Newbie MySQL/PHP question re output formatting
I can get this to work just fine:
?php
$number = 23999.39;
print $;
print number_format($number, 2);
?
Comes out $23,999.39
I'd like use the number_format() thingie on an array returned from a
mysql_query.
My current program snippet looks like:
$res = mysql_query(SELECT org, COUNT(*), SUM(annual_cost) AS
cost FROM a05
GROUP BY org ORDER BY cost DESC,$dbh);
if (!$res) {
echo mysql_errno().: . mysql_error ().;
return 0;
}
print table border=1;
while ($thearray = mysql_fetch_array($res)) {
printf(trtd {$thearray[org]} /td
td align=right {$thearray[COUNT(*)]} /td
td align=right $ {$thearray[cost]} /td/tr);
}
print /table;
and works fine -- see http://ibbmonitor.com/sked_1.php, 3rd block
of stuff down.
If I replace
{$thearray[cost]}
with
number_format({$thearray[cost]}, 2)
I get
$ number_format(7842554.24, 2)
in the cell where I would expect to get
$ 7,842,554.24
Any idea what I'm doing wrong?
Clearly, I don't understand arrays very well.
Thanks VERY much for any help on this.
bw
---
Bill Whitacre
[EMAIL PROTECTED]
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Re: Newbie MySQL/PHP question re output formatting
Bill Whitacre wrote:
printf(trtd {$thearray[org]} /td
td align=right {$thearray[COUNT(*)]} /td
td align=right $ {$thearray[cost]} /td/tr);
If I replace
{$thearray[cost]}
with
number_format({$thearray[cost]}, 2)
Although this is a MySQL mailing list and your problem is not MySQL
related, but a PHP question I'll give you a brief answer.
PHP does not evaluate functions inside a double quoted string, so you
should use:
trtd .number_format($thearray['cost'],2). /tdtd align...
Furthermore, using $thearray[cost] is not advisable; PHP will try to
find a constant named 'cost' and if it can't find one it will use the
string itself as the value. Use a real string (quoted) instead of
relying on this feature:
$thearray['cost'] or $thearray[cost]
Also you can use aliases in your query to avoid things like
$thearray[COUNT(*)];
Use something like this in your query:
SELECT COUNT(*) AS `count` FROM
and you can use $thearray['count'] instead
As a last pointer, printf() is pretty much useless here since you don't
use any variable formatting features of printf() here. print() or echo()
are more suitable in this case.
Regards, Jigal
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Re: Newbie MySQL/PHP question re output formatting
Bill Whitacre wrote:
I can get this to work just fine:
?php
$number = 23999.39;
print $;
print number_format($number, 2);
?
Comes out $23,999.39
I'd like use the number_format() thingie on an array returned from a
mysql_query.
My current program snippet looks like:
$res = mysql_query(SELECT org, COUNT(*), SUM(annual_cost) AS
cost FROM a05
GROUP BY org ORDER BY cost DESC,$dbh);
if (!$res) {
echo mysql_errno().: . mysql_error ().;
return 0;
}
print table border=1;
while ($thearray = mysql_fetch_array($res)) {
printf(trtd {$thearray[org]} /td
td align=right {$thearray[COUNT(*)]} /td
td align=right $ {$thearray[cost]} /td/tr);
}
print /table;
and works fine -- see http://ibbmonitor.com/sked_1.php, 3rd block of
stuff down.
If I replace
{$thearray[cost]}
with
number_format({$thearray[cost]}, 2)
I get
$ number_format(7842554.24, 2)
The issue is that PHP replaces $thearray[cost], with the contents of
that variable (that is an array, it doesn't matter). But in the second
case it replaces the same thing ($thearray[cost]), with the contents of
the variable, but you want to place there the result of the function.
To do date change the first line from
printf(trtd number_format({$thearray[cost]}, 2) /td
to
printf(trtd .number_format({$thearray[cost]}, 2). /td
in the cell where I would expect to get
$ 7,842,554.24
Any idea what I'm doing wrong?
Clearly, I don't understand arrays very well.
Thanks VERY much for any help on this.
bw
---
Bill Whitacre
[EMAIL PROTECTED]
--
Nuno Pereira
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Re: Newbie MySQL/PHP question re output formatting
Nuno Pereira wrote:
To do date change the first line from
printf(trtd number_format({$thearray[cost]}, 2) /td
to
printf(trtd .number_format({$thearray[cost]}, 2). /td
I think you mean:
print( trtd . number_format( $thearray['cost'], 2 ) . /td );
As stated by a previous poster to this thread, you shouldn't rely on
PHP's automagical conversion of nonexistent constants to strings. The
extra curly braces { } would also likely cause an error in the function
params, and are extraneous even if they don't.
--
Jasper Bryant-Greene
Freelance web developer
http://jasper.bryant-greene.name/
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