mysql - 为什么 mysql_fetch 的总和值加倍了？

Question

我有一个 2 数据库。

1为game_jnship_user，2为game_jnship_equip，数据如下:

这是game_jnship_user

+-----+-----------------+
| ID  | uid             |
+-----+-----------------+
| 1   | 50              |
+-----------------------+

这是game_jnship_equip

+-----+------------+---------------+--------------+
| ID  | uid        | ebonus        | etype        |
+-----+------------+---------------+--------------+
| 1   | 50         | 100           | 1            |
| 2   | 50         | 10            | 1            |
| 3   | 50         | 120           | 2            |
+-------------------------------------------------+

我查询如下:

$_G['uid'] = '50';
$ushuxing = DB::query("SELECT sum(t2.ebonus) AS equipatk, sum(t3.ebonus) AS equipdef FROM ".DB::table('game_jnship_user')." t1 LEFT JOIN ".DB::table('game_jnship_equip')." t2 ON (t1.uid = t2.uid AND t2.etype = '1') LEFT JOIN ".DB::table('game_jnship_equip')." t3 ON (t1.uid = t3.uid AND t3.etype = '2') WHERE t1.uid = $_G[uid]");
$rushuxing = DB::fetch($ushuxing);

之后，我的输出数据如下:

$rushuxing[equipatk] = 110;
$rushuxing[equipdef] = 240;

为什么 $rushuxing[equipdef] 输出是 240 而不是 120？

谢谢。

这是来自上面示例的原始 MySQL 查询:

SELECT sum(t2.ebonus) AS equipatk, sum(t3.ebonus) AS equipdef
FROM game_jnship_user t1
LEFT JOIN game_jnship_equip t2
    ON (t1.uid = t2.uid AND t2.etype = '1')
LEFT JOIN game_jnship_equip t3
    ON (t1.uid = t3.uid AND t3.etype = '2')
WHERE t1.uid = '50'

Answer 1

在 LEFT JOIN 操作中，etype = 2 的记录与 etype = 1 的 2 条记录相匹配，因此您得到 240 120。

您可以改用以下利用条件聚合的查询:

SELECT SUM(CASE WHEN etype = 1 THEN ebonus ELSE 0 END) AS equipatk,
       SUM(CASE WHEN etype = 2 THEN ebonus ELSE 0 END) AS equipdef
FROM game_jnship_equip
WHERE uid = 50

GROUP BY uid 子句不是必需的，因为您正在为特定 uid 值执行聚合。