[SOLVED] Number Order

[Schlo_50]

I want to display the dates (dat) from the database in order 1st, 5th, 15th etc.. How?

 

<? 
$result = mysql_query ($jan) or die ( mysql_error ()); 
while ($row = mysql_fetch_array ($result))
{
echo ( $row[dat]. ' , ' );
echo ( $row[message]. '<br>' );
}
?></p>

 

Thanks

[clown[NOR]]

try using ORDER BY maybe.. if i understood your question right

[Schlo_50]

Could you revise for me please..Not sure i understand you.

[Barand]

Assuming the date is in a DATE field so is formatted as 2007-05-01

 

echo "date ('jS M, Y', strtotime ($row['dat']));      //--> 1st May, 2007

 

http://www.php.net/date

[Schlo_50]

Its stored as a single number. Field type is INT.

[clown[NOR]]

$query = "SELECT * FROM `database` ORDER BY `dat`";
$result = mysql($query);
if (!$result) { die(mysql_error()); }

while ($row = mysql_fetch_assoc($result)) {
echo $row['dat'].' , '.$row['message'].'<br>';

 

EDIT: You should change the dbfield to DATETIME if you're handeling only dates

[Barand]

Its stored as a single number. Field type is INT.

 

So if it's like 1177974000 then you can leave out the strtotime function

 

echo "date ('jS M, Y', $row['dat']);      //--> 1st May, 2007

 

 

[Schlo_50]

link=topic=138678.msg588067#msg588067 date=1178044486]

$query = "SELECT * FROM `database` ORDER BY `dat`";
$result = mysql($query);
if (!$result) { die(mysql_error()); }

while ($row = mysql_fetch_assoc($result)) {
echo $row['dat'].' , '.$row['message'].'<br>';

 

EDIT: You should change the dbfield to DATETIME if you're handeling only dates

 

Cheers clown

[clown[NOR]]