SQL查询如何转换数据？
我偶然发现了这一点，并试图理解 考虑到实际正在转换数据 输入表SQL查询如何转换数据？,sql,sql-server,Sql,Sql Server,我偶然发现了这一点，并试图理解 考虑到实际正在转换数据 输入表 +---------+-------+---------------+ | user_id | State | Subscriptions | +---------+-------+---------------+ | 1 | LA | 4 | | 2 | LA | 4 | | 3 | LA | 12 | |
+---------+-------+---------------+
| user_id | State | Subscriptions |
+---------+-------+---------------+
| 1 | LA | 4 |
| 2 | LA | 4 |
| 3 | LA | 12 |
| 4 | LA | 12 |
| 5 | LA | 8 |
| 6 | LA | 3 |
| 7 | NY | 14 |
| 8 | NY | 15 |
| 9 | NY | 3 |
| 10 | NY | 2 |
| 11 | NY | 4 |
| 12 | NY | 12 |
| 13 | OH | 6 |
| 14 | OH | 8 |
| 15 | OH | 2 |
| 16 | OH | 3 |
+---------+-------+---------------+
+--------------------+----+----+----+
| Subscription_Range | LA | NY | OH |
+--------------------+----+----+----+
| 1 to 4 | 3 | 3 | 2 |
| 5 to 11 | 1 | 0 | 2 |
| 12 to 15 | 2 | 3 | 0 |
+--------------------+----+----+----+
+---------+-------+---------------+
| user_id | State | Subscriptions |
+---------+-------+---------------+
| 1 | LA | 4 |
| 2 | LA | 4 |
| 3 | LA | 12 |
| 4 | LA | 12 |
| 5 | LA | 8 |
| 6 | LA | 3 |
| 7 | NY | 14 |
| 8 | NY | 15 |
| 9 | NY | 3 |
| 10 | NY | 2 |
| 11 | NY | 4 |
| 12 | NY | 12 |
| 13 | OH | 6 |
| 14 | OH | 8 |
| 15 | OH | 2 |
| 16 | OH | 3 |
+---------+-------+---------------+
+--------------------+----+----+----+
| Subscription_Range | LA | NY | OH |
+--------------------+----+----+----+
| 1 to 4 | 3 | 3 | 2 |
| 5 to 11 | 1 | 0 | 2 |
| 12 to 15 | 2 | 3 | 0 |
+--------------------+----+----+----+
select（订阅时的大小写订阅仅为group by使用的键生成范围别名值时的第一部分大小写
洛杉矶，纽约，哦这三个部分
使用伪聚合函数模拟数据透视表
如果没有假聚合函数，每个值都放在不同的行中。.使用group by将具有相同范围的所有行减少为一行。.获得所需的方面
select (case when subscriptions <= 4 then '1 to 4'
when subscriptions <= 11 then '5 to 11'
when subscriptions <= 15 then '12 to 15'
end) as subscription_range,
sum(case when state = 'LA' then 1 else 0 end) as LA,
sum(case when state = 'NY' then 1 else 0 end) as NY,
sum(case when state = 'OH' then 1 else 0 end) as OH
from t
group by (case when subscriptions <= 4 then '1 to 4'
when subscriptions <= 11 then '5 to 11'
when subscriptions <= 15 then '12 to 15'
end)
order by min(subscriptions);
select（订阅时的情况在聚合之前对select
子句中的表达式进行求值时，您可以想象得到给定数据集的下表：
+--------------------+-----+----+----+
| subscription_range | LA | NY | OH |
+--------------------+-----+----+----+
| '1 to 4' | 1 | 0 | 0 |
| '1 to 4' | 1 | 0 | 0 |
| '12 to 15' | 1 | 0 | 0 |
| '12 to 15' | 1 | 0 | 0 |
| '5 to 11' | 1 | 0 | 0 |
| '1 to 4' | 1 | 0 | 0 |
| '12 to 15' | 0 | 1 | 0 |
| '12 to 15' | 0 | 1 | 0 |
| '1 to 4' | 0 | 1 | 0 |
| '1 to 4' | 0 | 1 | 0 |
| '1 to 4' | 0 | 1 | 0 |
| '12 to 15' | 0 | 1 | 0 |
| '5 to 11' | 0 | 0 | 1 |
| '5 to 11' | 0 | 0 | 1 |
| '1 to 4' | 0 | 0 | 1 |
| '1 to 4' | 0 | 0 | 1 |
+--------------------+-----+----+----+
这里，对于数据集中的每一行，第一个case
语句生成一个字符串，随后的case
语句生成一个1
或0
，具体取决于state
列是否验证测试表达式
聚合时，查询计算相同的subscription\u range
值集，并通过此数据对组进行分组，以使每个subscription\u range
都是不同的
随后各列中剩余的数值数据通过包含每个case
语句的sum
表达式对各组进行求和，得出：
+--------------------+----+----+----+
| subscription_range | LA | NY | OH |
+--------------------+----+----+----+
| 1 to 4 | 3 | 3 | 2 |
| 5 to 11 | 1 | 0 | 2 |
| 12 to 15 | 2 | 3 | 0 |
+--------------------+----+----+----+
要知道首先检查什么以及如何检查，需要深入了解您感兴趣的特定数据库引擎的源代码。我非常怀疑Microsoft是否会让您访问SQL Server的源代码，但也许您可以从众多免费使用的关系数据库引擎中提取源代码能够并开始挖掘。你可能会发现你的问题没有一个简单或通用的答案。感谢你花时间详细地写下答案！！感谢你花时间写下答案