Re: Code Structure/Errors
| From: | Dan Koken | Date: | Sun, 16 Jun 2002 07:30:42 +0000 |
| Subject: | Re: Code Structure/Errors | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-102381@lists.php.net to get a copy of this message | ||
Can simply set $i before the while.
I assume the $last_name is coming from the DB.
Not exactly sure what you want, but hope this helps.
good luck... Dan.
----------------------
$i = 0;
while ($row = mysql_fetch_array($result)) {
extract($row);
if (($i % 5) == 0) print "<tr>\n";
$i++;
@$last_name .= '';
if($last_name <> '')
<print stuff here>
else
<print other stuff>
...
if (($i % 5) == 0) print "</tr>\n";
}
Jason Soza wrote:
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