blocks|key|991118|text|试试那个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|991119|SELECT
++++c.firstname,
++++c.lastname,
++++c.CustomerId,
++++count(si.CustomerID)+as+'No+Of+Orders'
FROM
++++customer+c
LEFT+OUTER+JOIN
++++SalesInvoice+si+ON+si.CustomerID+=+c.CustomerID
GROUP+BY
++++c.CustomerID,
++++c.firstname,
++++c.lastname|code-block|syntax|javascript|991120|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 that one:

<pre><code>SELECT
 c.firstname,
 c.lastname,
 c.CustomerId,
 count(si.CustomerID) as 'No Of Orders'
FROM
 customer c
LEFT OUTER JOIN
 SalesInvoice si ON si.CustomerID = c.CustomerID
GROUP BY
 c.CustomerID,
 c.firstname,
 c.lastname
</code></pre>

blocks|key|994356|text|在OVER()子句中尝试此选项|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|994357|SELECT+c1.Customerid,+c1.firstname,+c1.lastname,+
+++++++COUNT(si.CustomerID)+OVER(PARTITION+BY+c1.CustomerID)+AS+'No+Of+Orders'
FROM+customer+c1+LEFT+JOIN+SalesInvoice+si+ON+si.CustomerID+=+c1.CustomerID|code-block|syntax|javascript|994358|为了提高性能，您需要此索引|994359|CREATE+INDEX+ix_Customerid_Customer+ON+Customer(Customerid)+INCLUDE(firstname,+lastname)|994360|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/ms189461.aspx^0|1|6|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Try this option with <a href="http://msdn.microsoft.com/en-us/library/ms189461.aspx" rel="nofollow">OVER()</a> clause

<pre><code>SELECT c1.Customerid, c1.firstname, c1.lastname, 
 COUNT(si.CustomerID) OVER(PARTITION BY c1.CustomerID) AS 'No Of Orders'
FROM customer c1 LEFT JOIN SalesInvoice si ON si.CustomerID = c1.CustomerID
</code></pre>

For improving performance you need this index

<pre><code>CREATE INDEX ix_Customerid_Customer ON Customer(Customerid) INCLUDE(firstname, lastname)
</code></pre>

blocks|key|5813739|text|有两种方法可以做到这点。一种方法是将所有非聚合列包括在group+by列表中，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5813740|SELECT
++++c.firstname,
++++c.lastname,
++++c.CustomerId,
++++count(si.CustomerID)+as+'No+Of+Orders'
FROM
++++customer+c
LEFT+OUTER+JOIN
++++SalesInvoice+si+ON+si.CustomerID+=+c.CustomerID
GROUP+BY
++++c.CustomerID,
++++c.firstname,
++++c.lastname|code-block|syntax|javascript|5813741|另一种方法是在选择列表中使用聚合函数(min或max应该可以)：|5813742|SELECT
++++MIN(c.firstname)+as+firstname,
++++MIN(c.lastname)+as+lastname,
++++c.CustomerId,
++++count(si.CustomerID)+as+'No+Of+Orders'
FROM
++++customer+c
LEFT+OUTER+JOIN
++++SalesInvoice+si+ON+si.CustomerID+=+c.CustomerID
GROUP+BY
++++c.CustomerID|5813743|我认为第一个查询可能在性能上更好。要提高整体性能，您需要创建一个包含索引或覆盖索引。|5813744|您可以通过以下链接了解所包含的索引：|5813745|5813746|Why+use+the+INCLUDE+clause+when+creating+an+index?|ordered-list-item|offset|length|5813747|http://msdn.microsoft.com/en-IN/library/ms190806.aspx|5813748|5813749|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1307990/why-use-the-include-clause-when-creating-an-index|1^0|0|0|0|0|0|0|0|0|1E|0|0|0|1H|1|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1A|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1D|8|@]|9|@$S|1E|T|1F|1|1G]]|A|$]]|$1|U|3|V|5|R|7|1H|8|@]|9|@$S|1I|T|1J|1|1K]]|A|$]]|$1|W|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14]]|15|$5|10|11|12|A|$13|V]]]]

There could be 2 ways to do it. 
One way is to include all non-aggregate columns into the group by list as below: 

<pre><code>SELECT
 c.firstname,
 c.lastname,
 c.CustomerId,
 count(si.CustomerID) as 'No Of Orders'
FROM
 customer c
LEFT OUTER JOIN
 SalesInvoice si ON si.CustomerID = c.CustomerID
GROUP BY
 c.CustomerID,
 c.firstname,
 c.lastname
</code></pre>

Another way to do it, would be to use an aggregate function (min or max should do) in the select list:

<pre><code>SELECT
 MIN(c.firstname) as firstname,
 MIN(c.lastname) as lastname,
 c.CustomerId,
 count(si.CustomerID) as 'No Of Orders'
FROM
 customer c
LEFT OUTER JOIN
 SalesInvoice si ON si.CustomerID = c.CustomerID
GROUP BY
 c.CustomerID
</code></pre>

I think the first query may be better in performance. To improve overall performance you will need to create an included or covering index.

You can have a look at the following links to give you an idea on included indexes:

<ol>
<li><a href="https://stackoverflow.com/questions/1307990/why-use-the-include-clause-when-creating-an-index">Why use the INCLUDE clause when creating an index?</a></li>
<li><a href="http://msdn.microsoft.com/en-IN/library/ms190806.aspx" rel="nofollow noreferrer">http://msdn.microsoft.com/en-IN/library/ms190806.aspx</a></li>
</ol>

Is there any way where I can avoid the inner join to get the customer firstname, and lastname from customer table and how can i optimize the execution plan?

<pre><code>SELECT c1.firstname, c1.lastname, t.*
FROM customer c1
INNER JOIN
 (select c.Customerid ,count(si.CustomerID) as 'No Of Orders'
 from Customer c
 LEFT OUTER join SalesInvoice si ON si.CustomerID = c.CustomerID
 GROUP by c.CustomerID) as t
ON c1.CustomerID = t.Customerid
</code></pre>

<img src="https://i.stack.imgur.com/OoGxS.png" alt="enter image description here">

How can i avoid an extra inner join to optimize this query?

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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有没有什么方法可以避免从customer表中获取客户名字和姓氏的内部连接，以及如何优化执行计划？SELECT c1.firstname, c1.lastname, t.*FROM customer c1INNER JOIN    (select c.Customerid ,count(si.CustomerID) as...

问如何避免额外的内连接来优化这个查询？
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问如何避免额外的内连接来优化这个查询？EN