blocks|key|1021788|text|decimal(9,6)在逗号右侧存储6位数字。是否显示尾随零是一个格式化决定，通常在客户端实现。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1021789|但是，由于SSMS格式化float时没有尾随零，因此可以通过将decimal转换为float来删除尾随零|1021790|select+
++++cast(123.4567+as+DECIMAL(9,6))
,+++cast(cast(123.4567+as+DECIMAL(9,6))+as+float)|code-block|syntax|javascript|1021791|打印：|1021792|123.456700++123,4567|1021793|(我的小数分隔符是逗号，但是SSMS的小数格式是用点表示的。显然是known+issue。)|1021794|entityMap|0|LINK|mutability|MUTABLE|url|http://connect.microsoft.com/SQLServer/feedback/details/458416/microsoft-sql-server-management-studio-doesnt-get-os-regional-setting^0|0|C|0|C|5|V|7|15|5|0|0|0|0|X|B|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1B|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1D|8|@]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|S|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

A <code>decimal(9,6)</code> stores 6 digits on the right side of the comma. Whether to display trailing zeroes or not is a formatting decision, usually implemented on the client side.

But since SSMS formats <code>float</code> without trailing zeros, you can remove trailing zeroes by casting the <code>decimal</code> to a <code>float</code>:

<pre><code>select 
 cast(123.4567 as DECIMAL(9,6))
, cast(cast(123.4567 as DECIMAL(9,6)) as float)
</code></pre>

prints:

<pre><code>123.456700 123,4567
</code></pre>

(My decimal separator is a comma, yet SSMS formats decimal with a dot. Apparently a <a href="http://connect.microsoft.com/SQLServer/feedback/details/458416/microsoft-sql-server-management-studio-doesnt-get-os-regional-setting" rel="noreferrer">known issue</a>.)

blocks|key|1021819|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1021820|select+Cast(+Cast(+(ROUND(+35.457514+,+2)+*100)+as+Int)+as+float+)+/100|code-block|syntax|javascript|1021821|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 this:

<pre><code>select Cast( Cast( (ROUND( 35.457514 , 2) *100) as Int) as float ) /100
</code></pre>

blocks|key|1236511|text|SELECT+REVERSE(ROUND(REVERSE(2.5500),1))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1236512|打印：|unstyled|1236513|2.55|1236514|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>SELECT REVERSE(ROUND(REVERSE(2.5500),1))</code></pre>

prints:

<pre><code>2.55</code></pre>

blocks|key|1236566|text|另一个选择..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1236567|我不知道这有多有效，但它似乎是有效的，并且不是通过float：|1236568|select+replace(rtrim(replace(
+++++++replace(rtrim(replace(cast(@value+as+varchar(40)),+'0',+'+')),+'+',+'0')
+++++++,+'.',+'+')),+'+',+'.')|code-block|syntax|javascript|1236569|中间一行去掉尾随空格，如果没有小数位，则外两行去掉小数点。|1236570|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|$]]

Another option...

I don't know how efficient this is but it seems to work and does not go via float:

<pre><code>select replace(rtrim(replace(
 replace(rtrim(replace(cast(@value as varchar(40)), '0', ' ')), ' ', '0')
 , '.', ' ')), ' ', '.')
</code></pre>

The middle line strips off trailing spaces, the outer two remove the point if there are no decimal digits

blocks|key|1021918|text|这个怎么样？假设进入函数的数据是@thisData：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1021919|BEGIN
++DECLARE+@thisText+VARCHAR(255)
++SET+@thisText+=+REPLACE(RTRIM(REPLACE(@thisData,+'0',+'+')),+'+',+'0')
++IF+SUBSTRING(@thisText,+LEN(@thisText),+1)+=+'.'
++++RETURN+STUFF(@thisText,+LEN(@thisText),+1,+'')
++RETURN+@thisText
END|code-block|syntax|javascript|1021920|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|$]]

