blocks|key|2915995|text|也许我不明白你问的问题有多复杂，但是...这不是应该做的吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2915996|SELECT+groupname,+SUM(value)
FROM+items
WHERE+groupname+IN+('a',+'b')
GROUP+BY+groupname|code-block|syntax|javascript|2915997|如果你不关心这个项目属于a或b中的哪一个，那么这就行了：|2915998|SELECT+SUM(value)
FROM+items
WHERE+groupname+IN+('a',+'b')|2915999|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Maybe I'm not understanding the complexity of what you're asking but... shouldn't this do?

<pre><code>SELECT groupname, SUM(value)
FROM items
WHERE groupname IN ('a', 'b')
GROUP BY groupname
</code></pre>

And if you don't care which of a or b the item belongs to then this will do:

<pre><code>SELECT SUM(value)
FROM items
WHERE groupname IN ('a', 'b')
</code></pre>

blocks|key|715527|text|你想要像这样的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|715528|SELECT+column,SUM(+column+)+FROM+table+GROUP+BY+column|code-block|syntax|javascript|715529|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You want something like

<pre><code>SELECT column,SUM( column ) FROM table GROUP BY column
</code></pre>

blocks|key|2916063|text|这就是(泰勒的答案)这个问题的意思，或者仅仅是这样；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2916064|SELECT+sum(item),+groupingField+FROM+someTable+GROUP+BY+groupingField|code-block|syntax|javascript|2916065|或者甚至是：|2916066|SELECT+count(*),+item+FROM+someTable+GROUP+BY+item|2916067|这将产生如下结果：|2916068|sum(item)++++%7C+++++groupingField
-------------%2B-----------------------
+71++++++++++%7C++++++++++A
+82++++++++++%7C++++++++++B|2916069|发问者，也许你能澄清你的意思，或者我是否过于简单化了？|2916070|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Is that (Tyler's answer) what the question meant, or is it simply this;

<pre><code>SELECT sum(item), groupingField FROM someTable GROUP BY groupingField
</code></pre>

or even:

<pre><code>SELECT count(*), item FROM someTable GROUP BY item
</code></pre>

which will produce results like this:

<pre><code>sum(item) | groupingField
-------------+-----------------------
 71 | A
 82 | B
</code></pre>

Questioner, perhaps you could clarify which you meant or if I'm oversimplifying?

blocks|key|3760474|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3760475|SELECT+B.[Group],+COUNT(*)+AS+GroupCount
FROM+Table1+A
LEFT+JOIN+Table2+B+ON+B.ItemType=A.ItemType
GROUP+BY+B.[Group]|code-block|syntax|javascript|3760476|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this: 

<pre><code>SELECT B.[Group], COUNT(*) AS GroupCount
FROM Table1 A
LEFT JOIN Table2 B ON B.ItemType=A.ItemType
GROUP BY B.[Group]
</code></pre>

blocks|key|710708|text|请参考此图像(1)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|710709|SELECT+category_id,SUM(amount)+AS+count++FROM+expense+GROUP+BY+category_id;|code-block|syntax|javascript|710710|entityMap|0|LINK|mutability|MUTABLE|url|https://i.stack.imgur.com/5R9ou.png^0|7|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Please refer this Image (<a href="https://i.stack.imgur.com/5R9ou.png" rel="nofollow noreferrer">1</a>)

<pre><code>SELECT category_id,SUM(amount) AS count FROM expense GROUP BY category_id;
</code></pre>

I have one field that I need to sum lets say named items
However that field can be part of group a or b
In the end I need to have all of the items summed for group a and group b

when I say grouped I mean there is a LEFT OUTER JOIN to another table the previous table has a type for the items and the one being joined has a group assigned for this item type

Sorry guys Im a little new to sql I am going to try out what you have given me an get back to you

Ok I feel like we are getting close just not yet allain's I can get them to separate but the issue I need to have both groups to sum on the same row which is difficult because I also have several LEFT OUTER JOIN's involved

Tyler's looks like it might work too so I am trying to hash that out real fast

Alain's seems to be the way to go but I have to tweek it a little more

Returning the sum of items depending on which type it is

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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