Looks to me like you'd be better off using an associative array, since
associating the count with the day of the week is the whole point. How
while (list($count, $day) = db_fetch_row($result))
$lastweek[$day] = $count;
You'd have to tweak the logic of the rest of the script to use the hash, but
it's a more natural way to do it. Just a suggestion...
> -----Original Message-----
> From: Tyrone Mills [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, April 24, 2001 3:58 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Can't get my head around this problem...
> Hello all...
> I've got what is probably a stupid question, it's a problem I can't seem
> get my head around it. Any help will be greatly appreciated.
> I'm pulling a count of connections from a MySQL database and the day of
> week it occurred on. Now the result set doesn't include empty values, so
> there were connections on only 2 days, I only get 2 rows returned, if
> were connections on 4 days, I get 4 rows returned. You get the idea...
> Here is the PHP code I am using the get the data:
> $sql = "SELECT COUNT(*), (DAYOFWEEK(dialer_login_date) -1) AS DAYOFWEEK
> dialer_login WHERE dialer_id = $dialer_id AND
> WEEK(date_format(dialer_login_date, '%Y-%m-%d'),0) = WEEK('$last_week',0)
> GROUP BY DAYOFWEEK";
> $result = db_query($sql);
> while ($daily_count = db_fetch_array($result))
> $lastweek = $daily_count["COUNT(*)"];
> $day = $daily_count["DAYOFWEEK"];
> For the sake of this question, let's say that there were connections on
> 2 days, so we get a result set with 2 rows. Something like this:
> COUNT(*) DAYOFWEEK
> 8 2
> 3 4
> That's great, but I need the array $lastweek to have the value 8 in
> 2 and the value 3 in element 4 and all other elements (0,1,3,5,6) as zero.
> As it sits, they are in elements 0 and 1 respectively.
> I get this:
> $lastweek = 8,3
> I need this:
> $lastweek = 0,0,8,0,3,0,0
> Now I've written all manner of wild and wonderful loops that iterate
> either $lastweek or $day and try and build a proper array. Invariably
> something doesn't work right. I'm sure I'm grossly over complicating this,
> but I am at my wits end. Any thoughts or ideas are most welcome.
> Tyrone Mills
> UNIX is user-friendly, it's just picky about who its friends are...
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