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 <karlar...@gmail.com>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
>
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Bastien
Cat, the other other white meat
