blocks|key|3165665|text|我想这就是你要找的。NEW_BAL是从余额中减去QTY的和：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3165666|SELECT+++master_table.ORDERNO,
+++++++++master_table.ITEM,
+++++++++SUM(master_table.QTY),
+++++++++stock_bal.BAL_QTY,
+++++++++(stock_bal.BAL_QTY+-+SUM(master_table.QTY))+AS+NEW_BAL
FROM+++++master_table+INNER+JOIN
+++++++++stock_bal+ON+master_bal.ITEM+=+stock_bal.ITEM
GROUP+BY+master_table.ORDERNO,
+++++++++master_table.ITEM|code-block|syntax|javascript|3165667|如果要使用新余额更新物料余额，请使用以下方法：|3165668|UPDATE+stock_bal
SET++++BAL_QTY+=+BAL_QTY+-+(SELECT+++SUM(QTY)
++++++++++++++++++++++++++++FROM+++++master_table
++++++++++++++++++++++++++++GROUP+BY+master_table.ORDERNO,
+++++++++++++++++++++++++++++++++++++master_table.ITEM)|3165669|这假设您向后提交了减法；它从余额中减去订单中的数量，这在不了解表的更多信息的情况下最有意义。如果我错了，只需交换这两个来更改它：|3165670|(SUM(master_table.QTY)+-+stock_bal.BAL_QTY)+AS+NEW_BAL|3165671|entityMap^0|A|7|O|3|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

I think this is what you're looking for. <code>NEW_BAL</code> is the sum of <code>QTY</code>s subtracted from the balance:

<pre><code>SELECT master_table.ORDERNO,
 master_table.ITEM,
 SUM(master_table.QTY),
 stock_bal.BAL_QTY,
 (stock_bal.BAL_QTY - SUM(master_table.QTY)) AS NEW_BAL
FROM master_table INNER JOIN
 stock_bal ON master_bal.ITEM = stock_bal.ITEM
GROUP BY master_table.ORDERNO,
 master_table.ITEM
</code></pre>

If you want to update the item balance with the new balance, use the following:

<pre><code>UPDATE stock_bal
SET BAL_QTY = BAL_QTY - (SELECT SUM(QTY)
 FROM master_table
 GROUP BY master_table.ORDERNO,
 master_table.ITEM)
</code></pre>

This assumes you posted the subtraction backward; it subtracts the quantities in the order from the balance, which makes the most sense without knowing more about your tables. Just swap those two to change it if I was wrong:

<pre><code>(SUM(master_table.QTY) - stock_bal.BAL_QTY) AS NEW_BAL
</code></pre>

blocks|key|2951920|text|使用子查询的简单复制和粘贴示例，请注意，两个查询都应返回一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2951921|select
(select+sum(items_1)+from+items_table_1+where+...)
-
(select+count(items_2)+from+items_table_1+where+...)+

as+difference|code-block|syntax|javascript|2951922|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|$]]

Simple copy &amp; paste example with subqueries,
Note, that both queries should return 1 row:

<pre><code>select
(select sum(items_1) from items_table_1 where ...)
-
(select count(items_2) from items_table_1 where ...) 

as difference
</code></pre>

blocks|key|259018|text|我不确定你到底想要什么，但我想大概是这样的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|259019|SELECT+`Item`,+`qty`-`BAL_QTY`+as+`qty`+FROM+((SELECT+Item,+SUM(`QTY`)+as+qty+FROM+`master_table`+GROUP+BY+`ITEM`)+as+A+NATURAL+JOIN+`stock_table`)+as+B|code-block|syntax|javascript|259020|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|$]]

I'm not sure exactly what you want, but I think it's along the lines of:

<pre><code>SELECT `Item`, `qty`-`BAL_QTY` as `qty` FROM ((SELECT Item, SUM(`QTY`) as qty FROM `master_table` GROUP BY `ITEM`) as A NATURAL JOIN `stock_table`) as B
</code></pre>

blocks|key|261291|text|啊作业。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|261292|因此，等等，您需要从已订购的项目总数中扣除库存项目的余额？我必须告诉你，这听起来有点倒退。一般来说，我认为人们会以相反的方式做这件事。从余额中扣除订购的商品总数。|261293|如果你真的需要这样做的话...假设该项在stock_bal中是唯一的...|261294|SELECT+s.ITEM,+SUM(m.QTY)+-+s.QTY+AS+result
FROM+stock_bal+s
INNER+JOIN+master_table+m+ON+m.ITEM+=+s.ITEM
GROUP+BY+s.ITEM,+s.QTY|code-block|syntax|javascript|261295|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

ah homework...

So wait, you need to deduct the balance of items in stock from the total number of those items that have been ordered? I have to tell you that sounds a bit backwards. Generally I think people do it the other way round. Deduct the total number of items ordered from the balance.

If you really need to do that though...
Assuming that ITEM is unique in stock_bal...

