Question

这是sql：

$sql = "SELECT DISTINCT o.order_id, cgd.name as customer_group, CONCAT(o.firstname, ' ', o.lastname) AS customer_name,
email, telephone, payment_firstname,  payment_lastname, payment_address_1, payment_address_2, payment_city, payment_zone, payment_postcode, GROUP_CONCAT(DISTINCT op.name) as prod_name, op.model as prod_model, GROUP_CONCAT(op.quantity) as prod_quantity,
CONCAT(o.shipping_firstname, ' ', o.shipping_lastname,',',o.shipping_address_1,',',o.shipping_address_2,',', o.shipping_city,'-',o.shipping_postcode) AS shipping_address,
 o.payment_method, o.shipping_method, o.total, o.currency_code,o.currency_value,
 o.date_added, oos.name as order_status, o.custom_field as agentcode
FROM oc_order o
LEFT JOIN oc_customer_group_description cgd ON (o.customer_group_id = cgd.customer_group_id)
LEFT JOIN oc_order_product op ON (o.order_id = op.order_id)
LEFT JOIN oc_order_status oos ON (o.order_status_id = oos.order_status_id) WHERE cgd.language_id = 1 GROUP BY o.order_id";

我有一个数据库，其结果具有相同ID的不同数组

Array ( [order_id] => 1 [prod_name] => prod1 [qty] => 1)
Array ( [order_id] => 1 [prod_name] => prod2 [qty] => 1)
Array ( [order_id] => 2 [prod_name] => prod1 [qty] => 1)
Array ( [order_id] => 3 [prod_name] => prod1 [qty] => 1)

我希望上面合并为：

    Array ( [order_id] => 1 [prod_name] => prod1,prod2 [qty] => 1,1)
    Array ( [order_id] => 2 [prod_name] => prod1 [qty] => 1)
    Array ( [order_id] => 3 [prod_name] => prod1 [qty] => 1)

在我添加GROUP_CONCAT后，它会输入：

Array ( [order_id] => 1 [prod_name] => prod1,prod1,prod2,prod2 [qty] => 1,1,1,1)
    Array ( [order_id] => 2 [prod_name] => prod1 [qty] => 1)
    Array ( [order_id] => 3 [prod_name] => prod1 [qty] => 1)

数据库结构是：

o (order): 
    1   order_id    int(11) Primary Primary
op (order_product): 
    1   order_product_id    int(11) 
    2   order_id    int(11) 
    3   product_id  int(11)
    4   name    varchar(255)
    5   quantity    int(4)
    6   price   decimal(15,4)
    7   total   decimal(15,4)


cgd (customer_group_description):
   1    customer_group_id   int(11) 
    2   language_id int(11)     
    3   name    varchar(32)     
    4   description text

Answer 1

在php中你可以循环执行：

$merge_arr = [];

foreach( $arr as $key => $row){

    if( isset( $merge_arr[ $row['order_id'] ] ) ) {
        $merge_arr[ $row['order_id'] ][ 'order_id' ] .= "," . $arr[ $key ][ 'prod_name' ];  
        $merge_arr[ $row['order_id'] ][ 'prod_name' ] .= "," . $arr[ $key ][ 'prod_name' ];
        $merge_arr[ $row['order_id'] ][ 'qty' ] ,= "," $arr[ $row['order_id'] ][ 'qty' ];
    } else {
        $merge_arr[ $row['order_id'] ][ 'prod_name' ] = $arr[ $row['order_id'] ][ 'prod_name' ];
        $merge_arr[ $row['order_id'] ][ 'qty' ] = $arr[ $row['order_id'] ][ 'qty' ];
    }

}

SQL中的

：

SELECT order_id, GROUP_CONCAT(prod_name), GROUP_CONCAT(qty)
FROM your_table
GROUP_BY order_id

如果可以，请使用更快的SQL解决方案。

你的数据库结构是这样的吗？（在一个简单的版本中）

o (order): 
    order_id PK
    order_status_id FK
    customer_group_id FK

op (order_product): 
    order_product_id PK
    order_id FK
    quantity

os (order_status):
    order_status_id PK

cgd (customer_group_description):
    customer_group_id PK

合并具有相同元素id的两个数组

1 个答案: