Re: [PHP-DB] Can't get my head around this problem...
Perhaps...?
$x = 0;
while ($x 7)
{
if (!isset($myArray[$x]))
{
$myArray[$x] =0;
}
}
-Szii
- Original Message -
From: Tyrone Mills [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, April 24, 2001 3:58 PM
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
to
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
the
week it occurred on. Now the result set doesn't include empty values, so
if
there were connections on only 2 days, I only get 2 rows returned, if
there
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
FROM
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
only
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
element
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
through
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.
Thanks!!
Tyrone Mills
UNIX is user-friendly, it's just picky about who its friends are...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] Can't get my head around this problem...
That will zero fill any elements that don't have data, but it won't move the
data in the elements to the right positions. Each element is supposed to
represent a day of the week, starting at Sunday. So if I have data
representing calls that came in on Tuesday and Thursday, I need that data in
elements 2 and 4. The way it comes from my query, the data is at elements 0
and 1. So yes, I do need to add the zero's to the other 5 days, which your
code does (much neater than any of my attempts as well), but it doesn't
arrange the data the way I need.
Thanks for taking a shot at it though, I appreciate it!
-Original Message-
From: Szii [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, April 24, 2001 6:27 PM
To: Tyrone Mills; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Can't get my head around this problem...
Perhaps...?
$x = 0;
while ($x 7)
{
if (!isset($myArray[$x]))
{
$myArray[$x] =0;
}
}
-Szii
- Original Message -
From: Tyrone Mills [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Tuesday, April 24, 2001 3:58 PM
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
to
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
the
week it occurred on. Now the result set doesn't include empty values, so
if
there were connections on only 2 days, I only get 2 rows returned, if
there
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
FROM
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
only
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
element
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
through
either $lastweek or $day and try and build a proper array. Invariably
something doesn't work right. I'm sure I'm grossly over complicating this,
but I am at my wits end. Any thoughts or ideas are most welcome.
Thanks!!
Tyrone Mills
UNIX is user-friendly, it's just picky about who its friends are...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] Can't get my head around this problem...
Looks to me like you'd be better off using an associative array, since
associating the count with the day of the week is the whole point. How
about...
while (list($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...
-Alex
-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
to
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
the
week it occurred on. Now the result set doesn't include empty values, so
if
there were connections on only 2 days, I only get 2 rows returned, if
there
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
FROM
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
only
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
element
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
through
either $lastweek or $day and try and build a proper array. Invariably
something doesn't work right. I'm sure I'm grossly over complicating this,
but I am at my wits end. Any thoughts or ideas are most welcome.
Thanks!!
Tyrone Mills
UNIX is user-friendly, it's just picky about who its friends are...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP-DB] Can't get my head around this problem...
Thanks Alex!! I'll give that a go!
-Original Message-
From: Alexander Fordyce [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, April 24, 2001 8:07 PM
To: Tyrone Mills; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Can't get my head around this problem...
Looks to me like you'd be better off using an associative array, since
associating the count with the day of the week is the whole point. How
about...
while (list($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...
-Alex
-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
to
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
the
week it occurred on. Now the result set doesn't include empty values, so
if
there were connections on only 2 days, I only get 2 rows returned, if
there
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
FROM
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
only
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
element
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
through
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.
Thanks!!
Tyrone Mills
UNIX is user-friendly, it's just picky about who its friends are...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
