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View Full Version : Help php return undefineded value in actionscript



rocky86
08-06-2007, 11:34 AM
Hi I having problem here how come the uid is undefined value on the actionscripts? the uname is able to be display but the uid is undefinded but I don't know why it return undefinded value? uid is the id of the username like eg: 1,2,3


PHP Code:



$resultpostal=mysql_query("SELECT location.uname,location.uid FROM location,districts WHERE districts.districtno ='$postalno' AND
location.lat BETWEEN districts.startlat AND districts.endlat AND location.lng BETWEEN districts.startlng AND districts.endlng");

$counterx=0;
while($row=mysql_fetch_array($resultpostal)){
foreach($row as $col_value){
$temp[$counterx]=$col_value;
$counterx++;
}


$report.="uid"."=".$temp[0]."&";
$report.="uname"."=".$temp[1]."&";



echo $report;

Anybody know?

Fumigator
08-06-2007, 09:09 PM
That is a really poor way to get results out of a query. Why are you using a foreach() array, assigning each individual array value to another array, then referring to this second array? Why not just use the first array?

Also you may as well use a "for" loop, as it takes care of the iterator for you (makes the code cleaner than a "while" loop).



for ($counterx = 0; $counterx < mysql_num_rows($resultpostal); $counterx++) {
$row[$counterx] = mysql_fetch_array($resultpostal);
}

//Array is ready for you to do whatever you want with it
foreach($row as $rowval) {
$report.="uid"."=".$rowval['uid']."&uname"."=".$rowval['uname']."&";
}

echo $report;



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