Code Structure/Errors
| From: | Jason Soza | Date: | Sun, 16 Jun 2002 05:31:45 +0000 |
| Subject: | Code Structure/Errors | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-102377@lists.php.net to get a copy of this message | ||
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