php - MySQL-每天休息两次，增加营业时间

该餐厅的营业时间为星期一的上午10点至星期二03：00，餐厅在星期二的上午10点至星期三03：00再次营业。

例如：
表格：business_hour

id | weedDay | openDay | openTime | closeDay | closeTime
-----------------------------------------------------------
1  | Monday  | Monday  | 10:00:00 | Tuesday  | 03:00:00
2  | Tuesday | Tuesday | 10:00:00 | Wednday  | 03:00:00

如何使显示的MySQL SELECT语句恢复正常？

<?php

require_once "pdo/pdo.php";

    try {
    $stmt = $pdo->prepare("
            SELECT
             CASE WHEN
                exists (
                  SELECT 1 FROM business_hour WHERE
                  (openday = dayname(now()) AND opentime <= time(now()))
                  OR
                  (closeday = dayname(now()) AND closetime >= time(now()))
                )
                THEN 'Open'
                ELSE 'Closed'
              END status
            ");
    $stmt->execute();
    while ($row = $stmt->fetch(PDO::FETCH_ASSOC))
    {
        echo $row['status'];
    }


    } catch( PDOException $e ) {
        echo $e.'Not selected'; // error message
    }
?>

最佳答案

我想您想知道餐厅现在是否营业：

select
  case when
    exists (
      select 1 from business_hour where
      (openday = dayname(now()) and opentime <= time(now()))
      or
      (closeday = dayname(now()) and closetime >= time(now()))
    )
    then 'Open'
    else 'Closed'
  end status;

请参见demo
仅当表business_hour中有有效的日期名称时，此方法才有效，因此，像Wednday这样的错字而不是Wednesday会使它失败。

关于php - MySQL-每天休息两次，增加营业时间，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55199635/