blocks|key|287099|text|select+month,+sum(count)+from+table+group+by+rollup(month,+count);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|287100|如果你想避免使用month+is+not+null，它也会给出净合计。|unstyled|287101|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>select month, sum(count) from table group by rollup(month, count);
</code></pre>

It will give net total as well, if you want to avoid it use month is not null.

blocks|key|3161417|text|其中一种方法是使用rollup。但是，因为您需要原始数据，所以此方法通过rownum对汇总执行“假”聚合：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3161418|select+Month,+sum(Count),
+++++++(case+when+rn+is+null+then+'SUM'+end)+as+status
from+(select+rownum+as+rn,+t.*
++++++from+t
+++++)+t
group+by+Month,+rollup(rn)|code-block|syntax|javascript|3161419|我不确定您在status列中提到的A和B是什么意思。|3161420|entityMap^0|9|6|0|0|H|1|J|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

One way to do this is with <code>rollup</code>. However, because you want the original data, this approach does a "fake" aggregation by rownum for the rollup:

<pre><code>select Month, sum(Count),
 (case when rn is null then 'SUM' end) as status
from (select rownum as rn, t.*
 from t
 ) t
group by Month, rollup(rn)
</code></pre>

I'm not sure what you mean by the <code>A</code> and <code>B</code> in the status column.

blocks|key|3161449|text|原始表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3161450|CREATE+TABLE+test
(
++++month+CHAR(3),+
++++num+INT,
++++status+CHAR(3)
);

INSERT+INTO+test+(month,+num,+status)+VALUES
++++('Jan',+5,+'A'),
++++('Jan',+7,+'B'),
++++('Feb',+1,+'A'),
++++('Feb',+2,+'B');|code-block|syntax|javascript|3161451|查询：|3161452|SELECT+month,+sum(num)+AS+num,+'SUM'+AS+status
FROM+test
GROUP+BY+month
UNION
SELECT+month,+num,+status
FROM+test
ORDER+BY+month+DESC,+status|3161453|结果：|3161454|MONTH+++NUM+++STATUS
Jan+++++5+++++A
Jan+++++7+++++B
Jan+++++12++++SUM
Feb+++++1+++++A
Feb+++++2+++++B
Feb+++++3+++++SUM|3161455|Here是SQLfiddle。|offset|length|3161456|entityMap|0|LINK|mutability|MUTABLE|url|http://sqlfiddle.com/#!6/8980e/2^0|0|0|0|0|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@$Q|17|R|18|1|19]]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

The original table:

<pre><code>CREATE TABLE test
(
 month CHAR(3), 
 num INT,
 status CHAR(3)
);

INSERT INTO test (month, num, status) VALUES
 ('Jan', 5, 'A'),
 ('Jan', 7, 'B'),
 ('Feb', 1, 'A'),
 ('Feb', 2, 'B');
</code></pre>

The query:

<pre><code>SELECT month, sum(num) AS num, 'SUM' AS status
FROM test
GROUP BY month
UNION
SELECT month, num, status
FROM test
ORDER BY month DESC, status
</code></pre>

Result:

<pre><code>MONTH NUM STATUS
Jan 5 A
Jan 7 B
Jan 12 SUM
Feb 1 A
Feb 2 B
Feb 3 SUM
</code></pre>

<a href="http://sqlfiddle.com/#!6/8980e/2" rel="nofollow">Here</a> is the SQLfiddle.

blocks|key|2201441|text|Oracle+have+MODEL子句:如果您的表中有status列，则只需执行简单的查询：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2201442|Model+Clause+Link+Oracle+Documentation:|2201443|SELECT+month,+count+,+status
FROM+my_test2
MODEL
PARTITION+BY+(MONTH)
DIMENSION+BY+(status)
MEASURES(count)
RULES(COUNT['Sum']+=+count['A']%2B+count['B'])
order+by+month+desc,+status;


MONTH+++++++++++++++++++++COUNT+STATUS+++++++++++++
--------------------+----------+--------------------
Jan+++++++++++++++++++++++++++5+A++++++++++++++++++++
Jan+++++++++++++++++++++++++++7+B++++++++++++++++++++
Jan++++++++++++++++++++++++++12+Sum++++++++++++++++++
Feb+++++++++++++++++++++++++++1+A++++++++++++++++++++
Feb+++++++++++++++++++++++++++2+B++++++++++++++++++++
Feb+++++++++++++++++++++++++++3+Sum++++++++++++++++++

+6+rows+selected+|code-block|syntax|javascript|2201444|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/cd/B28359_01/server.111/b28313/sqlmodel.htm^0|C|5|0|0|13|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@$9|Y|A|Z|1|10]]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Oracle have <code>MODEL</code> clause for the above:
If you've the column status in your table, you have simple query:

<a href="http://docs.oracle.com/cd/B28359_01/server.111/b28313/sqlmodel.htm" rel="nofollow">Model Clause Link Oracle Documentation:</a>

<pre><code>SELECT month, count , status
FROM my_test2
MODEL
PARTITION BY (MONTH)
DIMENSION BY (status)
MEASURES(count)
RULES(COUNT['Sum'] = count['A']+ count['B'])
order by month desc, status;


MONTH COUNT STATUS 
-------------------- ---------- --------------------
Jan 5 A 
Jan 7 B 
Jan 12 Sum 
Feb 1 A 
Feb 2 B 
Feb 3 Sum 

 6 rows selected 
</code></pre>

I want to sum of values of one column group by another column.
e.g

<pre><code>Month Count
Jan 5
Jan 7
Feb 1
Feb 2
</code></pre>

I want

<pre><code>Month Count Status
Jan 5 A
Jan 7 B
Jan 12 SUM
Feb 1 A
Feb 2 B
Feb 3 SUM
</code></pre>

Getting sum of rows based on column in oracle

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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我想将一列的值按另一列分组求和。e.gMonth     CountJan         5Jan         7Feb         1Feb         2我想要Month     Count   StatusJan         5       AJan         7       BJan  ...

问根据oracle中的列获取行的总和
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问根据oracle中的列获取行的总和EN