title: SQL command to sum by day/month
body: <p>I have the following table:</p>
  
  <pre><code>+----+---------------------+-------------+-----------------+
  | id | stat_time           |       reads |          writes |
  +----+---------------------+-------------+-----------------+
  |  1 | 2013-07-18 20:00:00 |    42614543 |         1342129 |
  |  2 | 2013-07-18 21:00:00 |    23085319 |          326139 |
  |  3 | 2013-07-25 12:00:00 |           0 |           39639 |
  |  4 | 2013-07-25 13:00:00 |      754166 |           39639 |
  |  5 | 2013-07-25 14:00:00 |      693382 |          295323 |
  |  6 | 2013-07-25 15:00:00 |     1334462 |               0 |
  |  7 | 2013-07-25 16:00:00 |    10748367 |          261489 |
  |  9 | 2013-07-25 17:00:00 |     4337294 |               0 |
  | 10 | 2013-07-25 18:00:00 |     3002796 |               0 |
  | 11 | 2013-07-25 20:00:00 |     3002832 |               0 |
  | 12 | 2013-07-25 23:00:00 |           0 |          333468 |
  | 13 | 2013-07-26 17:00:00 |    10009585 |               0 |
  | 15 | 2013-07-26 18:00:00 |     6005752 |               0 |
  | 17 | 2013-07-26 21:00:00 |      333663 |               0 |
  +----+---------------------+-------------+-----------------+
  </code></pre>
  
  <p>I would like to perform something like this:</p>
  
  <p><code>SELECT stat_time, SUM(reads), SUM(writes) from this_table GROUP BY stat_time HAVING 'the same day'..</code></p>
  
  <p>So a command which outputs three rows (one for 2013-07-18, second one for 2013-07-25 and third one for 2013-07-26) and in every such row in the column reads/writes there is a sum of reads/writes in this day.</p>
  
  <p>Thank you,
  David</p>