How about this? Assuming data coming into your function as @thisData:

<pre><code>BEGIN
 DECLARE @thisText VARCHAR(255)
 SET @thisText = REPLACE(RTRIM(REPLACE(@thisData, '0', ' ')), ' ', '0')
 IF SUBSTRING(@thisText, LEN(@thisText), 1) = '.'
 RETURN STUFF(@thisText, LEN(@thisText), 1, '')
 RETURN @thisText
END
</code></pre>

blocks|key|1236611|text|case+when+left(replace(ltrim(rtrim(replace(str(XXX,+38,+10),+'0',++'+'))),+'+',+'0'),+1)+=+'.'
then+'0'+
else+''
end+%2B

replace(ltrim(rtrim(replace(str(XXX,+38,+10),+'0',++'+'))),+'+',+'0')+%2B

case+when+right(replace(ltrim(rtrim(replace(str(XXX,+38,+10),+'0',++'+'))),+'+',+'0'),+1)+=+'.'
then+'0'+
else+''
end|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1236612|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>case when left(replace(ltrim(rtrim(replace(str(XXX, 38, 10), '0', ' '))), ' ', '0'), 1) = '.'
then '0' 
else ''
end +

replace(ltrim(rtrim(replace(str(XXX, 38, 10), '0', ' '))), ' ', '0') +

case when right(replace(ltrim(rtrim(replace(str(XXX, 38, 10), '0', ' '))), ' ', '0'), 1) = '.'
then '0' 
else ''
end
</code></pre>

blocks|key|1236656|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1236657|SELECT+REPLACE(TRIM(REPLACE(20.5500,+"0",+"+")),+"+",+"0")|code-block|syntax|javascript|1236658|给出20.55|1236659|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try this : 

<pre><code>SELECT REPLACE(TRIM(REPLACE(20.5500, "0", " ")), " ", "0")
</code></pre>

Gives 20.55

blocks|key|1021998|text|Cast(20.5500+as+Decimal(6,2))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1021999|应该做这件事。|unstyled|1022000|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>Cast(20.5500 as Decimal(6,2))
</code></pre>

should do it.

blocks|key|1236745|text|我遇到了类似的问题，但也需要删除不存在小数的小数点，这是我的解决方案，它将小数拆分为其组成部分，并将小数点字符串中的字符数基于小数部分的长度(不使用大小写)。更有趣的是，我的数字被存储为一个不带小数的浮点数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1236746|DECLARE+@MyNum+FLOAT
SET+@MyNum+=+700000
SELECT+CAST(PARSENAME(CONVERT(NUMERIC(15,2),@MyNum/10000),2)+AS+VARCHAR(10))+
%2B+SUBSTRING('.',1,LEN(REPLACE(RTRIM(REPLACE(CAST(PARSENAME(CONVERT(NUMERIC(15,2),@MyNum/10000),1)+AS+VARCHAR(2)),'0','+')),'+','0')))+
%2B+REPLACE(RTRIM(REPLACE(CAST(PARSENAME(CONVERT(NUMERIC(15,2),@MyNum/10000),1)+AS+VARCHAR(2)),'0','+')),'+','0')+|code-block|syntax|javascript|1236747|结果是痛苦的，我知道，但在上面的答案的帮助下，我做到了。|1236748|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I had a similar issue, but was also required to remove the decimal point where no decimal was present, here was my solution which splits the decimal into its components, and bases the number of characters it takes from the decimal point string on the length of the fraction component (without using CASE). To make matters even more interesting, my number was stored as a float without its decimals.

