[PHP] Code Structure/Errors

2002-06-15 Thread Jason Soza

I'm just curious if there's a way to restructure my code so as to avoid
getting Undefined Variable errors. I keep getting them and I know they're
nothing to worry about for the most part, I'd like to get rid of them
without turning off error reporting if possible. In the following code, I
get an undefined variable error for $i and for $last_name, is there anything
I can do to actually define them? $last_name is a variable produced by my
MySQL query, $i is just a counter:

while ($row = mysql_fetch_array($result)) {
extract($row);
$i++;
if($i==1) {
print tr\n;
}

if($last_name) {
print stuff here
} else {
print other stuff
}
...
if ($i==5) {
 print /tr\n;
 $i=0;
 }
}

Thanks!

Jason Soza


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Re: [PHP] Code Structure/Errors

2002-06-15 Thread Pushkar Pradhan

Quick soln. is to put a  symbol in front of the error generating line.
E.g.@$i++;
actual soln. would be to do some initialization like
$i = 0; outside your while loop and $last_name=;
 I'm just curious if there's a way to restructure my code so as to avoid
 getting Undefined Variable errors. I keep getting them and I know they're
 nothing to worry about for the most part, I'd like to get rid of them
 without turning off error reporting if possible. In the following code, I
 get an undefined variable error for $i and for $last_name, is there anything
 I can do to actually define them? $last_name is a variable produced by my
 MySQL query, $i is just a counter:

 while ($row = mysql_fetch_array($result)) {
   extract($row);
   $i++;
   if($i==1) {
   print tr\n;
   }

   if($last_name) {
   print stuff here
   } else {
   print other stuff
   }
 ...
   if ($i==5) {
print /tr\n;
$i=0;
}
   }

 Thanks!

 Jason Soza


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 To unsubscribe, visit: http://www.php.net/unsub.php


-Pushkar S. Pradhan


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