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Posted by "sonu gill" on 08/03/05 06:26
Sorry clarification required to my previous answer...
I sort of answered you in a hurry, and I made another assumption that I
didn't define.
I wanted to clarify that I'm assuming the array is setup as follows
arr[day]=>report.
arr[1]=>45
arr[2]=>56
arr[4]=>78
etc...
$y = 31;
for($x=0; $x < $y; $x++)
{
echo "day ". ($x+1) ." ----- report " . ((isset($arr[$x+1]))?
$arr[$x+1]:"0") . "<br />"; //$x+1 is required because you can never have
day 0
}
""sonu gill"" <sonugill@sonugill.com> wrote in message
news:AE.A3.04646.34330F24@pb1.pair.com...
> From what I gathered you simply want to print all days and the "reporte"
> field info and if there isn't any then 0.
>
> Assuming you used SQL and retrieved your information from the db into an
> associated Array
>
> $arr //the results from the DB...
>
> $y = 31; //Assuming you are working with 31 days...
>
> for($x=0; $x<$y; $x++)
> {
> echo "day ". ($x+1) ." ----- report " . ((isset($arr[$x+1]))?
> $arr[$x+1]:"0") . "<br />"; //$x+1 is required because you can never have
> day 0
> }
>
> hope this helps...
>
> -sonu
>
>
> ""Jesϊs Alain Rodrνguez Santos"" <wmaster@cfg.jovenclub.cu> wrote in
> message
> news:1974.192.168.6.61.1123038296.squirrel@correo.cfg.jovenclub.cu...
>> Hi, i have in my mysql db one table columm, some think like this:
>>
>> day reporte
>> --- -------
>> 1 45
>> 2 56
>> 4 78
>> 6 89
>> 7 90
>> etc...
>>
>> How can i print all the value from 'reporte' columm depending the 'day'
>> columm, and if there is not exist a day print 0 value, for example:
>>
>> day 1 ----- reporte 45
>> day 2 ----- reporte 56
>> day 3 ----- 0
>> day 4 ----- reporte 78
>> day 5 ----- 0
>> etc...
>>
>> Sorry for my english
>>
>>
>> --
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