dscbxou1900343 2015-09-18 09:03
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lastInsertId()在数据库中创建2个重复记录

So I've got a PHP page that inserts a booking into my database, and I am using the method lastInsertId() to present the user with their Booking ID (which is set to AUTO_INCREMENT in the database).

The problem is, two duplicate records are being added when the code runs, and the method is returning the second Booking ID. Can someone shed some light on why this is happening?

This is the code:

<?php

...

$stmt = $conn->prepare('INSERT bookings (DateFrom, DateTo, GuestHouseID_FK, GuestID_FK, RoomID_FK) VALUES (:DateFrom, :DateTo, :GuestHouseID_FK, :GuestID_FK, :RoomID_FK)'); 
$stmt->bindParam(':DateFrom', $DateFrom, PDO::PARAM_STR); 
$stmt->bindParam(':DateTo', $DateTo, PDO::PARAM_STR); 
$stmt->bindParam(':GuestHouseID_FK', $GuestHouseID_FK, PDO::PARAM_INT); 
$stmt->bindParam(':GuestID_FK', $GuestID_FK, PDO::PARAM_STR);
$stmt->bindParam(':RoomID_FK', $RoomID_FK, PDO::PARAM_STR);
$stmt->execute(); 
echo '<H3>'; 
echo 'Thank you. Your booking has been completed!'; 
echo '</H3>'; 
if($stmt->execute()){
    $booking_id = $conn->lastInsertId();
    echo "Your Booking ID is " . $booking_id;
    echo '<hr>';
}
echo '<p>Click here to return to the main menu</p>';
echo '<form action="@AT2_main_menu.html">';
echo '<input type="submit" value="Main menu">';
echo '</form>';
}
catch(PDOException $e) 
{
  echo "Something's gone wrong. Please try again or contact the admin"; 
}
$conn = null;
?>
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1条回答 默认 最新

  • douzinei6926 2015-09-18 09:06
    关注

    You're running $stmt->execute() twice. When you call it inside the if statement you're calling it a second time. Remove your first call to the function.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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