php Open daily day rage - php

Question

I building restaurant service and currently i'm stuck figuring out how to display open daily restaurant like google Business.
I have table design
+-----+----------+------+----------+----------+
| ref | id_resto | day | open | close |
+-----+----------+------+----------+----------+
| 2 | 5 | 1 | 13:00:00 | 21:00:00 |
| 5 | 5 | 2 | 13:00:00 | 21:00:00 |
| 7 | 5 | 3 | 13:00:00 | 21:00:00 |
| 9 | 5 | 4 | 13:00:00 | 21:00:00 |
| 10 | 5 | 6 | 13:00:00 | 22:00:00 |
| 11 | 5 | 7 | 14:00:00 | 21:00:00 |
+-----+----------+------+----------+----------+
day 7 = sunday, 1 = monday, 2 = tuesday, 3 = Wednesday and soo on
according to my table it should be display
Monday-Thursday 13:00 - 21:00
Saturday 13:00 - 22:00
Sunday 14:00 - 21:00
i've try some algorithm, but noting happened -_-
any idea?
Thanks

Answer 1

With a small change to your table structure, it should be simple to implement.
Here is my code to create the tables:
CREATE TABLE `daytype` (
`id` int(11) NOT NULL,
`description` varchar(45) DEFAULT NULL,
PRIMARY KEY (`id`)
);
INSERT INTO `daytype` VALUES (1,'Monday - Thursday'),(2,'Saturday'),(3,'Sunday');
CREATE TABLE `hours` (
`ref` int(11) NOT NULL,
`id_resto` int(11) DEFAULT NULL,
`id_daytypes` int(11) DEFAULT NULL,
`open` char(5) DEFAULT NULL,
`close` char(5) DEFAULT NULL,
PRIMARY KEY (`ref`)
);
INSERT INTO `hours` VALUES (2,5,1,'13:00','21:00'),(5,5,2,'13:00','21:00'),(7,5,3,'14:00','21:00');
From this you can then easily get the data you want with a simple join:
SELECT d.description, CONCAT(open, ' - ', close) as hours FROM test.hours AS h INNER JOIN daytype AS d ON (d.id = h.id_daytypes);
Output:
This structure has the added benefit of allowing you to set up different breaks, and different displays for each of the restaurants, if needed. For example, if a restaurant had different hours on Monday, you could add an entry to the daytypes table for Monday, and Tuesday-Thursday.
Overall, this should save some space since you don't need to enter duplicate info for each of the days that are the same.
Hopefully this will help you get started.
UPDATE: Per request, here is an example of how to check if something is open today.
You could have a php function that returns a comma delimited list of all valid daytypes for the current day. Something like this:
<?php
/*
1 Monday - Thursday
2 Saturday
3 Sunday
4 Friday
5 Monday
*/
function getDayType()
{
$dayOfWeek = date('l');
switch($dayOfWeek)
{
case "Monday":
return "1, 5"
case "Tuesday":
case "Wednesday":
case "Thursday":
return "1";
case "Friday":
return "4":
case "Saturday":
return "2";
case "Sunday":
return "3"
}
}
Then is is simple to determine the shops that are open today.
For example, like this query which will return all the restaurants that are open.
SELECT
h.id_resto
FROM
test.hours AS h
INNER JOIN
daytype AS d ON (d.id = h.id_daytypes)
WHERE
id_daytypes IN (1 , 5)
AND (
(HOUR(NOW()) > HOUR(open) AND HOUR(NOW()) < HOUR(close))
OR
(HOUR(NOW()) = HOUR(open) AND MINUTE(NOW()) >= MINUTE(open))
OR
(HOUR(NOW()) = HOUR(close) AND MINUTE(NOW()) < MINUTE(close))
);