使用php从数据库表中以人类可读的格式打印餐厅营业时间

Question

我有一张桌子，上面列出了餐馆的营业时间。这些列是id、eateries_id、day_of_week、start_time和end_time。每家餐馆在表中多次出现，因为每一天都有单独的条目。有关更多详细信息，请参阅前面的问题：determine if a restaurant is open now (like yelp does) using database, php, js

我现在想知道如何从这个表中提取数据，并以人类可读的格式打印出来。例如，不是说“M1-3，T1-3，W1-3，Th1-3，F1-8”，我想说"M -Th1-3，F1-8“。同样，我想要“M1-3，5-8”而不是“M1-3，M5-8”。如果没有大量if语句的强力方法，我该如何做到这一点？

谢谢。

Answer 1

我想我会大吃一惊的。

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

按天合并营业时间的视图定义

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`

查询以查找具有相同开放时间的连续日期的“岛”(基于Itzik Ben Gan的查询)

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

您希望将每天的一组间隔合并起来。坚持24小时格式(实际上我猜是先把它转换成秒)，直到你不得不把它转换成更友好的格式。

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

问题是当你允许秒数的时候。提前1秒关门的餐厅将被错过:(也许您需要允许15或5个-minute增量。如果有必要，对DB中的数据进行四舍五入。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。现在颠倒字典。不是将天映射到时间间隔，而是将时间间隔映射到天。现在，找到一种方法来智能地表示这些天数组。例如，set(1,2,3)可以显示为"M-W"，所以我建议:对于集合{1,2,3,4,5,6,7} (或{1,2,3,4,5})的每个幂集合，(手动)找到最佳的人类表示。现在将这个逻辑硬编码到字典中，字典将排序的字符串(这一点很重要)映射到人类表示(如"M-W，F“)。显示1-3，5-8很容易，一旦你使用一个间隔对象，如上面的链接所述。祝好运!让我知道你遇到了什么问题。

编辑：

这不是它们的最佳示例(没有显示重叠间隔的并集)，但是您需要注意"|“运算符

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])

您可以自己实现这个类，但它可能容易出现bug。