entityMap|blocks|key|5ihn5|text|使查询2成为子查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|dttl5|++++SELECT+ap.PLN_ID
+++++++++,+ap.ASSORTMENT_GROUP
+++++++++,+ap.STORE
+++++++++,+ap.PLANOGRAM
+++++++++,+ap.STATUS
+++++++++,+sq.cnt
+++++++++,+sq.width+
++++++FROM+ACN_PLANOGRAMS++ap
++++++JOIN+(
++++++++++++++++++SELECT+PLN_ID
+++++++++++++++++++++++,+count(*)++++++++++AS+cnt
+++++++++++++++++++++++,+sum(WIDTH)++++++++AS+width
++++++++++++++++++++FROM+ACN_FIXEL+
+++++++++++++++++++WHERE+type='0'
++++++++++++++++GROUP+BY+PLN_ID
++++++++++++++++++HAVING+count(*)+>+1
+++++++++++)+sq
++++++++ON+(+sq.PLN_ID+=+ap.PLN_ID+)
+++++WHERE+ap.PLANOGRAM+not+like+'%25<Untitled>%25'
+++++++++;|code-block|syntax|javascript|f7vj7^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Make Query 2 a subquery:

<pre><code> SELECT ap.PLN_ID
 , ap.ASSORTMENT_GROUP
 , ap.STORE
 , ap.PLANOGRAM
 , ap.STATUS
 , sq.cnt
 , sq.width 
 FROM ACN_PLANOGRAMS ap
 JOIN (
 SELECT PLN_ID
 , count(*) AS cnt
 , sum(WIDTH) AS width
 FROM ACN_FIXEL 
 WHERE type='0'
 GROUP BY PLN_ID
 HAVING count(*) &gt; 1
 ) sq
 ON ( sq.PLN_ID = ap.PLN_ID )
 WHERE ap.PLANOGRAM not like '%&lt;Untitled&gt;%'
 ;
</code></pre>

entityMap|blocks|key|frjel|text|有几种方法可以解决这个问题。在不了解您的数据模型或业务逻辑的情况下，我提供了一个最简单的解决方案，即派生表(内联视图)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2qvkh|SELECT+p.PLN_ID
+++++,+p.ASSORTMENT_GROUP
+++++,+p.STORE
+++++,+p.PLANOGRAM
+++++,+p.STATUS+
+++++,+f.fixel_count
+++++,+f.fixel_width
++FROM+ACN_PLANOGRAMS+p
+++++++inner+join+(SELECT+PLN_ID
++++++++++++++++++++++++++,+count(*)+as+fixel_count
++++++++++++++++++++++++++,+(sum(WIDTH))++++++AS+fixel_width
++++++++++++++++++++FROM+ACN_FIXEL+
++++++++++++++++++++WHERE+type='0'
++++++++++++++++++++GROUP+BY+PLN_ID
++++++++++++++++++++HAVING+count(*)+>+1+)+f+
+++++++on+f.pln_id+=+p.pln_id
+WHERE+p.PLANOGRAM+not+like+'%25<Untitled>%25'
+++++;++|code-block|syntax|javascript|c7tpg|此解决方案仅在两个结果集中返回PLN_ID的结果。如果您有不同的逻辑，则可能需要使用左外部连接。|1tun7^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|K|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|L|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|M|9|@]|A|@]|B|$]]|$2|J|4|-4|6|7|8|N|9|@]|A|@]|B|$]]]]

There are several ways to solve this. Without understanding your data model or business logic I offer the simplest solution, a derived table (inline view):

<pre><code>SELECT p.PLN_ID
 , p.ASSORTMENT_GROUP
 , p.STORE
 , p.PLANOGRAM
 , p.STATUS 
 , f.fixel_count
 , f.fixel_width
 FROM ACN_PLANOGRAMS p
 inner join (SELECT PLN_ID
 , count(*) as fixel_count
 , (sum(WIDTH)) AS fixel_width
 FROM ACN_FIXEL 
 WHERE type='0'
 GROUP BY PLN_ID
 HAVING count(*) &gt; 1 ) f 
 on f.pln_id = p.pln_id
 WHERE p.PLANOGRAM not like '%&lt;Untitled&gt;%'
 ; 
</code></pre>

This solution only returns results for PLN_ID in both result sets. If you have a different logic you may need to use LEFT OUTER JOIN instead.

entityMap|blocks|key|4n7r9|text|请更改加入您想要的内容。尝试此查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5pe1s|++++SELECT+
++++++++DISTINCT(a.PLN_ID),
++++++++(SUM(a.WIDTH))+AS+width,
++++++++b.PLN_ID,
++++++++b.ASSORTMENT_GROUP,
++++++++b.STORE,
++++++++b.PLANOGRAM,
++++++++b.STATUS+
++++FROM+
++++++++ACN_FIXEL+a+
++++INNER+JOIN+
++++++++ACN_PLANOGRAMS+b+ON+a.PLN_ID+=+b.PLN_ID
++++WHERE+
++++++++a.type+=+'0'+
++++++++AND+b.PLANOGRAM+NOT+LIKE+'%25<Untitled>%25'
++++GROUP+BY+
++++++++a.PLN_ID,
b.PLN_ID,
++++++++b.ASSORTMENT_GROUP,
++++++++b.STORE,
++++++++b.PLANOGRAM,
++++++++b.STATUS
++++HAVING+
++++++++COUNT(*)+>+1|code-block|syntax|javascript|a9qem^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Please changes join what you want you. Try this query:

<pre><code> SELECT 
 DISTINCT(a.PLN_ID),
 (SUM(a.WIDTH)) AS width,
 b.PLN_ID,
 b.ASSORTMENT_GROUP,
 b.STORE,
 b.PLANOGRAM,
 b.STATUS 
 FROM 
 ACN_FIXEL a 
 INNER JOIN 
 ACN_PLANOGRAMS b ON a.PLN_ID = b.PLN_ID
 WHERE 
 a.type = '0' 
 AND b.PLANOGRAM NOT LIKE '%&lt;Untitled&gt;%'
 GROUP BY 
 a.PLN_ID,
b.PLN_ID,
 b.ASSORTMENT_GROUP,
 b.STORE,
 b.PLANOGRAM,
 b.STATUS
 HAVING 
 COUNT(*) &gt; 1
</code></pre>

someone, please help to join the below queries.

I have tried my best but not able to join with the condition.

PLN_ID is the common column on both the tables.

Query 1-

<pre><code>SELECT PLN_ID
 , ASSORTMENT_GROUP
 , STORE
 , PLANOGRAM
 , STATUS 
 FROM ACN_PLANOGRAMS 
 WHERE PLANOGRAM not like '%&lt;Untitled&gt;%'
 ;
</code></pre>

​
Query 2

<pre><code> SELECT distinct(PLN_ID)
 , count(*)
 , (sum(WIDTH)) AS width
 FROM ACN_FIXEL 
 WHERE type='0'
 GROUP BY PLN_ID
 HAVING count(*) &gt; 1
 ;
</code></pre>

Join 2 oracle queries

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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 有人，请帮助加入以下查询。 我已经尽了最大的努力，但还是不能适应这种情况。 PLN_ID是这两个表上的公共列。 问题1- SELECT PLN_ID     , ASSORTMENT_GROUP     , STORE     , PLANOGRAM     , STATUS   FROM ACN_PLANOGRAM...

问Join 2 oracle查询
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