how can i print a returned array's multiple results

[rotaryfreak]

Hi,


in my database table, i have inserted the following data after creating the appropriate tables:

insert into Transaction values(0001, null, 'BransenEnt', 1, 'Mastercard', 99.99, '2010-06-05', '10:00:00');

insert into Transaction values(0002, 'blackbear62', null, null, 'Visa', 29.99, '2010-01-03', '13:47:00');

insert into Transaction values(0003, null, 'JoeRod', 2, 'Visa', 99.99, '2009-07-22', '17:15:00');

im building section of an application which needs to check for all transactions between two dates. I do this by:

$query = "SELECT * FROM Transaction WHERE Date between \"2009-06-12\" AND \"2010-07-05\" ";

When i run this straight from the mysql command line, i get the 3 above results. When i run the same query in my application, with some added sql functions:

$query = "SELECT * FROM Transaction WHERE Date between \"2009-06-12\" AND \"2010-07-05\" ";

$result = mysql_query($query);
$array = mysql_fetch_assoc($result);
print_r($array);

i only get the first result. Im pretty sure that it has to do with the array being returned only having 1 row so how do i use PHP to print out the other rows?

i havent been very successful with google :(

thank you

[johny10151981]

The reason is
mysql_fetch_ass oc($result)
always return the first un-accessed row.

If you want to get through the all row you need to do it until mysql_fetch_ass oc return null. Check the below code.

while($array=mysql_fetch_assoc($result))
{
 print_r($array);
}

[rotaryfreak]

awesome, thanks a lot!

[Dormilich]

you can also try PDO's or MySQLi's fetchAll() method