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"" <[EMAIL PROTECTED]> wrote in message 
news:[EMAIL PROTECTED]
> 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
>
>
> ""Jesús Alain Rodríguez Santos"" <[EMAIL PROTECTED]> wrote in 
> message 
> news:[EMAIL PROTECTED]
>> 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
>>
>>
>> -- 
>> Este mensaje ha sido analizado por MailScanner
>> en busca de virus y otros contenidos peligrosos,
>> y se considera que está limpio. 

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