<pre><code>SELECT s.ITEM, SUM(m.QTY) - s.QTY AS result
FROM stock_bal s
INNER JOIN master_table m ON m.ITEM = s.ITEM
GROUP BY s.ITEM, s.QTY
</code></pre>

blocks|key|261306|text|下面给出一个减法的例子：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|261307|Select+value1+-+(select+value2+from+AnyTable1)+from+AnyTable2|code-block|syntax|javascript|261308|value1+&+value2可以是count、sum、average+output等，但数值必须是相容的|offset|length|style|BOLD|261309|entityMap^0|0|0|0|1H|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

An example for subtraction is given below:

<pre><code>Select value1 - (select value2 from AnyTable1) from AnyTable2
</code></pre>

value1 &amp; value2 can be count,sum,average output etc. But the values should be comapatible

blocks|key|3165691|text|我已经尝试过这种技术。将从数据中减去的值乘以(-1)，然后求和()，得到减去的量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3165692|--+Loan+Outstanding
++++select++'Loan+Outstanding'+as+Particular,+sum(Unit),+sum(UptoLastYear),+sum(ThisYear),+sum(UptoThisYear)
++++from
++++(
++++++++select+
++++++++++++sum(laod.dr)+as+Unit,
++++++++++++sum(if(lao.created_at+<=+'2014-01-01',laod.dr,0))+as+UptoLastYear,
++++++++++++sum(if(lao.created_at+between+'2014-01-01'+and+'2015-07-14',laod.dr,0))+as+ThisYear,
++++++++++++sum(if(lao.created_at+<=+'2015-07-14',laod.dr,0))+as+UptoThisYear
++++++++from+loan_account_opening+as+lao
++++++++inner+join+loan_account_opening_detail+as+laod+on+lao.id=laod.loan_account_opening_id
++++++++where+lao.organization+=+3
++++++++union
++++++++select
++++++++++++sum(lr.installment)*-1+as+Unit,
++++++++++++sum(if(lr.created_at+<=+'2014-01-01',lr.installment,0))*-1+as+UptoLastYear,
++++++++++++sum(if(lr.created_at+between+'2014-01-01'+and+'2015-07-14',lr.installment,0))*-1+as+ThisYear,
++++++++++++sum(if(lr.created_at+<=+'2015-07-14',lr.installment,0))*-1+as+UptoThisYear
++++++++from+loan_recovery+as+lr
++++++++inner+join+loan_account_opening+as+lo+on+lr.loan_account_opening_id=lo.id
++++++++where+lo.organization+=+3
++++)+as+t3|code-block|syntax|javascript|3165693|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|$]]

I have tried this kind of technique. Multiply the subtract from data by (-1) and then sum() the both amount then you will get subtracted amount.

<pre><code>-- Loan Outstanding
 select 'Loan Outstanding' as Particular, sum(Unit), sum(UptoLastYear), sum(ThisYear), sum(UptoThisYear)
 from
 (
 select 
 sum(laod.dr) as Unit,
 sum(if(lao.created_at &lt;= '2014-01-01',laod.dr,0)) as UptoLastYear,
 sum(if(lao.created_at between '2014-01-01' and '2015-07-14',laod.dr,0)) as ThisYear,
 sum(if(lao.created_at &lt;= '2015-07-14',laod.dr,0)) as UptoThisYear
 from loan_account_opening as lao
 inner join loan_account_opening_detail as laod on lao.id=laod.loan_account_opening_id
 where lao.organization = 3
 union
 select
 sum(lr.installment)*-1 as Unit,
 sum(if(lr.created_at &lt;= '2014-01-01',lr.installment,0))*-1 as UptoLastYear,
 sum(if(lr.created_at between '2014-01-01' and '2015-07-14',lr.installment,0))*-1 as ThisYear,
 sum(if(lr.created_at &lt;= '2015-07-14',lr.installment,0))*-1 as UptoThisYear
 from loan_recovery as lr
 inner join loan_account_opening as lo on lr.loan_account_opening_id=lo.id
 where lo.organization = 3
 ) as t3
</code></pre>

I am using MySQL and I have two tables:

<code>master_table</code>

<ul>
<li>ORDERNO </li>
<li>ITEM</li>
<li>QTY</li>
</ul>

<code>stock_bal</code> 

<ul>
<li>ITEM </li>
<li>BAL_QTY</li>
</ul>

Master table has duplicate <code>ORDERNO</code> and <code>ITEM</code> values. I have get total <code>QTY</code> using SQL 'GROUP BY' clause.

I need to deduct/subtract <code>BAL_QTY</code> from SUM of <code>ITEM</code> (master_table). I've got SUM <code>QTY</code> value using query (actually there are many rows).

How to SUM and SUBTRACT using SQL?

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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我使用的是MySQL，我有两个表：master_tableITEMQTY ORDERNOstock_bal项目BAL_QTY主表具有重复的ORDERNO和ITEM值。我使用SQL 'GROUP BY‘子句获得了总QTY。我需要从ITEM的和(master_table)中减去BAL_QTY。我使用query得到了SUM QTY的值(实际上有很多行)。

问如何使用SQL进行加法和减法？
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