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($res)) {
printf("<tr><td> {$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("<tr><td> number_format({$thearray[cost]}, 2) </td>
to
printf("<tr><td> ".number_format({$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
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Nuno Pereira
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