使用JOIN
。
例
SELECT * FROM category LEFT JOIN
category_details ON category_details.cat_id = category.cat_id;
上面的查詢將返回所有的category
和相關category detail
。
OP評論響應
SELECT c.name,
IFNULL(sub_c.total, 0) num
FROM category c
LEFT JOIN (SELECT COUNT(*) total, cat_id
FROM category_details
GROUP BY cat_id
) sub_c ON (sub_c.cat_id = c.cat_id);
完整代碼
<?php
$query = "SELECT c.name,
IFNULL(sub_c.total, 0) num
FROM products_category c
LEFT JOIN (SELECT COUNT(*) total, cat_id
FROM product
GROUP BY cat_id
) sub_c ON (sub_c.cat_id = c.id)";
$result = mysql_query($query)or die(mysql_error());
echo "<table><tr><td>NameCount</td></tr>";
while($row = mysql_fetch_assoc($result))
{
echo "<tr><td>".$row['name']."(".$row['num'].")"."</td></tr>";
}
echo "</table>";
?>
MySQL表
產品分類表
CREATE TABLE IF NOT EXISTS `products_category` (
`id` int(11) NOT NULL,
`name` varchar(200) COLLATE utf8_unicode_ci NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
--
-- Dumping data for table `products_category`
--
INSERT INTO `products_category` (`id`, `name`) VALUES
(1, 'Erasmus'),
(2, 'Preston'),
(3, 'Ulric'),
(4, 'Gray'),
(5, 'Joseph'),
(6, 'Merrill'),
(7, 'Alan'),
(8, 'Jeremy'),
(9, 'Solomon'),
(10, 'Andrew'),
(11, 'Galvin'),
(12, 'Craig'),
(13, 'Cameron'),
(14, 'Omar'),
(15, 'Addison');
產品表
CREATE TABLE IF NOT EXISTS `product` (
`id` int(11) NOT NULL,
`name` varchar(200) COLLATE utf8_unicode_ci NOT NULL,
`cat_id` int(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;
--
-- Dumping data for table `product`
--
INSERT INTO `product` (`id`, `name`, `cat_id`) VALUES
(1, 'Wesley', 1),
(2, 'Graiden', 2),
(3, 'Cruz', 5),
(4, 'Hayden', 5),
(5, 'Kennedy', 6),
(6, 'Uriah', 8),
(7, 'Alan', 8),
(8, 'Cade', 1),
(9, 'Ryan', 5),
(10, 'Brody', 7);
上面會輸出
Erasmus(2)
Preston(1)
Ulric(0)
Gray(0)
Joseph(3)
Merrill(1)
Alan(1)
Jeremy(2)
Solomon(0)
Andrew(0)
Galvin(0)
Craig(0)
Cameron(0)
Omar(0)
Addison(0)
你在做什麼? – Neo
「我可以這樣使用」...是的 –
如何?你能告訴我嗎? – Bee