如何在SQLite中透视，即选择宽格式的长格式存储的表？

我想得到一个表，其中存储了学生的数据以及他们在一个查询中获得的所有科目的分数

这是我的表格结构：

表：

markdetails

## studid ## ## subjectid ##  ## marks ##
     A1            3                50
     A1            4                60
     A1            5                70
     B1            3                60
     B1            4                80
     C1            5                95

表：

学生信息

## studid ## ## name## ## subjectid_3 ## ## subjectid_4 ## ## subjectid_5 ##
      A1        Raam        50                60                 70
      B1        Vivek       60                80                null
      c1        Alex       null              null                95

实际结构：

## studid ##  ## name ##
      A1          Raam
      B1          Vivek
      c1          Alex

作为数据透视的结果，我希望结果集具有以下结构：

表：

学生信息

## studid ## ## name## ## subjectid_3 ## ## subjectid_4 ## ## subjectid_5 ##
      A1        Raam        50                60                 70
      B1        Vivek       60                80                null
      c1        Alex       null              null                95

---

如何在SQLite中实现这一点？

首先，您需要将当前表更改为临时表：

alter table student_info rename to student_name

drop table student_temp

然后，您需要重新创建

学生信息

：

create table student_info add column (
    stuid VARCHAR(5) PRIMARY KEY,
    name VARCHAR(255),
    subjectid_3 INTEGER,
    subjectid_4 INTEGER,
    subjectid_5 INTEGER
)

insert into student_info
select
    u.stuid,
    u.name,
    s3.marks as subjectid_3,
    s4.marks as subjectid_4,
    s5.marks as subjectid_5
from
    student_temp u
    left outer join markdetails s3 on
        u.stuid = s3.stuid
        and s3.subjectid = 3
    left outer join markdetails s4 on
        u.stuid = s4.stuid
        and s4.subjectid = 4
    left outer join markdetails s5 on
        u.stuid = s5.stuid
        and s5.subjectid = 5

然后，填充

学生信息

：

create table student_info add column (
    stuid VARCHAR(5) PRIMARY KEY,
    name VARCHAR(255),
    subjectid_3 INTEGER,
    subjectid_4 INTEGER,
    subjectid_5 INTEGER
)

insert into student_info
select
    u.stuid,
    u.name,
    s3.marks as subjectid_3,
    s4.marks as subjectid_4,
    s5.marks as subjectid_5
from
    student_temp u
    left outer join markdetails s3 on
        u.stuid = s3.stuid
        and s3.subjectid = 3
    left outer join markdetails s4 on
        u.stuid = s4.stuid
        and s4.subjectid = 4
    left outer join markdetails s5 on
        u.stuid = s5.stuid
        and s5.subjectid = 5

现在，只需放下临时表：

alter table student_info rename to student_name

drop table student_temp

这就是如何快速更新表的方法

---

SQLite缺少一个

pivot

函数，所以最好是硬编码一些左连接。

左联接

将在其联接条件中匹配任何行，并对第一个表或左表中不符合第二个表联接条件的任何行返回

null

。

因为作者不够友好，无法提供SQL来创建架构，这是为任何想尝试@Eric解决方案的人准备的

create table markdetails (studid, subjectid, marks);
create table student_info (studid, name);

insert into markdetails values('A1', 3, 50);
insert into markdetails values('A1', 4, 60);
insert into markdetails values('A1', 5, 70);
insert into markdetails values('B1', 3, 60);
insert into markdetails values('B1', 4, 80);
insert into markdetails values('C1', 5, 95);

insert into student_info values('A1', 'Raam');
insert into student_info values('B1', 'Vivek');
insert into student_info values('C1', 'Alex');

下面是一个使用

案例

和

分组方式

的替代解决方案

select
    si.studid,
    si.name,
    sum(case when md.subjectid = 3 then md.marks end) subjectid_3,
    sum(case when md.subjectid = 4 then md.marks end) subjectid_4,
    sum(case when md.subjectid = 5 then md.marks end) subjectid_5
from student_info si
join markdetails md on
        md.studid = si.studid
group by si.studid, si.name
;

