Re: help with a bucle
| From: | sonu gill | Date: | Wed, 03 Aug 2005 03:01:11 +0000 |
| Subject: | Re: help with a bucle | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-219878@lists.php.net to get a copy of this message | ||
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"" <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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