我试图进行SELECT查询....而不是UPDATE或INSERT或DELETE。
我有三张桌子。
我想运行一个查询，向我显示每张发票。每张发票只能有一个客户和多个商品......因此存在invoice_items
我当前的查询看起来像这样
SELECT i.order_date, c.name, thedata.info from invoices i inner join customers c ON (i.customer = c.id) right join ( select x.order, group_concat( concat(x.itemname,' ', x.itemdesc) separator "\n" ) as info from invoice_items x ) thedata on (i.id = thedata.order)
当我运行此查询时，我会收到一行，其中包含一个客户，一个发票以及每个项目的列表，无论发票ID或客户是谁...... ???
+---------------------+--------------+---------------------------------------------------------------------------------------------------------------------------------+
| order_date | name | info |
+---------------------+--------------+---------------------------------------------------------------------------------------------------------------------------------+
| 2014-01-23 20:39:20 | Joe Customer | Boxes for boxing
Shoes for shining
2" Hermosa Plank for bobblin
Boxes for boxing
bobbles for bobblin
Lot 297 Woodale Carmel Oak |
+---------------------+--------------+---------------------------------------------------------------------------------------------------------------------------------+
我的目标是获得相同的列表，但显示所有客户以及他们的项目。 我做错了什么？
以下是那些需要它们的模式。
客户
+---------------+------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+---------------+------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | text | NO | | NULL | |
| ship_address | text | NO | | NULL | |
| ship_address2 | text | NO | | NULL | |
| ship_city | text | NO | | NULL | |
| ship_state | text | NO | | NULL | |
| ship_zip | int(6) | NO | | NULL | |
| bill_address | text | NO | | NULL | |
| bill_address2 | text | NO | | NULL | |
| bill_city | text | NO | | NULL | |
| bill_state | text | NO | | NULL | |
| bill_zip | text | NO | | NULL | |
| phone | bigint(20) | NO | | NULL | |
| email | text | NO | | NULL | |
+---------------+------------+------+-----+---------+----------------+
发票
+-------------+----------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-------------+----------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| customer | int(11) | NO | | NULL | |
| order_date | datetime | NO | | NULL | |
| status | text | NO | | NULL | |
| freightcost | double | NO | | NULL | |
+-------------+----------+------+-----+---------+----------------+
Invoice_items
+-----------+---------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+-----------+---------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| order | int(11) | NO | | NULL | |
| qty | int(11) | NO | | NULL | |
| itemname | text | NO | | NULL | |
| itemdesc | text | NO | | NULL | |
| itemprice | double | NO | | NULL | |
+-----------+---------+------+-----+---------+----------------+
答案 0 :(得分：0)
尝试以下查询，如果使用GROUP_CONCAT（），则需要使用GROUP BY。
SELECT i.order_date,
c.name,
group_concat( concat(x.itemname,' ', x.itemdesc) separator "\n" ) as info
FROM invoices i
INNER JOIN customers c ON i.customer = c.id
LEFT JOIN invoice_items x ON i.id = x.order
GROUP BY i.order_date,c.name