blocks|key|1247644|text|首先，您需要将当前表更改为临时表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1247645|alter+table+student_info+rename+to+student_name|code-block|syntax|javascript|1247646|然后，您需要重新创建student_info|offset|length|style|CODE|1247647|create+table+student_info+add+column+(
++++stuid+VARCHAR(5)+PRIMARY+KEY,
++++name+VARCHAR(255),
++++subjectid_3+INTEGER,
++++subjectid_4+INTEGER,
++++subjectid_5+INTEGER
)|1247648|然后，填充student_info|1247649|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|1247650|现在，删除您的临时表：|1247651|drop+table+student_temp|1247652|这就是您可以快速更新表的方法。|1247653|SQLite缺少pivot函数，所以你能做的最好的事情就是硬编码一些左连接。left+join将匹配其联接条件中的任何行，并为第一个表或左表中不满足第二个表的联接条件的任何行返回null。|1247654|entityMap^0|0|0|A|C|0|0|5|C|0|0|0|0|0|8|5|12|9|2H|4|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@$I|19|J|1A|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1F|8|@$I|1G|J|1H|K|L]|$I|1I|J|1J|K|L]|$I|1K|J|1L|K|L]]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|11|$]]

First you need to change the current table to a temp table:

<pre><code>alter table student_info rename to student_name
</code></pre>

Then, you'll want to recreate <code>student_info</code>:

<pre><code>create table student_info add column (
 stuid VARCHAR(5) PRIMARY KEY,
 name VARCHAR(255),
 subjectid_3 INTEGER,
 subjectid_4 INTEGER,
 subjectid_5 INTEGER
)
</code></pre>

Then, populate <code>student_info</code>:

<pre><code>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
</code></pre>

Now, just drop your temp table:

<pre><code>drop table student_temp
</code></pre>

And that's how you can quickly update your table.

SQLite lacks a <code>pivot</code> function, so the best you can do is hard-code some left joins. A <code>left join</code> will bring match any rows in its join conditions and return <code>null</code> for any rows from the first, or left, table that don't meet the join conditions for the second table.

blocks|key|187228|text|由于作者没有给出SQL来创建模式，所以这里是为任何想要尝试@Eric解决方案的人准备的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|187229|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');|code-block|syntax|javascript|187230|这是一个使用case和group+by的替代解决方案。|offset|length|style|CODE|187231|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
;|187232|为了进行比较，下面是@Eric的解决方案中相同的select语句：|187233|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
;|187234|当有大量数据时，哪一个会表现得更好，这将是一件有趣的事情。|187235|entityMap^0|0|0|6|4|B|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

Since the author was not kind enough to give the SQL to create the schema, here it is for anyone who wants to try the solution from @Eric.

<pre><code>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');
</code></pre>

Here is an alternative solution using <code>case</code> with <code>group by</code>. 

<pre><code>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
;
</code></pre>

For comparison, here is the same select statement from @Eric's solution:

<pre><code>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
;
</code></pre>

It will be interesting to see which one would perform better when there is a lot of data.

blocks|key|4464536|text|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1wire-interface+DS18B20温度传感器数据，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4464537|CREATE+TABLE+temps+(Timestamp+DATETIME,+sensorID+TEXT,+temperature+NUMERIC);|code-block|syntax|javascript|4464538|示例：|4464539|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|4464540|使用SUBSTR()命令，我将时间戳“规范化”为10分钟周期。使用JOIN，使用查找表‘SensorName’将传感器更改为传感器|4464541|CREATE+VIEW+[TempsSlot10min]+AS
SELECT+SUBSTR(datetime(timestamp),1,15)%7C%7C'0:00'+AS+TimeSlot,
SensorName,
temperature+FROM
temps+JOIN+sensors+USING+(sensorID,+sensorID);|4464542|4464543|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|4464544|现在，魔术发生在上面提到的案例指令中。|4464545|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;|4464546|4464547|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|4464548|这里剩下的唯一问题是sensorName+'T1‘...'T10‘现在被硬编码到视图PivotTemps10min中，而不是从查找表中获取。|4464549|尽管如此，非常感谢你在这篇文章中的回答！|4464550|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|18|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|O|3|H|5|6|7|1A|8|@]|9|@]|A|$]]|$1|P|3|Q|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|R|3|S|5|6|7|1C|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|V|3|H|5|6|7|1E|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1G|8|@]|9|@]|A|$]]|$1|10|3|11|5|6|7|1H|8|@]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]]|13|$]]

great appendix! helped me to solve a similar problem with low effort and system load. I am using a Raspberry Pi to obtain 1wire-interface DS18B20 temperature sensor data as follows:

