blocks|key|2417563|text|您将得到2000，因为当您将table2中的一行连接到table1中的两行时，t2中的单行会复制自身以匹配t1中的两行。如果您随后将添加的重复行求和为额外的500|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2417564|解决这些问题的最简单方法是在加入它们之前对它们进行分组和求和：|2417565|SELECT+t1.sum+%2B+t2.sum+as+total
FROM
++(SELECT+pr_id,+sum(cl_one)+as+sum+FROM+table1+GROUP+BY+pr_id)+t1++++
++INNER+JOIN+
++(SELECT+pr_id,+sum(cl_2)+as+sum+FROM+table2+GROUP+BY+pr_id)+t2++++
++ON+t1.pr_id+=+t2.pr_id|code-block|syntax|javascript|2417566|这样，t1中的一行与t2中的一行匹配，没有重复...但是我不知道你如何在codeigniter/不管你正在使用的ORM中如何做到这一点，但我希望这能帮助/很容易地被翻译|2417567|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<p>You're getting 2000 because when you join one row in table2, to two rows in table1, the single row from t2 duplicates itself to match up to the two rows from t1. If you then sum the duplicated row you've added in an extra 500</p>

<p>The easiest way to resolve these things is to group and sum them before you join them:</p>

<pre><code>SELECT t1.sum + t2.sum as total
FROM
  (SELECT pr_id, sum(cl_one) as sum FROM table1 GROUP BY pr_id) t1    
  INNER JOIN 
  (SELECT pr_id, sum(cl_2) as sum FROM table2 GROUP BY pr_id) t2    
  ON t1.pr_id = t2.pr_id
</code></pre>

<p>This way one row from t1 matches one row from t2, no duplicating...
But I've no idea how you'd do this in codeigniter/whatever ORM you're using, but I hope this can help/be easily translated</p>


blocks|key|507250|text|终于成功了……|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|507251|修复此的简单查询|507252|SELECT+pr_id,+SUM(sum_v)+AS+total_amount,+COUNT(sum_v)+AS+total_counts+FROM(+
SELECT+pr_id,+cl_one+AS+sum_v+FROM+table1++UNION+ALL+
SELECT+pr_id,+cl2+AS+sum_v+FROM+table2)x+GROUP+BY+pr_id|code-block|syntax|javascript|507253|entityMap^0|0|7|0|0|3|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

<p><strong>Finally made it...</strong></p>

<p><strong>A simple query to fix this</strong> </p>

<pre><code>SELECT pr_id, SUM(sum_v) AS total_amount, COUNT(sum_v) AS total_counts FROM( 
SELECT pr_id, cl_one AS sum_v FROM table1  UNION ALL 
SELECT pr_id, cl2 AS sum_v FROM table2)x GROUP BY pr_id
</code></pre>


blocks|key|2419429|text|Codeigniter查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2419430|$res1+=+$this->db->select('SUM(cl_one)+as+total,+COUNT(cl_one)+as+count').from('Table1').get().row();
$res2+=+$this->db->select('SUM(cl_2)+as+total,+COUNT(cl_2)+as+count').from('Table2').get().row();
$total_amount+=+$res1[0]['total']+%2B+$res2[0]['total'];
$total_count+=+$res1[0]['count']+%2B+$res2[0]['count'];|code-block|syntax|javascript|2419431|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|$]]

<p>Codeigniter query: </p>

<pre><code>$res1 = $this-&gt;db-&gt;select('SUM(cl_one) as total, COUNT(cl_one) as count').from('Table1').get().row();
$res2 = $this-&gt;db-&gt;select('SUM(cl_2) as total, COUNT(cl_2) as count').from('Table2').get().row();
$total_amount = $res1[0]['total'] + $res2[0]['total'];
$total_count = $res1[0]['count'] + $res2[0]['count'];
</code></pre>


<p>I want to find the sum of this two tables</p>

<p>Please help me to fix this issue</p>

<h1>Table 1</h1>

<pre><code>id  |  pr_id  |  cl_one
========================
 1  |    2    |   500
------------------------
 2  |    2    |   500
</code></pre>

<h1>Table 2</h1>

<pre><code>id  |  pr_id  |   cl2
======================
 1  |    2    |   500
</code></pre>

<h1>codeigniter query</h1>

<pre><code>$this-&gt;db-&gt;select('SUM(s.cl_one)+SUM(l.cl2) AS total_amount, COUNT(s.cl_one)+SUM(l.cl2) AS total_counts')
                      -&gt;from('Table1 AS s')
                      -&gt;join('Table2 AS l', 's.pr_id = l.pr_id')
                      -&gt;get()-&gt;row();
</code></pre>

<h1>Result</h1>

<pre><code>total_amount = 2000;

total_counts = 4;
</code></pre>

<h1>I need a result like this</h1>

<pre><code>total_amount = 1500;

total_count = 3;
</code></pre>


I want to find the sum of this two tables

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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我想找出这两张表的总和请帮我解决这个问题表1id  |  pr_id  |  cl_one======================== 1  |    2    |   500------------------------ 2  |    2    |   500表2id  |  pr_id  |   cl2==...

问我想找出这两张表的总和
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问我想找出这两张表的总和EN