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
> bw@stripped
>
>
--
Nuno Pereira
