Грешка при извеждане на категории

php_mysql

Registered
Пробвам се да извадя категории те от таблицата categories:

Код:
<?php
 
	echo '<div class="moduletable-dark1"><h3>Жанров каталог</h3>
	<table cellspacing="0" cellpadding="0" width="100%" border="0">';
	$categories1 = mysql_query("SELECT * FROM categories ORDER BY id"); 
	while($r = mysql_fetch_array($categories1)) {
		echo '<tr align="left">
			        <td>
      		<a href="#"></a>
		<a href="#" class="mainlevel-dark1">'.$r[cat_title].' <span class="small">(0)</span></a></span>
			        </td>
			</tr>';
		} 
		echo '</table>
		</div>';
?>

Код:
CREATE TABLE  categories` (
  `cat_id` mediumint(8) unsigned NOT NULL auto_increment,
  `cat_title` varchar(100) default NULL,
  PRIMARY KEY  (`cat_id`)
) ENGINE=MyISAM ;

`Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in

ред:
Код:
while($r = mysql_fetch_array($categories1)) {
 
<?php

echo '<div class="moduletable-dark1"><h3>Жанров каталог</h3>
<table cellspacing="0" cellpadding="0" width="100%" border="0">';
$categories1 = mysql_query("SELECT * FROM categories ORDER BY id") or die(mysql_error());
while($r = mysql_fetch_assoc($categories1)) {
echo '<tr align="left">
<td>
<a href="#"></a>
<a href="#" class="mainlevel-dark1">'.$r[cat_title].' <span class="small">(0)</span></a></span>
</td>
</tr>';
}
echo '</table>
</div>';
?>
 
lam3r4370 каза:
<?php

echo '<div class="moduletable-dark1"><h3>Жанров каталог</h3>
<table cellspacing="0" cellpadding="0" width="100%" border="0">';
$categories1 = mysql_query("SELECT * FROM categories ORDER BY id") or die(mysql_error());
while($r = mysql_fetch_assoc($categories1)) {
echo '<tr align="left">
<td>
<a href="#"></a>
<a href="#" class="mainlevel-dark1">'.$r[cat_title].' <span class="small">(0)</span></a></span>
</td>
</tr>';
}
echo '</table>
</div>';
?>

Благодаря , имам още 1 въпрос значи виждаш това (0) това искам да показва колко заглавия има в тази категория.
Иначе +1 :)
 

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