如何在php mysql的外鍵表中插入多於1行？

Question

表 1：通知表2：用戶通知

$notice = "SELECT id FROM notification
    WHERE noti_user ='Pro'";
$noticeqry = mysqli_query($con, $notice); 

$sql = "INSERT INTO `user_notification` (`notif_id`, `user_id`) 
  values while($noticerow = mysqli_fetch_array($noticeqry)){("$noticerow['id']", "$univuid"}";
  $query= mysqli_query($con,$sql);

不工作並出現錯誤解析錯誤：語法錯誤，意外變量“$noticerow”

Answer 1

在INSERT語法是路要走，而除了這個事實存在SQL注入的問題，我相信你正在試圖做的是：

while ($noticerow = mysqli_fetch_array($noticeqry)) {
    $sql = "INSERT INTO `user_notification` (`notif_id`, `user_id`)
            VALUES ({$noticerow['id']}, $univuid)";
    mysqli_query($con, $sql);
}

但是，我強烈建議轉為使用參數化查詢來避免 SQL 注入問題，這可以將上述處理替換為：

$sql = 'INSERT INTO `user_notification` (`notif_id`, `user_id`) VALUES (?, ?)';
$stmt = mysqli_prepare($con, $sql);

while ($noticerow = mysqli_fetch_array($noticeqry)) {
    mysqli_stmt_bind_param($stmt, 'ii', $noticerow['id'], $univuid);
    mysqli_stmt_execute($stmt);
}

Answer 2

也許您可以嘗試通過 for 循環插入，如下所示。

$notice = "SELECT id FROM notification WHERE noti_user ='Pro'";
$noticeqry = mysqli_query($con, $notice); 
$row = $noticeqry->fetch_assoc();
$count_id = mysqli_num_rows($noticeqry);

for($i=0; $i < $count_id; $i++){

    $noticerow = $row['id'][$i];

    $sql = "INSERT INTO `user_notification` (`notif_id`, `user_id`) values ('$noticerow', '$univuid')";
    $query= mysqli_query($con,$sql);

}

if($quey){echo "New record created successfully";}else{ echo "Error: " . $sql . "<br>" . mysqli_error($con);}