php - Showing Featured Item From the Database -

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i have database table below.

id, bungalow_name, type, address, featured

a bungalow can featured in home page. if bungalow featured, featured column has value 1. have 50 bungalows in tables , 5-7 bungalows featured @ given time.

let's assume featured bungalow names below.

bungalow 1, bungalow 2, bungalow 3, .........., bungalow 6

what i'm trying show featured bungalow in home page each day. , want loop below below each month. given don't want show bungalow randomly each page load. want show per day 1 bungalow basis. 

today              -> bungalow 1 tomorrow           -> bungalow 2 day after tomorrow -> bungalow 3 ... after bungalow 6, bungalow 1 shown on next day.

how can it? possible sql/php?

you use mysql query:

select * bungalows id = (   select b1.id       bungalows b1 left join bungalows b2     on b1.id>b2.id , b2.featured=1       b1.featured=1   group     b1.id   having     count(b2.id) = (select                       datediff(curdate(), '2013-05-06') mod                       (select count(*) bungalows featured=1))   )

please see fiddle here. '2013-05-06' day when want start show first featured bungalow. shown ordered id, strarting '2013-05-06'.

edit

the following query return number of elapsed days since 2013-05-06:

select datediff(curdate(), '2013-05-06')

the mod function return integer remainder of division of number of elapsed day number of featured rows:

select datediff(curdate(), '2013-05-06') mod                           (select count(*) bungalows featured=1)

if there 6 featured bungalows, return 0 first day,1 second,2,3,4,5, , 0,1,2...again.

mysql not have function return rank (number of row), have simulate somehow. simulated way:

select b1.id, count(b2.id)   bungalows b1 left join bungalows b2   on b1.id>b2.id , b2.featured=1   b1.featured=1 group   b1.id

i'm joining bungalows table itself. rank of bungalow id count of bungalows have id less (hence join b1.id>b2.id).

i'm selecting row have rank returned function above:

having count(b2.id) = (select datediff(curdate(), '2013-05-06') mod (select count(*) bungalows featured=1))

if use mysql, initial query simplified this:

select b1.*   bungalows b1 left join bungalows b2   on b1.id>b2.id , b2.featured=1   b1.featured=1 group   b1.id having   count(b2.id) = (select                     datediff(curdate(), '2013-05-06') mod                     (select count(*) bungalows featured=1))

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