php - 使用 php 以人类可读格式从数据库表中打印餐厅营业时间

Question

我有一张表格，列出了餐厅的营业时间。这些列是 id、eateries_id、day_of_week、start_time 和 end_time。每个餐馆在表中多次出现，因为每天都有一个单独的条目。有关详细信息，请参阅上一个问题: determine if a restaurant is open now (like yelp does) using database, php, js

我现在想知道如何从此表中获取数据并以人类可读的格式打印出来。例如，与其说“M 1-3、T 1-3、W 1-3、Th 1-3、F 1-8”，不如说“M-Th 1-3、F 1-8” .同样，我想要“M 1-3、5-8”而不是“M 1-3、M 5-8”。如果没有大量 if 语句的蛮力方法，我怎么能做到这一点？

谢谢。

Answer 1

我想我会很高兴的。

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

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`

查询以查找具有相同开放时间的连续日期的“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