Имам един скрипт:
Но, когато го стартирам, ми излиза: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /opt/lampp/htdocs/lessons.php on line 22
ем пробвах с array, пак не става.
Как ще е? :idea:
Код:
<?php
ob_start("ob_gzhandler");
require_once("include/inc.php");
dbconn();
$res = mysql_query("SELECT COUNT(*) FROM lessons");
$row = mysql_fetch_array($res);
$count = $row[0];
stdhead("Lessons");
?>
<label>There are <?=$count?> lessons.</label>
<?php
if($count) {
$query = "SELECT lessons.name, lessons.cat_id, lessons_cats.name, lessons_cats.id AS cat_idd, users.username FROM lessons LEFT JOIN lessons_cats ON cat_id = lessons_cats.id LEFT JOIN users ON lessons.owner = users.id ORDER BY id";
$res = mysql_query($query);
?>
<table class=box align="left" width="90%">
<tr><td class="colhead">Категория</td><td class="colhead">Автор</td><td class="colhead">Урок</td></tr>
<?php
while($array = mysql_fetch_assoc($res)) {
print("<tr><td><a class=\"lessons\" href=\"?cid=$array[cat_idd]\">$array[name]</a></td><td>$array[username]</td><td>$array[name]</td></tr>");
}
?>
</table>
<?
}
stdfoot();
?>Но, когато го стартирам, ми излиза: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /opt/lampp/htdocs/lessons.php on line 22
ем пробвах с array, пак не става.
Как ще е? :idea: