4

我有两张表 - 一张描述您可以购买的物品,一张存储“必需物品”以购买其他物品。第一个表中有一些项目需要其他项目作为一种贸易形式。有时您需要一定数量的所需物品。这是两个表的架构:

+---------------+-----------------+------+-----+---------+----------------+
| Field         | Type            | Null | Key | Default | Extra          |
+---------------+-----------------+------+-----+---------+----------------+
| storeID       | int(5) unsigned | NO   | PRI | NULL    | auto_increment |
| itemID        | int(10)         | NO   |     | NULL    |                |
| quantity      | int(10)         | NO   |     | NULL    |                |
| cost          | int(10)         | NO   |     | NULL    |                |
+---------------+-----------------+------+-----+---------+----------------+

+----------+---------+------+-----+---------+-------+
| Field    | Type    | Null | Key | Default | Extra |
+----------+---------+------+-----+---------+-------+
| parentID | int(10) | NO   |     | NULL    |       |
| itemID   | int(10) | NO   |     | NULL    |       |
| quantity | int(10) | NO   |     | NULL    |       |
+----------+---------+------+-----+---------+-------+

第二个表与第一个表链接storeID <-> parentID。商店中的物品可能有许多必需的物品。

问题:我需要运行一个返回商店中所有商品的查询,但还显示所有必需的商品及其数量。它应该看起来像这样:

4 x Item1,5 x Item2

我有这个当前的查询:

SELECT a.*, b.typeName, GROUP_CONCAT(CONCAT(c.quantity, " x ", d.typeName)) as `reqItems`
        FROM lpStore a 
        INNER JOIN typeNames b ON (a.typeID = b.typeID)
        INNER JOIN lpRequiredItems c ON (a.storeID = c.parentID)
        INNER JOIN typeNames ON (d.typeID = c.typeID)
        GROUP BY c.parentID
        ORDER BY a.cost ASC 

typeNames只是一个将 itemID 与其实际名称联系起来的表。

现在,此查询运行良好,但它似乎不包括HAVE NO REQUIRED ITEMS的项目。我怀疑它与 GROUP BY 一起是 JOINS 之一(它可能正在对该表中不存在的所有 NULL 值进行分组?),但是我还没有弄清楚。如果不需要,我需要它在该列中返回空白或 NULL。

编辑:样本数据

为简单起见,我删除了使用项目名称表连接表

此处提供示例数据:http ://www.sqlfiddle.com/#!2/d8dca/1

实际结果:

+---------+--------+------+----------+-----------------------------+
| storeID | typeID | cost | quantity | reqItems                    |
+---------+--------+------+----------+-----------------------------+
|       1 |   2514 | 2000 |        3 | 5 x 3668,10 x 4825          |
|       3 |   8785 | 5000 |        2 | 1 x 4875,15 x 1102,5 x 9813 |
|       4 |    579 | 1500 |        5 | NULL                        |
+---------+--------+------+----------+-----------------------------+

预期成绩:

+---------+--------+------+----------+-----------------------------+
| storeID | typeID | cost | quantity | reqItems                    |
+---------+--------+------+----------+-----------------------------+
|       1 |   2514 | 2000 |        3 | 5 x 3668,10 x 4825          |
|       2 |   3199 | 1000 |        1 | NULL                        |
|       3 |   8785 | 5000 |        2 | 1 x 4875,15 x 1102,5 x 9813 |
|       4 |    579 | 1500 |        5 | NULL                        |
+---------+--------+------+----------+-----------------------------+
4

3 回答 3

4

问题是GROUP BY parentId,将GROUP_CONCAT()放在子查询中,然后在LEFT JOIN子查询上使用 a 。当GROUP BY parentId等于null这些值时,将被忽略,因此您不返回任何结果:

SELECT  a.`storeID`, 
        a.`typeID`, 
        a.`cost`, 
        a.`quantity` , 
        `reqItems` 
FROM    lpStore a
LEFT JOIN
(
  select  parentID,
     GROUP_CONCAT( CONCAT( quantity,  " x ", typeID ) ) AS  `reqItems` 
  from lpRequiredItems
  group by parentID
) c 
  ON ( a.storeID = c.parentID ) 
ORDER   BY  a.`storeID`, 
        a.`typeID`, 
        a.`cost`, 
        a.`quantity`
LIMIT 0 , 30

请参阅带有演示的 SQL Fiddle

结果是:

| STOREID | TYPEID | COST | QUANTITY |                    REQITEMS |
--------------------------------------------------------------------
|       1 |   2514 | 2000 |        3 |          5 x 3668,10 x 4825 |
|       2 |   3199 | 1000 |        1 |                      (null) |
|       3 |   8785 | 5000 |        2 | 5 x 9813,1 x 4875,15 x 1102 |
|       4 |    579 | 1500 |        5 |                      (null) |
于 2013-01-28T17:27:58.960 回答
0

问题是“内连接”,它只返回匹配的值,即“必需”表中的一个条目。请尝试使用左连接。

SELECT a.*, b.typeName, GROUP_CONCAT(CONCAT(c.quantity, " x ", d.typeName)) as `reqItems`
        FROM lpStore a 
        INNER JOIN typeNames b ON (a.typeID = b.typeID)
        LEFT JOIN lpRequiredItems c ON (a.storeID = c.parentID)
        INNER JOIN typeNames ON (d.typeID = c.typeID)
        GROUP BY c.parentID
        ORDER BY a.cost ASC 
于 2013-01-28T06:57:02.560 回答
0

你只需要使用LEFT JOIN而不是INNER JOIN

SELECT  a.`storeID`, 
        a.`typeID`, 
        a.`cost`, 
        a.`quantity` , 
        GROUP_CONCAT( CONCAT( c.quantity,  " x ", c.typeID ) ) AS  `reqItems` 
FROM    lpStore a
        LEFT JOIN lpRequiredItems c 
            ON ( a.storeID = c.parentID ) 
GROUP   BY c.parentID
ORDER   BY  a.`storeID`, 
        a.`typeID`, 
        a.`cost`, 
        a.`quantity` 
LIMIT 0 , 30
于 2013-01-28T08:32:04.313 回答