为了进行比较，这里是@Eric的解决方案中相同的select语句：

select
    u.stuid,
    u.name,
    s3.marks as subjectid_3,
    s4.marks as subjectid_4,
    s5.marks as subjectid_5
from
    student_info u
    left outer join markdetails s3 on
        u.stuid = s3.stuid
        and s3.subjectid = 3
    left outer join markdetails s4 on
        u.stuid = s4.stuid
        and s4.subjectid = 4
    left outer join markdetails s5 on
        u.stuid = s5.stuid
        and s5.subjectid = 5
;

---

当有很多数据时，看看哪一个会表现得更好会很有趣。

很棒的附录！帮助我以较低的工作量和系统负载解决了类似的问题。我使用Raspberry Pi获取1线接口DS18B20温度传感器数据，如下所示：

CREATE TABLE temps (Timestamp DATETIME, sensorID TEXT, temperature NUMERIC);

例如：

sqlite> .headers on
sqlite> .mode column
sqlite> select * from temps where timestamp > '2014-02-24 22:00:00';

Timestamp            sensorID         temperature
-------------------  ---------------  -----------
2014-02-24 22:00:02  28-0000055f3f10  19.937
2014-02-24 22:00:03  28-0000055f0378  19.687
2014-02-24 22:00:04  28-0000055eb504  19.937
2014-02-24 22:00:05  28-0000055f92f2  19.937
2014-02-24 22:00:06  28-0000055eef29  19.812
2014-02-24 22:00:07  28-0000055f7619  19.625
2014-02-24 22:00:08  28-0000055edf01  19.687
2014-02-24 22:00:09  28-0000055effda  19.812
2014-02-24 22:00:09  28-0000055e5ef2  19.875
2014-02-24 22:00:10  28-0000055f1b83  19.812
2014-02-24 22:10:03  28-0000055f3f10  19.937
2014-02-24 22:10:04  28-0000055f0378  19.75
2014-02-24 22:10:04  28-0000055eb504  19.937
2014-02-24 22:10:05  28-0000055f92f2  19.937

sqlite> select * from TempsSlot10min where timeslot >= '2014-02-24 22:00:00';

TimeSlot             SensorName  temperature
-------------------  ----------  -----------
2014-02-24 22:00:00  T1          19.937
2014-02-24 22:00:00  T2          19.687
2014-02-24 22:00:00  T3          19.937
2014-02-24 22:00:00  T4          19.937
2014-02-24 22:00:00  T5          19.812
2014-02-24 22:00:00  T6          19.625
2014-02-24 22:00:00  T10         19.687
2014-02-24 22:00:00  T9          19.812
2014-02-24 22:00:00  T8          19.875
2014-02-24 22:00:00  T7          19.812
2014-02-24 22:10:00  T1          19.937
2014-02-24 22:10:00  T2          19.75
2014-02-24 22:10:00  T3          19.937
2014-02-24 22:10:00  T4          19.937
2014-02-24 22:10:00  T5          19.875

select * from PivotTemps10min where timeslot >= '2014-02-24 22:00:00';

TimeSlot             T1          T2              T10
-------------------  ----------  ---------- ...  ----------
2014-02-24 22:00:00  19.937      19.687          19.687
2014-02-24 22:10:00  19.937      19.75           19.687
2014-02-24 22:20:00  19.937      19.75           19.687
2014-02-24 22:30:00  20.125      19.937          19.937
2014-02-24 22:40:00  20.187      20.0            19.937
2014-02-24 22:50:00  20.25       20.062          20.062
2014-02-24 23:00:00  20.25       20.062          20.062

使用SUBSTR（）命令，我将时间戳“规范化”为10分钟。使用JOIN，使用查找表“传感器”将传感器ID更改为传感器名称

CREATE VIEW [TempsSlot10min] AS
SELECT SUBSTR(datetime(timestamp),1,15)||'0:00' AS TimeSlot,
SensorName,
temperature FROM
temps JOIN sensors USING (sensorID, sensorID);

例如：

sqlite> .headers on
sqlite> .mode column
sqlite> select * from temps where timestamp > '2014-02-24 22:00:00';

