使用php从数据库表以可读格式打印餐厅营业时间

我有一张列有餐馆营业时间的桌子。列为id、eateries、day、start和end。每个餐厅在表中都有多次表示，因为每天都有一个单独的条目。有关更多详细信息，请参见上一个问题：

我现在想知道如何从这个表中获取数据，并以人类可读的格式打印出来。例如，我想说的不是“m1-3，t1-3，w1-3，th1-3，f1-8”，而是“m1-th1-3，f1-8”。类似地，我想要“m1-3，5-8”而不是“m1-3，m5-8”。如果没有大量if语句的蛮力方法，我怎么做呢

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谢谢。

我想我会在这里狂欢一下

试验台

CREATE TABLE `opening_hours` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `eateries_id` int(11) DEFAULT NULL,
  `day_of_week` int(11) DEFAULT NULL,
  `start_time` time DEFAULT NULL,
  `end_time` time DEFAULT NULL,
  PRIMARY KEY (`id`)
) 

测试数据

INSERT INTO `test`.`opening_hours`
(
`eateries_id`,
`day_of_week`,
`start_time`,
`end_time`)
SELECT 2 AS eateries_id, 1 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 2 AS eateries_id, 1 AS day_of_week, '17:00' AS start_time, '20:00' as end_time union all
SELECT 2 AS eateries_id, 2 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 2 AS eateries_id, 2 AS day_of_week, '17:00' AS start_time, '20:00' as end_time union all
SELECT 2 AS eateries_id, 3 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 2 AS eateries_id, 4 AS day_of_week, '13:00' AS start_time, '20:00' as end_time union all
SELECT 2 AS eateries_id, 5 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 2 AS eateries_id, 6 AS day_of_week, '13:00' AS start_time, '20:00' as end_time union all
SELECT 2 AS eateries_id, 7 AS day_of_week, '13:00' AS start_time, '21:00' as end_time
                                                                       union all
SELECT 3 AS eateries_id, 1 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 3 AS eateries_id, 2 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 3 AS eateries_id, 3 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 3 AS eateries_id, 4 AS day_of_week, '13:00' AS start_time, '20:00' as end_time union all
SELECT 3 AS eateries_id, 5 AS day_of_week, '13:00' AS start_time, '15:00' as end_time union all
SELECT 3 AS eateries_id, 6 AS day_of_week, '13:00' AS start_time, '20:00' as end_time union all
SELECT 3 AS eateries_id, 7 AS day_of_week, '13:00' AS start_time, '21:00'  as end_time

查看定义以按天合并开放时间

CREATE VIEW `test`.`groupedhours` 
AS 
  select `test`.`opening_hours`.`eateries_id` AS `eateries_id`,
         `test`.`opening_hours`.`day_of_week` AS `day_of_week`,
         group_concat(concat(date_format(`test`.`opening_hours`.`start_time`,'%l'),' - ',date_format(`test`.`opening_hours`.`end_time`,'%l %p')) order by `test`.`opening_hours`.`start_time` ASC separator ', ') AS `OpeningHours` 
         from `test`.`opening_hours` 
         group by `test`.`opening_hours`.`eateries_id`,`test`.`opening_hours`.`day_of_week`

查询以查找连续几天中开放时间相同的“岛屿”（基于伊兹克·本·甘的查询）

---

您希望为每天合并一组时间间隔。坚持使用24小时格式（我想实际上是先将其转换为秒），直到您必须将其转换为更人性化的格式

问题是当你允许几秒钟。。。提前一秒关闭的餐厅将被错过：（可能您需要允许15或5分钟的增量。如果有必要，请将数据舍入到DB中。因此，方法是：使用间隔数据结构，将给定日期的所有间隔合并在一起。现在反转字典。而不是将天映射到间隔，而是将间隔映射到天。现在找到一种智能地表示这些天组的方法。Fo例如，

set（1,2,3）

可以显示为“M-W”，因此我建议：对于集合的每个幂集

{1,2,3,4,5,6,7}

（或

{1,2,3,4,5}

），找到最佳的人类表示（手动）。现在硬编码此逻辑-将其保存到映射排序的字符串（这很重要）的字典中，例如“1235”对于“M-W，F”这样的人类表示。一旦您使用上面链接中描述的间隔对象，显示1-3，5-8就很容易了。祝您好运！让我知道您遇到了什么问题

编辑：

这不是它们的最佳示例（不显示重叠间隔的并集），但您关心的是“|”运算符

---

您可以自己实现这个类，但它可能容易出现bug。

它是如何存储在db中的“m1-3”的？还是时间戳？这是一个有趣的问题…我将尝试思考一下。它确实让我想起了一些电路简化，包括卡诺图：）我假设您关心高级方法，并且可以处理与您的精确表不匹配的答案。顺便问一下，是否只允许整小时？最早可能的小时和最晚可能的小时是什么时候？我将首先创建一个哈希表，将一天映射到小时，例如

