Re: [PHP] Finding next recored in a array

[Jim Lucas]

Richard Kurth wrote:

include ("includes/location.php");
$query = "    SELECT    day
       FROM    emailcampaign
       where    campaign_id = '1'
       AND    member_id = '8'
   ";
$DB_Change_Campaign_Results = safe_query($query);
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
  $Campaign_array[] = $row['day'];
}
if ( isset($Campaign_array[($k + 1)])) {
   echo $Campaign_array[($k + 1)];
} else {
   echo 'Not found';
}
var_dump($Campaign_array);


This is what I get now when I run this

1

*array*
 0 => string '0' /(length=1)/
 1 => string '1' /(length=1)/
 2 => string '3' /(length=1)/
 3 => string '6' /(length=1)/
 4 => string '9' /(length=1)/
 5 => string '12' /(length=2)/
 6 => string '15' /(length=2)/
 7 => string '20' /(length=2)/
 8 => string '25' /(length=2)/
 9 => string '30' /(length=2)/

Are there going to be wholes in the date range?

if so, you will have to do that last bit like this.


<?php

include ("includes/location.php");

#
# Setting $k
# Make sure that $k is an integer, not a string.
# hence, no quotes
$k = 5;

$query = " SELECT  day
                FROM    emailcampaign
                WHERE   campaign_id = '1'
                AND     member_id = '8'
        ";
$DB_Change_Campaign_Results = safe_query($query);
$Campaign_array = array();
while ( $row = mysql_fetch_assoc($DB_Change_Campaign_Results) ) {
        $Campaign_array[] = $row['day'];
}

sort($Campaign_array);

foreach ( $Campaign_array AS $day ) {
        if ( $day <= $k ) {
                $day = next($Campaign_array);
                break;
        }
}

echo $day;

var_dump($Campaign_array);

?>



--
Jim Lucas

   "Some men are born to greatness, some achieve greatness,
       and some have greatness thrust upon them."

Twelfth Night, Act II, Scene V
    by William Shakespeare

--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php