SQL查询SELECT，ORDER

Question

SQL Query SELECT,ORDER date

Hi, I need to select the date from my table, the problem is that I have above 130 rows and it will grow every day. I want to short it and in the query to select the date that equal for today date=date("Y-m-d") and the date that bigger from the date of today.

It can be a variable the date of today because I'm using a date selection to paginated between the dates. I know I can use timestamp but I dont know how to implement it in the query.

  $r = mysql_query("SELECT  date FROM yoman  ORDER BY date ASC,time ASC  ") 
            or die(mysql_error()); 

I hope you will unserstand, http://attiatech.com/projects/tours/index.php?db=cmVsXzIwMTI= this is the link, try to pick the 1.10.2012 and see how many rows are printed, i have the second option of view paginated but i want in this query to filter it too.

Answer 1

I assume you store the date as string "Y-m-d".

It might be better to use this though as table column type:

http://dev.mysql.com/doc/refman/5.1/en/datetime.html

So e.g.:

$startdate= date("Y-m-d", $day); 
$enddate= date("Y-m-d",strtotime("+1 day", $day)) - 1; 

SQL to fetch one day goes like this:

SELECT * 
FROM yoman 
WHERE date BETWEEN $startdate AND $enddate 
ORDER BY date ASC,time ASC