Fwd: Re: [PHP-DB] need help with foreach()
Here is the complete function I am using.
I returned, for testing i commented out the foreach loop and returned
$staff, then $tips both arrays returned NULL when i did a
var_dump(pointvalue($startdate));
can anyone see how this could be solved?
Cheers,
dave
==
function pointvalue($start){
$query = SELECT * FROM Tips WHERE date = $start and date = ($start +
INTERVAL 6 DAY) ;
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$date = $row[Date];
$tips[$date] = $row[TotalTips];
}
$query = SELECT * FROM Rota WHERE date = $start and date = ($start +
INTERVAL 6 DAY) ;
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
$date = $row[Date];
if (isset($staff[$date])){
$staff[$date] = $staff[$date] + 1;
}
else{
$staff[$date] = 1;
}
}
return $staff;
}
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Re: Fwd: Re: [PHP-DB] need help with foreach()
On Monday 10 March 2003 17:56, David Rice wrote:
Here is the complete function I am using.
I returned, for testing i commented out the foreach loop and returned
$staff, then $tips both arrays returned NULL when i did a
var_dump(pointvalue($startdate));
I haven't been following this thread so I'm not sure what your objective is.
But ...
function pointvalue($start){
$query = SELECT * FROM Tips WHERE date = $start and date = ($start +
INTERVAL 6 DAY) ;
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
$date = $row[Date];
You should really be using:
$date = $row['Date'];
Also what is $date supposed to be storing? Each iteration of the while-loop
it's being overwritten with the 'latest' date.
$tips[$date] = $row[TotalTips];
As above you should put single-quotes around your array subscripts.
}
$query = SELECT * FROM Rota WHERE date = $start and date = ($start +
INTERVAL 6 DAY) ;
$result = mysql_query($query);
while ($row = mysql_fetch_array($result)){
$date = $row[Date];
Again, see above comments.
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[PHP-DB] need help with foreach()
Okay, i have two arrays, $tips and $staff
$tips has a key date (which is the date of the first day of the week, the
second result in the array has the key of the second day of the week etc...
) and the value is a floating point decimal.
$staff also has a key date and the value is an integer, the number of
staff working in one day.
To find out the ammount of tips every staff member is to get we have to
divide the value of $tips by the value of $staff (when the dates are the
same)
what i was trying to get this to work is below,
any help would be great,
cheers
dave
=
foreach($tips as $key = $value){
$pointvalue[$key] = $value / current($staff) ;
next($staff);
}
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Re: [PHP-DB] need help with foreach()
Hey thank's for the ideas but neither of them work, doh...
Okay fredrik I know your idea won't work cos list only works with numericaly
indexed arrays, both the arrays that i am using are indexed by date.
(it produces a parse error when run)
=
while(list($d, $t) = each($tips)){
$res = $t / $staff[$d];
// Do what you need with $res ...
}
=
Janet your idea i would have thought would work but i can't get it to,
for some reason
whenever I run the script it just gives me...
Warning: Invalid argument supplied for foreach() in
/home/filterseveuk/public_html/project/tips.php on line 56
Not sure as to what it is talking abotu here as far as I know, this should
work.
=
foreach($tips as $key = $value){
$pointvalue[$key] = $value / $staff[$key] ;
}
==
Does anyone else have a solution!?
Cheers,
Dave
In a message dated 3/9/2003 2:07:26 PM Pacific Standard Time,
[EMAIL PROTECTED] writes:
Okay, i have two arrays, $tips and $staff
$tips has a key date (which is the date of the first day of the week, the
second result in the array has the key of the second day of the week etc...
) and the value is a floating point decimal.
$staff also has a key date and the value is an integer, the number of
staff working in one day.
To find out the ammount of tips every staff member is to get we have to
divide the value of $tips by the value of $staff (when the dates are the
same)
what i was trying to get this to work is below,
any help would be great,
cheers
dave
=
foreach($tips as $key = $value){
$pointvalue[$key] = $value / current($staff) ;
next($staff);
}
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Re: [PHP-DB] need help with foreach()
[EMAIL PROTECTED] (David Rice) writes:
Hey thank's for the ideas but neither of them work, doh...
Okay fredrik I know your idea won't work cos list only works with
numericaly indexed arrays, both the arrays that i am using are indexed
by date.
You really had me wondering there for a moment. I've done this dozens
of times and list() does work with other than numerical values.
(it produces a parse error when run)
Ok, I hadn't tested the code. For some reason you have to group $t /
$staff[$d] in parantheses, why I really don't know, it shouldn't be
necessary.
=
while(list($d, $t) = each($tips)){
$res = ($t / $staff[$d]);
^ ^
// Do what you need with $res ...
}
=
The code below is tested and works.
---
$tips = array(foo = bar,
bar = foo);
$staff = array(foo = foobar,
bar = barfoo);
while(list($d, $t) = each($tips)){
printf(%s - %s - %sbr\n, $d, $t, $staff[$d]);
}
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Re: [PHP-DB] need help with foreach()
On Monday 10 March 2003 08:10, David Rice wrote:
Hey thank's for the ideas but neither of them work, doh...
Okay fredrik I know your idea won't work cos list only works with
numericaly indexed arrays, both the arrays that i am using are indexed by
date. (it produces a parse error when run)
Parse error simply means the code has a syntax error. Not necessarily that the
idea doesn't work.
Janet your idea i would have thought would work but i can't get it to,
for some reason
whenever I run the script it just gives me...
Warning: Invalid argument supplied for foreach() in
/home/filterseveuk/public_html/project/tips.php on line 56
Not sure as to what it is talking abotu here as far as I know, this should
work.
=
foreach($tips as $key = $value){
$pointvalue[$key] = $value / $staff[$key] ;
}
What does print_r($tips) give you? While you're at it do a var_dump($tips) as
well.
--
Jason Wong - Gremlins Associates - www.gremlins.biz
Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
--
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