1

结构

我有 3 张桌子：

第一个是 [设备] 表：

+----------+----------+-----------+-------------+----------+---------+
| equip_id | chara_id | weapon_id | headgear_id | armor_id | ring_id |
+----------+----------+-----------+-------------+----------+---------+
|        3 |        1 |         3 |           3 |        5 |       6 |
|        4 |        2 |         1 |           2 |        3 |       4 |
+----------+----------+-----------+-------------+----------+---------+

第二个是 [item] 表：

+---------+--------------+-----------+----------+----------+----------+-------------------------------+-----------+------------+
| item_id | item_name    | item_type | item_atk | item_def | item_atr | item_img                      | item_desc | item_price |
+---------+--------------+-----------+----------+----------+----------+-------------------------------+-----------+------------+
|       0 | Halberd      |         1 |      220 |       20 |        0 | pics/weapons/halberd.png      |           |        400 |
|       1 | Axe          |         1 |      220 |      -10 |        0 | pics/weapons/axe.png          |           |        200 |
|       2 | Wooden Sword |         1 |       70 |        0 |        0 | pics/weapons/wooden-sword.png |           |        225 |
|       3 | Dagger       |         1 |       60 |        5 |        0 | pics/weapons/dagger.png       |           |         55 |
|       4 | Bow          |         1 |      120 |        1 |        0 | pics/weapons/bow.png          |           |          0 |
|       5 | Helmet       |         4 |        0 |       20 |        0 | pics/headgear/helmet.png      |           |        155 |
|       6 | Tunic        |         2 |        0 |       10 |        0 | pics/armors/tunic.png         |           |         50 |
|       7 | Armour       |         2 |        0 |       45 |        0 | pics/armors/armour.png        |           |          0 |
|       8 | Necklace     |         3 |       15 |        5 |        0 | pics/accessories/necklace.png |           |        199 |
+---------+--------------+-----------+----------+----------+----------+-------------------------------+-----------+------------+

最后一个是 [chara] 表：

+----------+------------+----------------+------------+
| chara_id | chara_name | chara_class_id | chara_gold |
+----------+------------+----------------+------------+
|        1 | Lawrence   |              1 |          0 |
|        2 | Testo      |              0 |          0 |
|        3 | Viscocent  |              2 |          0 |
|        4 | Piatos     |              1 |          0 |
|        5 | Hello      |              4 |          0 |
+----------+------------+----------------+------------+

我的障碍：

我想从选定的字符中显示每个、和（都来自设备表）的等价物item_name（来自项目表） 。例如：weapon_idarmor_idheadgear_idring_id

武器 ID：1 => 物品表中的物品
名称 Armor_id：3 => 物品表中的物品名称等。

预期结果：当我查询一个角色的装备时，它返回武器名称而不是武器 ID。
---------------
chara_id：1
武器：匕首
头饰：匕首（字符数据是样本，只要它返回 item_name 就没有关系）
盔甲：头盔
戒指：束腰外衣

这可以通过 4 个查询简单地解决，但如果有更好的解决方案，我不想使用 4 个查询。

4

3 回答 3

3

尝试

SELECT e.chara_id,
       e.weapon_id, i1.item_name weapon_name,
       e.headgear_id, i2.item_name headgear_name,
       e.armor_id, i3.item_name armor_name,
       e.ring_id, i4.item_name ring_name
  FROM equipment e LEFT JOIN
       item i1 ON e.weapon_id = i1.item_id LEFT JOIN
       item i2 ON e.headgear_id = i2.item_id LEFT JOIN
       item i3 ON e.armor_id = i3.item_id LEFT JOIN
       item i4 ON e.ring_id = i4.item_id 
 WHERE e.chara_id = 1

输出

| CHARA_ID | WEAPON_ID | WEAPON_NAME | HEADGEAR_ID | HEADGEAR_NAME | ARMOR_ID | ARMOR_NAME | RING_ID | RING_NAME |
------------------------------------------------------------------------------------------------------------------
|        1 |         3 |      Dagger |           3 |        Dagger |        5 |     Helmet |       6 |     Tunic |

SQLFiddle

如果您根本不需要结果集中的 id，只需将它们从查询中删除

于 2013-04-23T03:23:52.033 回答
0

您可以使用连接

SELECT c.chara_name as name, i.item_name
FROM chara c
  LEFT JOIN equipment AS e ON c.chara_id = e.chara_id
  LEFT JOIN item AS i ON i.id = e.item_id
WHERE c.chara_id = 1

你可能会得到多行来处理..但这将是一个查询。

于 2013-04-23T03:24:35.053 回答
0

我想你正在寻找这个：

SELECT e.equip_id, w.item_name AS weapon, a.item_name AS armor, 
    h.item_name AS headgear, r.item_name AS ring
FROM equipment e
INNER JOIN item AS w ON (w.item_id = e.weapon_id)
INNER JOIN item AS a ON (w.item_id = e.armor_id)
INNER JOIN item AS h ON (w.item_id = e.headgear_id)
INNER JOIN item AS r ON (w.item_id = e.ring_id)

;-)

于 2013-04-23T03:25:43.133 回答