<pre><code>DECLARE @MyNum FLOAT
SET @MyNum = 700000
SELECT CAST(PARSENAME(CONVERT(NUMERIC(15,2),@MyNum/10000),2) AS VARCHAR(10)) 
+ SUBSTRING('.',1,LEN(REPLACE(RTRIM(REPLACE(CAST(PARSENAME(CONVERT(NUMERIC(15,2),@MyNum/10000),1) AS VARCHAR(2)),'0',' ')),' ','0'))) 
+ REPLACE(RTRIM(REPLACE(CAST(PARSENAME(CONVERT(NUMERIC(15,2),@MyNum/10000),1) AS VARCHAR(2)),'0',' ')),' ','0') 
</code></pre>

The result is painful, I know, but I got there, with much help from the answers above.

blocks|key|1022109|text|我知道这是一个古老的帖子，但我想提供我想出的SQL|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022110|DECLARE+@value+DECIMAL(23,3)
set+@value+=+1.2000
select+@value+original_val,+
++++SUBSTRING(++CAST(+@value+as+VARCHAR(100)),+
++++++++++++++++0,
++++++++++++++++PATINDEX('%25.%25',CAST(@value+as+VARCHAR(100)))
++++++++++++)
++++++%2B+CASE+WHEN+ROUND(+
++++++++++++++++++++++++REVERSE(+SUBSTRING(+CAST(@value+as+VARCHAR(100)),
++++++++++++++++++++++++++++++++++++++++PATINDEX('%25.%25',CAST(@value+as+VARCHAR(100)))%2B1,
++++++++++++++++++++++++++++++++++++++++LEN(CAST(@value+as+VARCHAR(100)))
++++++++++++++++++++++++++++++++++++++++)
++++++++++++++++++++++++++++++++)
++++++++++++++++++++,1)+>+0+THEN+
++++++++++++'.'+
++++++++++++%2B++REVERSE(ROUND(REVERSE(SUBSTRING(+CAST(@value+as+VARCHAR(100)),
++++++++++++++++++++++++++++++++++++++++++++++++PATINDEX('%25.%25',CAST(@value+as+VARCHAR(100)))%2B1,
++++++++++++++++++++++++++++++++++++++++++++++++LEN(CAST(@value+as+VARCHAR(100)))
++++++++++++++++++++++++++++++++++++++++++++++++)
++++++++++++++++),1))
++++++++ELSE+''+END++AS+modified_val|code-block|syntax|javascript|1022111|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|$]]

I understand this is an old post but would like to provide SQL that i came up with

<pre><code>DECLARE @value DECIMAL(23,3)
set @value = 1.2000
select @value original_val, 
 SUBSTRING( CAST( @value as VARCHAR(100)), 
 0,
 PATINDEX('%.%',CAST(@value as VARCHAR(100)))
 )
 + CASE WHEN ROUND( 
 REVERSE( SUBSTRING( CAST(@value as VARCHAR(100)),
 PATINDEX('%.%',CAST(@value as VARCHAR(100)))+1,
 LEN(CAST(@value as VARCHAR(100)))
 )
 )
 ,1) &gt; 0 THEN 
 '.' 
 + REVERSE(ROUND(REVERSE(SUBSTRING( CAST(@value as VARCHAR(100)),
 PATINDEX('%.%',CAST(@value as VARCHAR(100)))+1,
 LEN(CAST(@value as VARCHAR(100)))
 )
 ),1))
 ELSE '' END AS modified_val
</code></pre>

blocks|key|1022155|text|我知道这个帖子很旧，但是对于那些不使用SQL+Server+2012或更高版本的人，或者因为任何原因不能使用格式化功能，那么下面的工作就可以了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022156|而且，如果数字小于1(例如0.01230000)，许多解决方案都不起作用。|1022157|请注意，以下内容不适用于负数。|1022158|DECLARE+@num+decimal(28,14)+=+10.012345000
SELECT+PARSENAME(@num,2)+%2B+REPLACE(RTRIM(LTRIM(REPLACE(@num-PARSENAME(@num,2),'0','+'))),'+','0')+

set+@num+=+0.0123450000
SELECT+PARSENAME(@num,2)+%2B+REPLACE(RTRIM(LTRIM(REPLACE(@num-PARSENAME(@num,2),'0','+'))),'+','0')+|code-block|syntax|javascript|1022159|分别返回10.012345和0.012345。|1022160|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I know this thread is very old but for those not using SQL Server 2012 or above or cannot use the FORMAT function for any reason then the following works.

