array problem

[Jamfool]

<?php require_once('Connections/jxpDREAM.php'); ?>

<?php
mysql_select_db($database_jxpDREAM, $jxpDREAM);
$query_data_grab = "SELECT `data` FROM datadump";
$data_grab = mysql_query($query_data_grab, $jxpDREAM) or die(mysql_error());

while ($row = mysql_fetch_array($data_grab, MYSQL_BOTH)) {
printf ($row["data"]);
}

mysql_free_result($data_grab);
?>

ok this uses mysql_fetch_array to grab the data and stick it into an array.(then prints it just to make sure its there) However how do I go about assigning it to an array variable?
like this-->

$vCht4 = array(60,40,20,34,26,52,41,20,34,43,64,40);

so that my sql data is assigned to the $vcht4 array variable as above...
This doesnt work-->
$vCht4 = mysql_fetch_array($data_grab);

(Any links to array handling php/sql greatly received)

 

[bastienk]

while ($row = mysql_fetch_array($data_grab, MYSQL_BOTH)) { 
       $myarray[] = $row["data"]; 
   }

will place the data into an array called $myarray



Bastien

Cat, the other other white meat

 

[DRJ478]

Remember that a result from a SQL query returns record by record. The decision you have to make is if you want to use a temporary variable to store the column values of each record or if you can actually write your code that you process one row at a time.
If you are planning on iterating the array you create then that's a waste of resource and time, since you can just step through the result set without having to use any temporary variable.