Timestamp            sensorID         temperature
-------------------  ---------------  -----------
2014-02-24 22:00:02  28-0000055f3f10  19.937
2014-02-24 22:00:03  28-0000055f0378  19.687
2014-02-24 22:00:04  28-0000055eb504  19.937
2014-02-24 22:00:05  28-0000055f92f2  19.937
2014-02-24 22:00:06  28-0000055eef29  19.812
2014-02-24 22:00:07  28-0000055f7619  19.625
2014-02-24 22:00:08  28-0000055edf01  19.687
2014-02-24 22:00:09  28-0000055effda  19.812
2014-02-24 22:00:09  28-0000055e5ef2  19.875
2014-02-24 22:00:10  28-0000055f1b83  19.812
2014-02-24 22:10:03  28-0000055f3f10  19.937
2014-02-24 22:10:04  28-0000055f0378  19.75
2014-02-24 22:10:04  28-0000055eb504  19.937
2014-02-24 22:10:05  28-0000055f92f2  19.937

sqlite> select * from TempsSlot10min where timeslot >= '2014-02-24 22:00:00';

TimeSlot             SensorName  temperature
-------------------  ----------  -----------
2014-02-24 22:00:00  T1          19.937
2014-02-24 22:00:00  T2          19.687
2014-02-24 22:00:00  T3          19.937
2014-02-24 22:00:00  T4          19.937
2014-02-24 22:00:00  T5          19.812
2014-02-24 22:00:00  T6          19.625
2014-02-24 22:00:00  T10         19.687
2014-02-24 22:00:00  T9          19.812
2014-02-24 22:00:00  T8          19.875
2014-02-24 22:00:00  T7          19.812
2014-02-24 22:10:00  T1          19.937
2014-02-24 22:10:00  T2          19.75
2014-02-24 22:10:00  T3          19.937
2014-02-24 22:10:00  T4          19.937
2014-02-24 22:10:00  T5          19.875

select * from PivotTemps10min where timeslot >= '2014-02-24 22:00:00';

TimeSlot             T1          T2              T10
-------------------  ----------  ---------- ...  ----------
2014-02-24 22:00:00  19.937      19.687          19.687
2014-02-24 22:10:00  19.937      19.75           19.687
2014-02-24 22:20:00  19.937      19.75           19.687
2014-02-24 22:30:00  20.125      19.937          19.937
2014-02-24 22:40:00  20.187      20.0            19.937
2014-02-24 22:50:00  20.25       20.062          20.062
2014-02-24 23:00:00  20.25       20.062          20.062

现在，魔术发生在上面提到的案例指令中

CREATE VIEW [PivotTemps10min] AS
SELECT TimeSlot,
AVG(CASE WHEN sensorName = 'T1' THEN temperature END) AS T1,
AVG(CASE WHEN sensorName = 'T2' THEN temperature END) AS T2,
...
AVG(CASE WHEN sensorName = 'T10' THEN temperature END) AS T10
FROM TempsSlot10min
GROUP BY TimeSlot;

例如：

sqlite> .headers on
sqlite> .mode column
sqlite> select * from temps where timestamp > '2014-02-24 22:00:00';

Timestamp            sensorID         temperature
-------------------  ---------------  -----------
2014-02-24 22:00:02  28-0000055f3f10  19.937
2014-02-24 22:00:03  28-0000055f0378  19.687
2014-02-24 22:00:04  28-0000055eb504  19.937
2014-02-24 22:00:05  28-0000055f92f2  19.937
2014-02-24 22:00:06  28-0000055eef29  19.812
2014-02-24 22:00:07  28-0000055f7619  19.625
2014-02-24 22:00:08  28-0000055edf01  19.687
2014-02-24 22:00:09  28-0000055effda  19.812
2014-02-24 22:00:09  28-0000055e5ef2  19.875
2014-02-24 22:00:10  28-0000055f1b83  19.812
2014-02-24 22:10:03  28-0000055f3f10  19.937
2014-02-24 22:10:04  28-0000055f0378  19.75
2014-02-24 22:10:04  28-0000055eb504  19.937
2014-02-24 22:10:05  28-0000055f92f2  19.937

sqlite> select * from TempsSlot10min where timeslot >= '2014-02-24 22:00:00';

