使用php以人类可读格式从数据库表打印餐厅营业时间

Question

我有一张桌子列出了餐馆的营业时间。 列是id，eateries_id，day_of_week，start_time和end_time。 每个餐馆在表格中多次出现，因为每天都有一个单独的条目。 有关更多详细信息，请参阅此前一个问题： 使用数据库，php，js确定餐厅现在是否已打开（如yelp）

我现在想知道如何从该表中获取数据并以人类可读的格式打印出来。 例如，我不想说“M 1-3，T 1-3，W 1-3，Th 1-3，F 1-8”，而是想说“M-Th 1-3，F 1-8” 。 类似地，我想要“M 1-3,5-8”而不是“M 1-3，M 5-8”。 如果没有大量if语句的强力方法，我怎么能这样做呢？

谢谢。

Answer 1

以为我会在这方面受到打击。

测试表

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

查看定义以合并每天的营业时间

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`

查询以相同的开放时间查找连续日的“岛屿”（基于Itzik Ben Gan的一个）

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

结果

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

Answer 2

你想每天结合一堆间隔。 坚持24小时格式（实际上我将其转换为秒），直到你必须将其转换为人性化的格式。

http://pyinterval.googlecode.com/svn/trunk/html/index.html

麻烦的是，当你允许几秒钟时...一个提前1秒钟关闭的餐馆将被遗漏:(也许你需要允许15或5分钟的增量。如果你需要，可以在DB中舍入数据。所以，方法是：使用区间数据结构，将给定日期的所有区间合并在一起。现在反转字典。而不是将天数映射到区间，将区间映射到天。现在找到一种方法来智能地表示这些天组。例如， set(1,2,3)可以显示为“MW”，所以我建议：对于集合的每个幂集{1,2,3,4,5,6,7} （或{1,2,3,4,5} ）找到最好的人类表示（手工）。现在硬编码这个逻辑 - 将它保存到一个字典中，该字典将排序的字符串（这很重要）（例如“1235”）映射到人类表示，例如“MW” ，F“。显示1-3,5-8很容易，一旦你使用间隔对象，如上面的链接所述。祝你好运！让我知道你遇到了什么问题。

编辑：

这不是他们拥有的最好的例子（不显示重叠间隔的联合），但是你关心“|” 操作者

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])

你可以自己实现这个类，但它可能容易出错。