Re: help with a bucle

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 > > > ""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 >> >> >> -- >> Este mensaje ha sido analizado por MailScanner >> en busca de virus y otros contenidos peligrosos, >> y se considera que está limpio.