{'M'：set（1,2,3,5,6,7,8），'T'：set（1,2,3）}

或

{'M'：'yyyyyyy'}

（指示是否每小时（此处从1开始））是可用的。这是确定哪两天有相同的时间表的第一步，也是计算和简化开放时间联盟的第一步。为什么在您的示例中所有的时间都以1开始？这是一家晚餐厅吗？您关心24小时制吗？您希望AM和PM在其中吗？是否允许半小时增量？回答问题：时间以时间格式存储，如12:34:05。日期以整数形式存储在单独的列中（例如，2代表星期一）。允许任何时间到秒，尽管我99.9%确信没有餐厅的营业时间会涉及秒：）。最早的时间是00:00:00，最晚的时间是23:59:59，所以是24小时格式。当我为用户打印时间时，我希望使用am/pm，尽管在我计算出剩余时间后，这种转换应该不会很困难。高级方法很好，我只是不知道从哪里开始解决这个问题！MySQL。这就是你要问的，对吧？这很有趣。我的意见是，SQL应该让您快速获得数据，因为这是它最擅长的。我认为最好用更高级的语言来按摩它。您的SQL令人印象深刻。这里我假设Mon=1，Tue=2。。。Sun=7，因为这是一些国家的惯例——一周从周一开始，周日结束。我想我明白你的意思。。不过，难道我不需要硬编码一百多天的组合吗？不完全是，2^5=32；2^7=128:）但是，这些是可以生成的。

SET @rownum = NULL;
SET @rownum2 = NULL;



SELECT S.eateries_id, 
concat(CASE WHEN 
S.day_of_week <> E.day_of_week 
    THEN 
    CONCAT(CASE S.day_of_week 
             WHEN 1 THEN 'Su'
             WHEN 2 THEN 'Mo'     
             WHEN 3 THEN 'Tu'     
             WHEN 4 THEN 'We'
             WHEN 5 THEN 'Th'    
             WHEN 6 THEN 'Fr'    
             WHEN 7 THEN 'Sa'  
            End, ' - ')
    ELSE ''        
END,
CASE E.day_of_week 
     WHEN 1 THEN 'Su'
     WHEN 2 THEN 'Mo'     
     WHEN 3 THEN 'Tu'     
     WHEN 4 THEN 'We'
     WHEN 5 THEN 'Th'    
     WHEN 6 THEN 'Fr'    
     WHEN 7 THEN 'Sa'  
End, ' ', S.OpeningHours) AS `Range`
FROM (

SELECT 
    A.day_of_week,
    @rownum := IFNULL(@rownum, 0) + 1  AS rownum,
    A.eateries_id,
    A.OpeningHours
FROM `test`.`groupedhours` as A
WHERE NOT EXISTS(SELECT * FROM `test`.`groupedhours` B
                 WHERE A.eateries_id = B.eateries_id
                  AND A.OpeningHours = B.OpeningHours
                  AND B.day_of_week = A.day_of_week -1) 
ORDER BY eateries_id,day_of_week) AS S

JOIN (
SELECT 
    A.day_of_week,
    @rownum2 := IFNULL(@rownum2, 0) + 1  AS rownum,
    A.eateries_id,
    A.OpeningHours
FROM `test`.`groupedhours` as A 
WHERE NOT EXISTS(SELECT * FROM `test`.`groupedhours` B
                 WHERE A.eateries_id = B.eateries_id
                  AND A.OpeningHours = B.OpeningHours
                  AND B.day_of_week = A.day_of_week + 1)
ORDER BY eateries_id,day_of_week) AS E

ON  S.eateries_id = E.eateries_id AND
    S.OpeningHours = S.OpeningHours AND 
    S.rownum = E.rownum

eateries_id             Range
2                Su - Mo 1 - 3 PM, 5 - 8 PM
2                Tu 1 - 3 PM
2                We 1 - 8 PM
2                Th 1 - 3 PM
2                Fr 1 - 8 PM
2                Sa 1 - 9 PM
3                Su - Tu 1 - 3 PM
3                We 1 - 8 PM
3                Th 1 - 3 PM
3                Fr 1 - 8 PM
3                Sa 1 - 9 PM

unioned:

>>> interval[1, 4] | interval[2, 5]
interval([1.0, 5.0])

>>> interval[1, 2] | interval[4, 5]
interval([1.0, 2.0], [4.0, 5.0])