在单个查询中以总和的百分比汇总表

Posted 2023-05-07

【中文标题】在单个查询中以总和的百分比汇总表

【英文标题】：Summarize tables with percent of sum in single query

【发布时间】：2017-03-20 23:38:40

【问题描述】：

我有一个ACTIVE_TRANSPORTATION 表：

+--------+----------+--------+
| ATN_ID |   TYPE   | LENGTH |
+--------+----------+--------+
|      1 | SIDEWALK |   20.6 |
|      2 | SIDEWALK |   30.1 |
|      3 | TRAIL    |   15.9 |
|      4 | TRAIL    |   40.4 |
|      5 | SIDEWALK |   35.2 |
|      6 | TRAIL    |   50.5 |
+--------+----------+--------+

它通过ATN_ID 与INSPECTION 表相关：

+---------+--------+------------------+
| INSP_ID | ATN_ID | LENGTH_INSPECTED |
+---------+--------+------------------+
|     101 |      2 |             15.2 |
|     102 |      3 |              5.4 |
|     103 |      5 |             15.9 |
|     104 |      6 |             20.1 |
+---------+--------+------------------+

我想把信息总结如下：

+----------+--------+-------------------+
|   TYPE   | LENGTH | PERCENT_INSPECTED |
+----------+--------+-------------------+
| SIDEWALK |   85.9 |               36% |
| TRAIL    |  106.8 |               23% |
+----------+--------+-------------------+

如何在单个查询中做到这一点？

【参考方案1】：

这里是使用 ACCESS 2010 更新的答案。注意LENGTH在 ACCESS 中是保留的，所以需要改成LENGTH_

SELECT 
       TYPE,
       SUM(LENGTH) as LENGTH_,
       SUM(IIF(ISNULL(LENGTH_INSPECTED),0, LENGTH_INSPECTED))/SUM(LENGTH) as  PERCENT_INSPECTED
FROM 
ACTIVE_TRANSPORTATION A
LEFT JOIN INSPECTION B 
    ON A.ATN_ID = B.ATN_ID
GROUP BY TYPE

这是我最初在 SQL SERVER 2014 中使用 T-SQL 的答案

SELECT SUM(LENGTH) as LENGTH,
       SUM(ISNULL(LENGTH_INSPECTED,0))/SUM(LENGTH) as PERCENT_INSPECTED,
       TYPE
FROM 
ACTIVE_TRANSPORTATION A
LEFT JOIN INSPECTION B 
    ON A.ATN_ID = B.ATN_ID
GROUP BY TYPE

如果您需要将其转换为百分比、四舍五入等，请告诉我，但我猜这部分对您来说很容易。

【讨论】：

你还没有勇气尝试检查它是否在 Access 中运行？