我有三个表:产品(ID,名称),类别(id,name)和 products_categories (product_id,category_id)。
每个产品属于一个或多个类别。
我想要检索所有产品,并显示哪些产品已经属于类别" X"。 我的div是这样的:
<span>Category "X"</span>
[some php / fetch_assoc() / … ]
<div class="product">Product A</div>
<div class="product selected">Product B</div>
<div class="product">Product B</div>
目前,它使用两个查询:一个用于获取所有产品,另一个用于检查产品是否在products_categories中。所以在第一个问题中使用php进行了很多小问题。
$getAllProducts = "SELECT products.name as productName, products.id as productID FROM products";
$resultProduct=$mysqli->query($getAllProducts);
while($product=$resultProduct->fetch_assoc()){
$reqChecked = "SELECT * FROM products_categories
WHERE product_id=" . $product["productID"] ."
AND category_id=" . $category["id"]; //$category["id"] is given
$resultChecked = $mysqli->query($reqChecked);
$row = $resultChecked->fetch_row();
$selected = ""
if ( isset($row[0]) ) {
$selected = "selected";
}
只能使用一个查询进行此操作吗?我尝试使用左连接(产品上的products_categories),但列出了属于多个类别的产品他们所处的每个类别。
修改
以下是一些示例数据
产品表
+----+-----------+
| id | name |
+----+-----------+
| 1 | product_1 |
| 2 | product_2 |
| 3 | product_3 |
+----+-----------+
类别表
+----+------------+
| id | name |
+----+------------+
| 1 | category_1 |
| 2 | category_2 |
| 3 | category_3 |
+----+------------+
加入表格
+------------+-------------+
| product_id | category_id |
+------------+-------------+
| 1 | 1 |
| 1 | 2 |
| 2 | 2 |
+------------+-------------+
现在,让我们说我正在编辑category_2页面,我希望得到以下结果:
+------------+--------------+-------------+
| product_id | product_name | category_id |
+------------+--------------+-------------+
| 1 | product_1 | 2 | --product_1 belongs to category_1 and category_2, but I only need it one time.
| 2 | product_2 | 2 |
| 3 | product_3 | NULL | --product_3 belongs to nothing but I want to see it.
+------------+--------------+-------------+
答案 0 :(得分:4)
这只是一个简单的连接问题。我原本以为您可能需要一些查询魔术来显示产品是否属于给定类别。但是,如果您只是使用下面的查询,则可以检查PHP中每行的类别名称并采取相应的行动。
SELECT p.id,
p.name AS product,
c.name AS category -- check for value 'X' in your PHP code
FROM products p
INNER JOIN products_categories pc
ON p.id = pc.product_id
INNER JOIN categories c
ON c.id = pc.category_id
请注意,您当前的方法实际上是尝试在PHP代码本身中进行连接,这有很多原因。
<强>更新强>
SELECT t1.id AS product_id,
t1.name AS product_name,
CASE WHEN t2.productSum > 0 THEN '2' ELSE 'NA' END AS category_id
FROM products t1
LEFT JOIN
(
SELECT product_id,
SUM(CASE WHEN category_id = 2 THEN 1 ELSE 0 END) AS productSum
FROM products_categories
GROUP BY product_id
) t2
ON t1.id = t2.product_id