php 中的自定义累计总和

Question

我在 php mysql.

的累积总和上取得了成功

+---+---------+
| id| SomeNumt|
+---+---------+
|  1|       10|
|  2|       12|
|  3|        3|
|  4|       15|
|  5|       23|
+---+---------+

SELECT t1.id,
       t1.SomeNumt,
       SUM(t2.SomeNumt) AS SUM
FROM t t1
INNER JOIN t t2 ON t1.id >= t2.id
GROUP BY t1.id,
         t1.SomeNumt
ORDER BY t1.id

输出：

+---+---------+--------+
| id| SomeNumt|     SUM|
+---+---------+--------+
|  1|       10|      10|
|  2|       12|      22|
|  3|        3|      25|
|  4|       15|      40|
|  5|       23|      63|
+---+---------+--------+

现在我有一个小问题，当我自定义查询时(WHERE t1.SomeNumt>3)

...
INNER JOIN t t2 ON t1.id >= t2.id
WHERE t1.SomeNumt>3
...

输出错误

+---+---------+--------+
| id| SomeNumt|     SUM|
+---+---------+--------+
|  1|       10|      10|
|  2|       12|      22|
|  4|       15|      40|
|  5|       23|      63|
+---+---------+--------+

但我期待：

+---+---------+--------+
| id| SomeNumt|     SUM|
+---+---------+--------+
|  1|       10|      10|
|  2|       12|      22|
|  4|       15|      37|   -- 10+22+15
|  5|       23|      60|   -- 10+22+15+23
+---+---------+--------+

如何修复？

Answer 1

您需要对 t2 应用相同的条件：

SELECT t1.id,
       t1.SomeNumt,
       SUM(t2.SomeNumt) AS SUM
FROM t t1
INNER JOIN t t2 ON t1.id >= t2.id
WHERE t1.SomeNumt > 3 AND t2.SomeNumt > 3
GROUP BY t1.id,
         t1.SomeNumt
ORDER BY t1.id

或者您可以在子查询中应用条件：

SELECT t1.id,
       t1.SomeNumt,
       SUM(t2.SomeNumt) AS SUM
FROM (SELECT * FROM t WHERE SomeNumt > 3) t1
INNER JOIN (SELECT * FROM t WHERE SomeNumt > 3) t2 ON t1.id >= t2.id
GROUP BY t1.id,
         t1.SomeNumt
ORDER BY t1.id

输出（对于两个查询）：

id  SomeNumt    SUM
1   10          10
2   12          22
4   15          37
5   23          60

Demo on dbfiddle

如果您使用的是 MySQL 8+，则应按照@GordonLinoff 的回答中所述使用 window 函数。

Answer 2

如果你想要累加和使用window函数：

SELECT t.id, t.SomeNumt,
       SUM(t2.SomeNumt) OVER (ORDER BY t.id) AS SUM
FROM t
ORDER BY t.id;

如果您添加一个 WHERE 子句，那么这应该会执行您想要的操作。

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