44

我想要一张表格，以长格式存储学生数据以及他们在一个查询中为所有科目获得的分数。

这是我的表结构：

桌子：markdetails

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

桌子：student info

实际结构：

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

作为旋转的结果，我希望结果集具有以下宽格式结构：

桌子：Student Info

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

我怎样才能在 SQLite 中做到这一点？

4

5 回答 5

39

由于作者不太友好地提供 SQL 来创建模式，因此这里适用于任何想要尝试来自 @Eric 的解决方案的人。

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');

这是使用casewith的替代解决方案group by。

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
;

为了比较，这里是来自@Eric 解决方案的相同选择语句：

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
;

当有大量数据时，看看哪一个表现更好将会很有趣。

于 2011-11-24T02:52:18.043 回答
28

首先，您需要将当前表更改为临时表：

alter table student_info rename to student_name

然后，您需要重新创建student_info：

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

然后，填充student_info：

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，只需删除您的临时表：

drop table student_temp

这就是您可以快速更新表格的方式。

SQLite 缺少一个pivot函数，所以你能做的最好的就是硬编码一些左连接。Aleft join将匹配其连接条件中null的任何行，并返回第一个或左表中不满足第二个表的连接条件的任何行。

于 2009-08-06T05:14:15.567 回答
11

很棒的附录！帮助我以低工作量和系统负载解决了类似的问题。我正在使用 Raspberry Pi 获取 1wire-interface DS18B20 温度传感器数据，如下所示：

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

例子：

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

使用 SUBSTR() 命令我将时间戳“标准化”为 10 分钟。通过 JOIN，sensorID 使用查找表“sensors”更改为 SensorName

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

例子：

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

现在，神奇的事情发生在上面提到的 CASE 指令上。

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;

例子：

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 sensorName 'T1' ... 'T10' 现在被硬编码到 VIEW [PivotTemps10min] 中，而不是从查找表中获取。

尽管如此，非常感谢您在此主题中的答案！

于 2014-02-24T21:52:49.170 回答
5

感谢@pospec4444 的链接是@haridsv 很棒的答案的修改版本。它使用filter子句更加简洁

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
;
于 2019-12-31T05:19:21.490 回答
0

如果您有更简单的要求将同一字段中的孩子捆绑在一起，那么 group_concat 是您的朋友。

非常感谢来自此线程的 Simon Slaver：http: //sqlite.1065341.n5.nabble.com/Howto-pivot-in-SQLite-tp26766p26771.html

于 2016-08-26T16:18:07.973 回答