我的情况是，我想计算要在我制作的战斗模块上使用的玩家攻击力，但只是想知道我实际上有两个选择:
计算服务器造成的伤害。(我当前的选项)
- 使用 PHP 计算 DAMAGE DEALT 和 UPDATE 服务器数据库值。
- 传递 2 个字符 ID，然后计算查询中的所有内容并更新所有内容(这可能吗)。
问题:我可以在查询中执行此操作吗？(选项 B)
我当前的设置:
1 个角色有 4 个项目，我通过在客户端添加所有 4 个项目 atk 和角色基本 atk 来计算角色 atk。 (我认为这很容易出现安全漏洞)
然后更新服务器端的值。
这是我的表格:
角色:
+----------+------------+----------------+-------------+------------+----------+----------+----------+-----------+-----------+
| chara_id | chara_name | chara_class_id | chara_level | chara_gold | chara_hp | chara_mp | chara_xp | chara_atk | chara_def |
+----------+------------+----------------+-------------+------------+----------+----------+----------+-----------+-----------+
| 1 | LawrenceX | 1 | 5 | 230 | -175 | 1000 | 0 | 7 | 3 |
| 3 | Viscocent | 2 | 2 | 96 | -206 | 1100 | 1700 | 5 | 5 |
| 4 | Piatos | 1 | 1 | 120 | -60 | 1000 | 0 | 7 | 3 |
| 5 | Hello | 1 | 1 | 300 | -50 | 1000 | 200 | 2 | 8 |
| 6 | Sample | 3 | 2 | 251 | -85 | 900 | 0 | 9 | 1 |
| 8 | Sampuro | 2 | 1 | 170 | 895 | 1100 | 700 | 5 | 5 |
| 12 | fail | 2 | 3 | 481 | 1100 | 1300 | 0 | 21 | 9 |
| 13 | new | 1 | 1 | 1000 | -80 | 1000 | 0 | 5 | 5 |
+----------+------------+----------------+-------------+------------+----------+----------+----------+-----------+-----------+
项目:
+---------+-----------------+-----------+----------+----------+----------+---------------------------------+-------------------------------------------------+------------+
| 0 | None | 0 | 0 | 0 | 0 | pics/none.png | | 400 |
| 1 | Axe | 1 | 220 | 10 | 0 | pics/weapons/axe.png | Another lumberjack axe is another man's weapon. | 200 |
| 2 | Wooden Sword | 1 | 70 | 0 | 0 | pics/weapons/wooden-sword.png | A wooden sword, 99% made from wood | 225 |
| 3 | Dagger | 1 | 60 | 5 | 0 | pics/weapons/dagger.png | A Dagger, Cheap and Sharp | 55 |
| 4 | Bow | 1 | 120 | 1 | 0 | pics/weapons/bow.png | The basics and simplest of all bows. | 120 |
| 5 | Helmet | 4 | 0 | 50 | 0 | pics/headgears/helmet.png | iron helmet - made from an iron pot scraps. | 155 |
| 6 | Tunic | 2 | 10 | 10 | 0 | pics/armors/tunic.png | A peasants tunic. | 50 |
| 7 | Armour | 2 | 0 | 75 | 0 | pics/armors/armour.png | | 150 |
| 8 | Necklace | 3 | 25 | 15 | 0 | pics/accessories/necklace.png | | 199 |
| 9 | Studded Leather | 2 | 25 | 60 | 0 | pics/armors/studded-leather.png | | 240 |
+---------+-----------------+-----------+----------+----------+----------+---------------------------------+-------------------------------------------------+------------+
设备:
+----------+----------+-----------+-------------+----------+---------+
| equip_id | chara_id | weapon_id | headgear_id | armor_id | ring_id |
+----------+----------+-----------+-------------+----------+---------+
| 3 | 1 | 14 | 5 | 6 | 8 |
| 5 | 3 | 4 | 5 | 6 | 8 |
| 6 | 4 | 11 | 5 | 7 | 8 |
| 7 | 5 | 12 | 5 | 6 | 8 |
| 8 | 6 | 3 | 16 | 7 | 8 |
| 10 | 8 | 15 | 5 | 7 | 8 |
| 13 | 12 | 14 | 5 | 6 | 17 |
| 40 | 13 | 3 | 5 | 7 | 8 |
+----------+----------+-----------+-------------+----------+---------+
表关系:
1 chara = 1 equipment
1 weapon_id, armor_id, ring_id, headgear_id = 1 item (total of 4 items, headgear_id = 1 item).
我可以使用此查询获取角色的装备(KUDOS @JC):
SELECT i1.item_atk weapon_atk,i1.item_def weapon_def,
i2.item_atk headgear_atk,
i2.item_def headgear_def,
i3.item_atk armor_atk,
i3.item_def armor_def,
i4.item_atk ring_atk,
i4.item_def ring_def
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
结果:
+------------+------------+--------------+--------------+-----------+-----------+----------+----------+
| weapon_atk | weapon_def | headgear_atk | headgear_def | armor_atk | armor_def | ring_atk | ring_def |
+------------+------------+--------------+--------------+-----------+-----------+----------+----------+
| 275 | 25 | 0 | 50 | 10 | 10 | 25 | 15 |
+------------+------------+--------------+--------------+-----------+-----------+----------+----------+
现在我想合计该角色装备的 atk 和 def 并在该查询中返回它
预期结果:
+------------+------------+
| total_atk | total_def |
+------------+------------+
| 310 | 100 |
+------------+------------+
最佳答案
这是我能想到的最简单的方法。
SELECT IFNULL(W.item_atk, 0) + IFNULL(H.item_atk, 0) + IFNULL(A.item_atk, 0) + IFNULL(R.item_atk, 0) AS total_atk
, IFNULL(W.item_def, 0) + IFNULL(H.item_def, 0) + IFNULL(A.item_def, 0) + IFNULL(R.item_def, 0) AS total_def
FROM equipment E
LEFT JOIN item W ON W.item_id = E.weapon_id
LEFT JOIN item H ON H.item_id = E.headgear_id
LEFT JOIN item A ON A.item_id = E.armor_id
LEFT JOIN item R ON R.item_id = E.ring_id
WHERE E.chara_id = 1
我已重命名表的别名以轻松跟踪它们。我使用了 IFNULL 以防角色没有特定的设备。
================================================== =====================================
伙计，我刚刚又做了一个查询，我认为这比上面的查询要快。不过，我还没有测试过它们。
SELECT SUM(IFNULL(I.item_atk, 0)) AS total_atk
, SUM(IFNULL(I.item_def, 0)) AS total_def
FROM equipment E
LEFT JOIN item I ON I.item_id = E.weapon_id
OR I.item_id = E.headgear_id
OR I.item_id = E.armor_id
OR I.item_id = E.ring_id
WHERE E.chara_id = 1
GROUP BY E.chara_id
关于php - 从 2 个表中选择项目的总值并更新另一个表的值(复杂查询)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/16432378/