blocks|key|3987995|text|您可以根据值小于阈值的次数来定义组。定义了每一组。在此之后，您需要一个行号：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3987996|select+t.*,
+++++++(case+when+t.score+<+1+then+0
+++++++++++++else+row_number()+over+(partition+by+t.customerId,+grp,+score+order+by+t.attempt)
++++++++end)+as+ranking
from+(select+t.*,
+++++++++++++sum(case+when+score+<+1+then+1+else+0+end)+over+(partition+by+t.customerId+order+by+t.attempt)+as+grp
++++++from+t
+++++)+t;|code-block|syntax|javascript|3987997|Here是一个db<>fiddle。|offset|length|3987998|entityMap|0|LINK|mutability|MUTABLE|url|https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=a2e747a0a33c3a25e5f65fe275873e38^0|0|0|0|4|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

You can define a group by the number of times the value is less than your threshold. That defines each group. After that, you want a row number:

<pre><code>select t.*,
 (case when t.score &lt; 1 then 0
 else row_number() over (partition by t.customerId, grp, score order by t.attempt)
 end) as ranking
from (select t.*,
 sum(case when score &lt; 1 then 1 else 0 end) over (partition by t.customerId order by t.attempt) as grp
 from t
 ) t;
</code></pre>

<a href="https://dbfiddle.uk/?rdbms=sqlserver_2017&amp;fiddle=a2e747a0a33c3a25e5f65fe275873e38" rel="nofollow noreferrer">Here</a> is a db&lt;>fiddle.

blocks|key|3988000|text|DECLARE+@T+table+(CustomerID+int,+Attempt+int,+score+decimal(10,2)+)
INSERT+INTO+@T
VALUES
(111,+1,+1)
,(111,+2,+1)
,(111,+3,+1)
,(111,+4,+0.5)
,(111,+5,+1)
,(111,+6,+0)
,(222,+5,+0.5)
,(222,+6,+1)
,(222,+7,+0.5)
,(222,+8,+1)
,(222,+9,+1)
,(222,+110,+1)
select+t.*,
+++(case+when+t.score+<+1+then+0
+++++++++else+row_number()+over+(partition+by+t.customerId,+grp+order+by+t.attempt)
++++end)+as+ranking
from+(select+t.*,
+++++++++sum(case+when+score+<+1.00+then+1+else+0+end)+over+(partition+by+t.customerId+order+by+t.attempt)+as+grp
++from+@t+t
+)+t;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3988001|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>DECLARE @T table (CustomerID int, Attempt int, score decimal(10,2) )
INSERT INTO @T
VALUES
(111, 1, 1)
,(111, 2, 1)
,(111, 3, 1)
,(111, 4, 0.5)
,(111, 5, 1)
,(111, 6, 0)
,(222, 5, 0.5)
,(222, 6, 1)
,(222, 7, 0.5)
,(222, 8, 1)
,(222, 9, 1)
,(222, 110, 1)
select t.*,
 (case when t.score &lt; 1 then 0
 else row_number() over (partition by t.customerId, grp order by t.attempt)
 end) as ranking
from (select t.*,
 sum(case when score &lt; 1.00 then 1 else 0 end) over (partition by t.customerId order by t.attempt) as grp
 from @t t
 ) t;
</code></pre>

I've already tried looking on here for the answer to this question, and although have found similar queries, I haven't found this exact one.

I'm looking to count the instances of a customer getting a certain score, and if they get less than that score I'd like the count to be reset.

This is the data I have:

<img src="https://i.stack.imgur.com/CjXK6.png" alt="Data Before Running Total"> 

And this is the result I would like to produce:

<img src="https://i.stack.imgur.com/QcIaT.png" alt="With Running Total">

Any help would be appreciated, as well as an explanation of any advanced code used.

Resetting Running Total based on value in another column per customerID - SQL Server

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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我已经尝试在这里寻找这个问题的答案，尽管我找到了类似的查询，但我还没有找到确切的答案。我希望对客户获得某个分数的实例进行计数，如果他们得到的分数小于该分数，我希望该计数被重置。这是我拥有的数据：​这是我想要产生的结果：​​任何帮助，以及任何使用的高级代码的解释都将不胜感激。

问根据每个customerID的另一列中的值重置运行合计- SQL Server
EN

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问根据每个customerID的另一列中的值重置运行合计- SQL Server
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