Normally, what I do here is handle that in the loop to display the records ... so start by adding an order by clause to keep the dates together SELECT * FROM transportdokument WHERE dato >= '16/7/2013' AND dato <= '18/7/2013' order by dato $prior_date = ""; $sHTML = "<table>"; while($rows = mysql_fetch_array($result)){ if ($prior_date != $rows['dato']){ if($open_table){ $sHTML .= "</table><table>"; $prior_date = $rows['dato']; } } $sHTML .= "<tr>"; $sHTML .= "<td>". $rows['dato'] . "</td>"; $sHTML .= "<td>". $rows['some_field'] . "</td>"; $sHTML .= "<td>". $rows['another_field'] . "</td>"; $sHTML .= "<td>". $rows['third_field'] . "</td>"; $sHTML .= "</tr>"; } $sHTML .= "</table>"; On Thu, Jul 18, 2013 at 9:43 AM, Karl-Arne Gjersøyen <karlarneg@xxxxxxxxx>wrote: > Hello again. > In my program I have this: > > mysql> SELECT * FROM transportdokument WHERE dato >= '16/7/2013' AND dato > <= '18/7/2013'; > > This list all reccrds for 3 days. I need a way to split it up for every day > even when the requst is as above and don't know in what way I can do it. > > I like to have all records for day 16 in one table in PHP/HTML and all > records for day 17 in another table. > i.e, Day 16 have 5 rows and day 17th and 18th have 7 and 8 rows. > > I hope for your help and advice to do also this correct. > > Thank you for your time and effort! > > Karl > -- Bastien Cat, the other other white meat