservations (bookingdatetime, remoteip) 
                                  VALUES (:bookingdatetime, :remoteip)");

    $stmt->bindParam(':bookingdatetime', $bookingdatetime);
    $stmt->bindParam(':remoteip', $remoteip);

    // insert a row
    $stmt->execute();

    $bookingid = $conn->lastInsertId();

    echo json_encode(array("title" => "WE DID IT!", "body" => " The row # " . $bookingid . " was made!"))

} catch(PDOException $e) {
    echo json_encode(array("title" => "Database Error", "body" => $e->getMessage()));
}
$conn = null;

编辑：我应该说，即使没有成功创建该行，它仍然给出一个“我们做到了！”消息，并且行ID返回为0

编辑2：如果我在我的prepare语句列名称上输入类型，它仍将返回“ WE DID IT” ...这将是一个明显的错误。

Answer 1

有一个函数rowCount，它说明上一个查询插入/更新的行数

$stmt->execute();
$rowCount = $stmt->rowCount();

$bookingid = $conn->lastInsertId();

if($rowCount > 0)
    echo json_encode(array("title" => "WE DID IT!", "body" => " The row # " . $bookingid . " was made!"));
else 
    echo json_encode(array("title" => "Database Error", "body" =>"Insert Failed"));