undefined variable when using if ($var) {}
| From: | Martha S | Date: | Mon, 25 Mar 2002 16:44:25 +0000 |
| Subject: | undefined variable when using if ($var) {} | ||
| Groups: | php.general | ||
| Request: | Send a blank email to php-general+get-90057@lists.php.net to get a copy of this message | ||
I'm rather new to PHP, so this should be fairly easy to answer. I checked
the manual and FAQ already.
I'm using the following code, and I get the following message if $id has
nothing in the var (i have it set to a default of type int, not null in
mysql). Is there a way around this or something I'm missing?
Thanks :)
-- code---
if ($id) {
$result = mysql_query("SELECT * FROM entries WHERE id=$id",$db);
$myrow = mysql_fetch_array($result);
printf("<table border=1><tr>\n");
printf("<tr><td>%s</td></tr>", $myrow["title"]);
printf("<tr><td>%s | ", $myrow["id"]);
printf("%s</td></tr>", $myrow["posted"]);
printf("<tr><td>%s\n</td></tr>", $myrow["post"]);
printf("</table><tr>\n");
printf("<a href=\"$PHP_SELF\">all</a>");
} else {
do {
printf("%s <a href=\"%s?id=%s\">%s</a><br>\n",
$myrow["id"], $PHP_SELF,
$myrow["id"], $myrow["title"], $myrow["post"]);
} while ($myrow = mysql_fetch_array($result));
} else {
// no records to display
echo "Sorry, no records were found!";
}