Question

我有一个sql请求，使用mysqli_fetch_all检索数据库中具有其类别的所有产品，当一个产品具有两个类别，而我返回的结果是该产品的两倍，或者是该类别的n倍时，就会出现问题该产品具有该功能，如果有人用php或sql向我解释了如何做我希望的功能，则我无法在html模板中正确输出该产品。

products
--------
id
name
etc ...

categories
----------
id
name

category_product
----------------
category_id
product_id

$sql = "SELECT
          p.id, 
          p.name, 
          p.price, 
          categories.name as category_name 
          FROM products as p 
             JOIN category_product 
                ON p.id = category_product.product_id
             JOIN categories 
                ON category_product.category_id = categories.id";

output => [0] => Array ( 
                   [id] => 3 
                   [name] => product 3 
                   [price] => 1.00 
                   [category_name] => cat 4 ) 
          [1] => Array ( 
                   [id] => 3 
                   [name] => product 3 
                   [price] => 1.00 
                   [category_name] => cat 3 ) 

expected => [0] => Array ( 
                   [id] => 3 
                   [name] => product 3 
                   [price] => 1.00 
                   [category_name] => Array(
                                        [0] => cat 3
                                        [1] => cat 4 ]))

Answer 1

更改查询以生成类别名称的GROUP_CONCAT，然后在获取后可以在PHP中使用explode拆分该值。例如（假设您在连接MySQLi上使用$conn）：

$sql = "SELECT
          p.id, 
          p.name, 
          p.price, 
          GROUP_CONCAT(categories.name SEPARATOR '|') as category_name 
          FROM products as p 
          JOIN category_product ON p.id = category_product.product_id
          JOIN categories ON category_product.category_id = categories.id
          GROUP BY p.id, p.name, p.price";
$result = $conn->query($sql) or die($conn->error);
while ($row = $result->fetch_assoc()) {
    // split the category names into an array
    $row['category_name'] = explode('|', $row['category_name']);
    // do something with the row value
}

请注意，您需要为GROUP_CONCAT指定一个分隔符（在这种情况下，我使用了|），该分隔符不会出现在类别名称中。

如何在PHP / SQL中获取产品的所有类别

1 个答案: