blocks|key|260436|text|SELECT+*
FROM+++MyTable
WHERE++LEN(CONCATENATE(C.First,+C....+,+C.Eighth))+-+LEN(REPLACE(CONCATENATE(C.First,+C....+,+C.Eighth),+'A',+''))+=+1+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|260437|作为一个例子|unstyled|260438|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>SELECT *
FROM MyTable
WHERE LEN(CONCATENATE(C.First, C.... , C.Eighth)) - LEN(REPLACE(CONCATENATE(C.First, C.... , C.Eighth), 'A', '')) = 1 
</code></pre>

As an example

blocks|key|959625|text|尝试以下几点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|959626|select+
*+from+YourTable
where+REPLACE(TRIM('X'+FROM+col1),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col2),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col3),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col4),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col5),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col6),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col7),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col8),'A',1)
%2B+REPLACE(TRIM('X'+FROM+col9),'A',1)>+1|code-block|syntax|javascript|959627|我也使用过TRIM，它需要Server+2017或更高版本。|offset|length|style|CODE|959628|或如下：|959629|select+*
from+YourTable
where+replace(replace(concat(col1,col2,col3,col4,col5,col6,col7,col8,col9),+'X',+''),+'A',+1)+>+1|959630|请看演示。|959631|entityMap^0|0|0|5|4|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

try the following:

<pre><code>select 
* from YourTable
where REPLACE(TRIM('X' FROM col1),'A',1)
+ REPLACE(TRIM('X' FROM col2),'A',1)
+ REPLACE(TRIM('X' FROM col3),'A',1)
+ REPLACE(TRIM('X' FROM col4),'A',1)
+ REPLACE(TRIM('X' FROM col5),'A',1)
+ REPLACE(TRIM('X' FROM col6),'A',1)
+ REPLACE(TRIM('X' FROM col7),'A',1)
+ REPLACE(TRIM('X' FROM col8),'A',1)
+ REPLACE(TRIM('X' FROM col9),'A',1)&gt; 1
</code></pre>

I have used <code>TRIM</code> also which requires SQL Server 2017 or above.

or the following:

<pre><code>select *
from YourTable
where replace(replace(concat(col1,col2,col3,col4,col5,col6,col7,col8,col9), 'X', ''), 'A', 1) &gt; 1
</code></pre>

Please see the demo <a href="https://dbfiddle.uk/?rdbms=sqlserver_2019&amp;fiddle=1cd35738dacb2fc651d7ac15156f20f7" rel="nofollow noreferrer">here</a>.

blocks|key|959673|text|您可以使用APPLY|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|959674|SELECT+C.*
FROM+Criteria+C+CROSS+APPLY
+++++(SELECT+COUNT(*)+as+num_a_s
++++++FROM+(VALUES+(First),+(Second),+.+.+.++--+list+all+the+columns+here
+++++++++++)+V(x)
++++++WHERE+v.x+=+'A'
+++++)+v
WHERE+v.num_a_s+>=+2;|code-block|syntax|javascript|959675|注意:如果将这些值存储在列中而不是单独的行中，则数据模型可能有问题。|959676|entityMap^0|5|5|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can use <code>APPLY</code>:

<pre><code>SELECT C.*
FROM Criteria C CROSS APPLY
 (SELECT COUNT(*) as num_a_s
 FROM (VALUES (First), (Second), . . . -- list all the columns here
 ) V(x)
 WHERE v.x = 'A'
 ) v
WHERE v.num_a_s &gt;= 2;
</code></pre>

Note: Something is probably wrong with your data model if you are storing these values in columns rather than in separate rows.

blocks|key|259451|text|这里有一种方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|259452|++create+table+dbo.t1(x+int,+col1+varchar(10),+col2+varchar(10))


++insert+into+dbo.t1+values(1,'A','A')
++insert+into+dbo.t1+values(2,'A','')
++insert+into+dbo.t1+values(3,'','')
++insert+into+dbo.t1+values(4,'','X')





++select+x
+++++++++,count(case+when+n='A'+then+1+end)+as+cnt++++++
++++from+dbo.t1
cross+apply(values+('col1',col1)
++++++++++++++++++,('col2',col2)
++++++++++++++++++---repeat+this+for+columns+up+to+col9...
+++++++++++)x(m,n)
++group+by+x
++having+count(case+when+n='A'+then+1+end)>1|code-block|syntax|javascript|259453|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|$]]

Here is one way to do this

<pre><code> create table dbo.t1(x int, col1 varchar(10), col2 varchar(10))


 insert into dbo.t1 values(1,'A','A')
 insert into dbo.t1 values(2,'A','')
 insert into dbo.t1 values(3,'','')
 insert into dbo.t1 values(4,'','X')





 select x
 ,count(case when n='A' then 1 end) as cnt 
 from dbo.t1
cross apply(values ('col1',col1)
 ,('col2',col2)
 ---repeat this for columns up to col9...
 )x(m,n)
 group by x
 having count(case when n='A' then 1 end)&gt;1
</code></pre>

I have a SQL Server table that includes 9 columns used to indicated what things to include or exclude in part of the UI. Each column can have the value 'A', 'X', or blank. Each row should have at most 1 'A' in any of the columns. 

Due to an error many columns have multiple 'A' values. How can I write a query that returns every row that breaks this constraint? 

All I have is something like:

<pre><code>SELECT PrimaryKey
FROM Criteria C
WHERE (C.First = 'A' AND C.Second = 'A')
 OR (C.First = 'A' AND C.Third = 'A')
 OR (C.First = 'A' AND C.Fourth = 'A')
 ...
 OR (C.Eighth = 'A' AND C.Ninth = 'A')
</code></pre>

Is there any cleaner or more elegant way to write this code?

SQL determine if more than one column on a given row has the same value

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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我有一个Server表，其中包含9列，用于指示在UI的一部分中包含或排除哪些内容。每一列的值都可以是'A‘、'X’或空白。每一行在任何列中最多应有1 'A‘。由于一个错误，许多列有多个'A‘值。如何编写返回打破此约束的每一行的查询？我所拥有的只是：SELECT PrimaryKeyFROM Criteria CWHER...

问SQL确定给定行上是否有多个列具有相同的值。
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问SQL确定给定行上是否有多个列具有相同的值。EN