php - 如何使用 PHP 更新 MySQL 中的不同值

Question

我有一个问题不知道如何解决。

我想为 MySQL 中的所有“ID”更新不同的值。

我使用 PHP 来实现此目的，这是我的代码:

//First ID check 

$result = mysqli_query($conn,"SELECT subscription_end, today_date FROM users_manager WHERE id=62");

while($date = mysqli_fetch_array($result))
{
  $date['subscription_end'];
  $date['today_date'];

  if ($date['subscription_end'] <= $date['today_date']) {
    $update = "UPDATE users_manager SET status='of' WHERE id=62";
  } else {
    $update = "UPDATE users_manager SET status='on' WHERE id=62";
  }
}

//Second ID check
$result = mysqli_query($conn,"SELECT subscription_end, today_date FROM users_manager WHERE id=50");

while($date = mysqli_fetch_array($result))
{
  $date['subscription_end'];
  $date['today_date'];

  if ($date['subscription_end'] <= $date['today_date']) {
    $update = "UPDATE users_manager SET status='off' WHERE id=50";
  } else {
    $update = "UPDATE users_manager SET status='on' WHERE id=50";
  }
}

当我添加新用户并且该用户具有新 ID 示例 55 时，我必须再次添加一行代码来检查新用户。

你能告诉我如何随时查看所有这些以及新老用户吗？

Answer 1

试试这个，一切都记录在案..如果您没有得到任何信息，请问我。

/**
 * UserId:  This variable holds the userId,
 *          You have to bring it from database and referenced to this variable.
*/
$userId = 10;
// This variable represents the status column in the 'user_manager' Table.
$status = null;

$result = mysqli_query($conn,"SELECT subscription_end, today_date FROM 
users_manager WHERE id={$userId}");

while($date = mysqli_fetch_array($result))
{
    $date['subscription_end'];
    $date['today_date'];

    $status = ($date['subscription_end'] <= $date['today_date']) ? 'off' : 'on';

    $update = "UPDATE users_manager SET status={$status} WHERE id={$userId}";

}