2

我有这两个表，其中包含一些数据样本。我想提取每个类别中的分类广告数量。我试了一下，每一个都得到（2），这是不正确的。所以希望有人能帮助我。

CREATE TABLE IF NOT EXISTS `categories` (
  `id` int(255) NOT NULL AUTO_INCREMENT,
  `name` text COLLATE utf8_unicode_ci NOT NULL,
  `subcategory_id` int(2) NOT NULL DEFAULT '0',
  `parent_id` int(255) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=27 ;

--
-- Dumping data for table `categories`
--

INSERT INTO `categories` (`id`, `name`, `subcategory_id`, `parent_id`) VALUES
(1, 'Announcements', 0, 0),
(2, 'Employment', 0, 0),
(3, 'Items For Sale', 0, 0),
(4, 'Services', 0, 0),
(5, 'Garage Sales', 0, 0),
(6, 'Automobiles', 0, 0),
(7, 'Announcement1', 1, 1),
(8, 'Announcement2', 1, 1),

--
-- Table structure for table `classifieds`
--

CREATE TABLE IF NOT EXISTS `classifieds` (
  `classified_id` int(255) NOT NULL AUTO_INCREMENT,
  `title` text COLLATE utf8_unicode_ci NOT NULL,
  `description` text COLLATE utf8_unicode_ci NOT NULL,
  `category_id` int(10) NOT NULL,
  `name` text COLLATE utf8_unicode_ci NOT NULL,
  `authorized` int(10) NOT NULL DEFAULT '0',
  PRIMARY KEY (`adid`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=177 ;

--
-- Dumping data for table `classifieds`
--

INSERT INTO `classifieds` (`classified_id`, `title`, `description`, `category_id`, `name`, `authorized`) VALUES
(1, 'Test Classified', 'Here is the First Test classified listing.', 1, 1);

INSERT INTO `classifieds` (`classified_id`, `title`, `description`, `category_id`, `name`, `authorized`) VALUES
(2, 'GMC For Sell', 'Looks like new 1979 GMC.', 6, 1);

这里

$query = "SELECT category_id, COUNT(title) FROM classifieds GROUP BY category_id"; 
$result = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_array($result)
$num_items_in_category = $row['COUNT(title)'];
echo "<><a href='category-".$row['id'].".php' >".$row['name'].$num_items_in_category."</a></li>";

谢谢

4

2 回答 2

2

稍微改变一下 SQL，然后循环遍历结果？

$query = "SELECT c.id, c.name,
                 COUNT(cl.category_id) AS num_items_in_category
              FROM category_id c
                 LEFT JOIN aclassifieds cl ON cl.category_id=c.id
              GROUP BY c.id";  
$result = mysql_query($query) or die(mysql_error()); 

while ($row = mysql_fetch_array($result)) {
    echo "<li><a href='category-".$row['id'].".php' >".$row['name'].$row['num_items_in_category']."</."</a></li>"; 
}
于 2012-05-11T04:05:56.883 回答
0

以防万一其他人想使我受益：

enter $query = "SELECT c.id, c.name,
             COUNT(cl.title) AS num_items_in_category
          FROM categories c
             LEFT JOIN classifieds cl ON cl.category_id=c.id
          GROUP BY c.id";
  $result = mysql_query($query) or die(mysql_error());

while ($row = mysql_fetch_array($result)) { echo "

  • “.$row['name'].$row['num_items_in_category']。”
    • "; } 这里 谢谢

    于 2012-05-11T16:22:03.357 回答