mysql | 计算按时间分组的记录

Question

I have a code to get array of dates (This month - 12):

for ($i = 0; $i <= 11; $i++) {
    $months[] = date("Y-m", strtotime( date( 'Y-m-01' )." -$i months"));
}

To get records from DB by specified month I can use the code below:

$sql = 'SELECT COUNT(id) FROM `#__records` WHERE MONTH(date)=1 AND YEAR(date)=2010';

Hot to create my request to get data by each month? Thanks!

Answer 1

Try this

for ($i = 0; $i <= 11; $i++) {
    $months[] = date("Y-m", strtotime( date( 'Y-m-01' )." -$i months"));
}

$count = sizeof($months);

 for($i=0;$i<$count;$i++)
 {
    $sql = 'SELECT COUNT(id) AS id FROM `#__records` WHERE MONTH(date)='".$months[$i]."' AND YEAR(date)=2010';
    $result = mysql_query($sql);
    if($result)
{
$data_array = array();
    $data = mysql_fetch_assoc($result);
    if($data)
    {
        $data_array[] = array($data['id']);
    }
}
}
print_r($data_array);

Answer 2

You can do this directly in SQL with an appropriate GROUP BY query.

The trick is this little expression. Given any DATE or DATETIME`, it yields the first day of the month.

DATE(DATE_FORMAT(datevalue, '%Y-%m-01'))

The query uses that formula a lot to figure things out.

Try this:

SELECT COUNT(id) AS idcount,
       DATE(DATE_FORMAT(date, '%Y-%m-01')) AS month_beginning
  FROM table
 WHERE date >= DATE(DATE_FORMAT(NOW(), '%Y-%m-01')) - INTERVAL 12 MONTH
   AND date <  DATE(DATE_FORMAT(NOW(), '%Y-%m-01')) + INTERVAL 1 MONTH
  GROUP BY DATE(DATE_FORMAT(date, '%Y-%m-01'))
  ORDER BY DATE(DATE_FORMAT(date, '%Y-%m-01'))

This gives back one row for each of the most recent twelve months.

Here's a more extensive writeup on this kind of summary query. http://www.plumislandmedia.net/mysql/sql-reporting-time-intervals/