<pre><code>CREATE TABLE temps (Timestamp DATETIME, sensorID TEXT, temperature NUMERIC);
</code></pre>

example:

<pre><code>sqlite&gt; .headers on
sqlite&gt; .mode column
sqlite&gt; select * from temps where timestamp &gt; '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
</code></pre>

using the SUBSTR() command I am "normalizing" the Timestamps to 10 minutes periods. With JOIN the sensorID is changed into a SensorName using the lookup-table 'sensors'

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

example:

<pre><code>sqlite&gt; select * from TempsSlot10min where timeslot &gt;= '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
</code></pre>

now, the magic happens with the above mentioned CASE instruction. 

<pre><code>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;
</code></pre>

example:

<pre><code>select * from PivotTemps10min where timeslot &gt;= '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
</code></pre>

The only problem remaining here is that the sensorName 'T1' ... 'T10' is now hardcoded into the VIEW [PivotTemps10min] and not taken from the lookup table.

Nonetheless, thank you very much for the answers in this thead!

blocks|key|187264|text|如果您有一个更简单的需求，即将同一领域中的孩子捆绑在一起，那么group_concat将是您的好朋友。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|187265|非常感谢这个帖子中的Simon+Slaver：http://sqlite.1065341.n5.nabble.com/Howto-pivot-in-SQLite-tp26766p26771.html|offset|length|187266|entityMap|0|LINK|mutability|MUTABLE|url|http://sqlite.1065341.n5.nabble.com/Howto-pivot-in-SQLite-tp26766p26771.html^0|0|N|24|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

If you have a simpler requirement of bundling together the children in the same field, group_concat is your friend.

Huge thanks to Simon Slaver from this thread:
<a href="http://sqlite.1065341.n5.nabble.com/Howto-pivot-in-SQLite-tp26766p26771.html" rel="nofollow">http://sqlite.1065341.n5.nabble.com/Howto-pivot-in-SQLite-tp26766p26771.html</a>

blocks|key|187533|text|感谢@posecec4444的link，这里是@haridsv令人惊叹的答案的修改版本。它使用filter子句使其更简洁。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|187534|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
;|code-block|syntax|javascript|187535|entityMap|0|LINK|mutability|MUTABLE|url|https://modern-sql.com/use-case/pivot^0|1B|6|F|4|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Thanks to @pospec4444's <a href="https://modern-sql.com/use-case/pivot" rel="noreferrer">link</a> here is modified version of @haridsv's awesome answer. It uses <code>filter</code> clause to be little more concise

<pre><code>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
;
</code></pre>

I'd like to get a table which stores students data in <a href="https://en.wikipedia.org/wiki/Wide_and_narrow_data#Narrow" rel="noreferrer">long format</a> and the marks they receive for all of their subjects in one query.

This is my table structure:

Table: <code>markdetails</code>

<pre><code>## studid ## ## subjectid ## ## marks ##
 A1 3 50
 A1 4 60
 A1 5 70
 B1 3 60
 B1 4 80
 C1 5 95
</code></pre>

Table: <code>student info</code>

Actual Structure:

<pre><code>## studid ## ## name ##
 A1 Raam
 B1 Vivek
 c1 Alex
</code></pre>

I want the result set to have the following <a href="https://en.wikipedia.org/wiki/Wide_and_narrow_data#Wide" rel="noreferrer">wide format</a> structure as result of the pivotization:

Table: <code>Student Info</code>

<pre><code>## studid ## ## name## ## subjectid_3 ## ## subjectid_4 ## ## subjectid_5 ##
 A1 Raam 50 60 70
 B1 Vivek 60 80 null
 c1 Alex null null 95
</code></pre>

How can I accomplish this in SQLite?

How to pivot in SQLite or i.e. select in wide format a table stored in long format?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想要一个表，它以格式存储学生数据，并在一个查询中存储他们所有科目的分数。这是我的表结构：表：markdetails## studid ## ## subjectid ##  ## marks ##     A1            3                50     A1            4   ...

问如何在SQLite中透视或者以宽格式选择以长格式存储的表？
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问如何在SQLite中透视或者以宽格式选择以长格式存储的表？EN