TimeSlot             SensorName  temperature
-------------------  ----------  -----------
2014-02-24 22:00:00  T1          19.937
2014-02-24 22:00:00  T2          19.687
2014-02-24 22:00:00  T3          19.937
2014-02-24 22:00:00  T4          19.937
2014-02-24 22:00:00  T5          19.812
2014-02-24 22:00:00  T6          19.625
2014-02-24 22:00:00  T10         19.687
2014-02-24 22:00:00  T9          19.812
2014-02-24 22:00:00  T8          19.875
2014-02-24 22:00:00  T7          19.812
2014-02-24 22:10:00  T1          19.937
2014-02-24 22:10:00  T2          19.75
2014-02-24 22:10:00  T3          19.937
2014-02-24 22:10:00  T4          19.937
2014-02-24 22:10:00  T5          19.875

select * from PivotTemps10min where timeslot >= '2014-02-24 22:00:00';

TimeSlot             T1          T2              T10
-------------------  ----------  ---------- ...  ----------
2014-02-24 22:00:00  19.937      19.687          19.687
2014-02-24 22:10:00  19.937      19.75           19.687
2014-02-24 22:20:00  19.937      19.75           19.687
2014-02-24 22:30:00  20.125      19.937          19.937
2014-02-24 22:40:00  20.187      20.0            19.937
2014-02-24 22:50:00  20.25       20.062          20.062
2014-02-24 23:00:00  20.25       20.062          20.062

这里剩下的唯一问题是传感器名称“T1”…'T10'现在硬编码到视图[PivotTemps10min]中，而不是从查找表中获取

---

尽管如此，非常感谢你在这篇文章中给出的答案

如果您有一个更简单的要求，将同一领域的孩子捆绑在一起，那么group_concat就是您的朋友

非常感谢Simon Slaver在这篇文章中： 多亏了@pospec4444，这里是@haridsv的精彩答案的修改版。它使用

filter

子句更简洁一些

select
    si.studid,
    si.name,
    sum(md.marks) filter(where md.subjectid = 3) subjectid_3,
    sum(md.marks) filter(where md.subjectid = 4) subjectid_4,
    sum(md.marks) filter(where md.subjectid = 5) subjectid_5
from student_info si
join markdetails md on
        md.studid = si.studid
group by si.studid, si.name
;

---

谢谢Eric…查询可以很好地获取学生的全部详细信息。但是我想修改表中的内容和列。我想你没有收到我的问题。我想更改表。@arams:太好了，很高兴听到这个消息！如果它解决了你的问题，请投票/将其标记为答案！第一行可能是打字错误。也许它应该改为

alter table student\u info rename to student\u temp

？@Eric我的应用程序需要更多的列（确切地说是369列；而OP的应用程序需要3个主题）。但是sqlite语句不能执行超过64个联接。您能建议在此处进行编辑以完成64次以上的联接吗？或者有解决办法吗？“现在，放下你的临时表：

drop table student_temp

”，这张

student_temp

temp表是从哪里来的，Eric？我有机会在一张大约150000行的表上进行了排序测试。一个复杂的问题是我事先不知道列的数量，所以我必须做一些预处理来确定需要的列的数量。此外，并非所有行都有相同数量的数据。使用外部连接方法，我的电脑花了50秒。使用case-when方法时，需要15秒。使用Reformae2和plyr的组合（我使用R运行sqlite），大约花费了1040秒。不过，您的里程数可能会有所不同。@Chow，完全同意。我的桌子有280000行，花了大约20秒。这个答案应该在最上面。@haridsv这是一个不错的更好的答案。另外，值得一读：对于任何尝试使用字符串的人，请在pivot

CASE

行中使用

MAX

而不是

SUM

。否则，字符串将被解释为数字，结果可能很奇怪。是否添加了相应的索引？e、 g.（subjectid），（studid）和on（studid，name）：鼓励链接到外部资源，但请在链接周围添加上下文，以便您的其他用户了解它是什么以及为什么存在。总是引用一个重要链接中最相关的部分，以防目标站点无法访问或永久脱机。这正是我搜索的内容。非常感谢。我相信很多使用SQL的物联网爱好者都会提到这一点。我的申请几乎完全一样。