Also, a lot of the solutions did not work if the number was less than 1 (e.g. 0.01230000).

Please note that the following does not work with negative numbers.

<pre><code>DECLARE @num decimal(28,14) = 10.012345000
SELECT PARSENAME(@num,2) + REPLACE(RTRIM(LTRIM(REPLACE(@num-PARSENAME(@num,2),'0',' '))),' ','0') 

set @num = 0.0123450000
SELECT PARSENAME(@num,2) + REPLACE(RTRIM(LTRIM(REPLACE(@num-PARSENAME(@num,2),'0',' '))),' ','0') 
</code></pre>

Returns 10.012345 and 0.012345 respectively.

blocks|key|1022165|text|SELECT+CONVERT(DOUBLE+PRECISION,+[ColumnName])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1022166|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>SELECT CONVERT(DOUBLE PRECISION, [ColumnName])
</code></pre>

blocks|key|1236843|text|尝尝这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1236844|select+CAST(123.456700+as+float),cast(cast(123.4567+as+DECIMAL(9,6))+as+float)|code-block|syntax|javascript|1236845|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 this.

<pre><code>select CAST(123.456700 as float),cast(cast(123.4567 as DECIMAL(9,6)) as float)
</code></pre>

blocks|key|1022180|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022181|select+isnull(cast(floor(replace(rtrim(ltrim('999,999.0000')),',',''))+as+int),0)|code-block|syntax|javascript|1022182|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 this:

<pre><code>select isnull(cast(floor(replace(rtrim(ltrim('999,999.0000')),',','')) as int),0)
</code></pre>

blocks|key|1022224|text|最简单的方法是将值转换为浮点数，然后转换为字符串数据类型。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022225|CAST(CAST(123.456000+AS+FLOAT)+AS+VARCHAR(100))|code-block|syntax|javascript|1022226|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|$]]

The easiest way is to CAST the value as FLOAT and then to a string data type.

<pre><code>CAST(CAST(123.456000 AS FLOAT) AS VARCHAR(100))
</code></pre>

blocks|key|1022265|text|我遇到了类似的问题，需要从像xx0000,x00000,xxx000这样的数字中去掉尾随零|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1022266|我使用：|1022267|select+LEFT(code,LEN(code)%2B1+-+PATINDEX('%25[1-Z]%25',REVERSE(code)))+from+Tablename|code-block|syntax|javascript|1022268|Code是包含要修剪的数字的字段的名称。希望这对其他人有帮助。|1022269|entityMap^0|E|K|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

I had a similar problem, needed to trim trailing zeros from numbers like <code>xx0000,x00000,xxx000</code>

I used:

<pre><code>select LEFT(code,LEN(code)+1 - PATINDEX('%[1-Z]%',REVERSE(code))) from Tablename
</code></pre>

Code is the name of the field with the number to be trimmed. Hope this helps someone else.

blocks|key|1022364|text|我不愿意强制转换为浮点型，因为我的小数点中可能会有比浮点型更多的数字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022365|当FORMAT与标准.net格式字符串'g8‘一起使用时，将返回小数(例如1e-08)非常小的情况下的科学记数法，这也不适合|offset|length|style|CODE|1022366|使用自定义格式字符串(https://docs.microsoft.com/en-us/dotnet/standard/base-types/custom-numeric-format-strings)可以实现我想要的功能：|1022367|DECLARE+@n+DECIMAL(9,6)+=1.23;
SELECT+@n
-->+1.230000
SELECT+FORMAT(@n,+'0.######')
-->+1.23|code-block|syntax|javascript|1022368|如果您希望您的数字至少有一个尾随零，因此2.0不会变成2，请使用像0.0#####这样的格式字符串|1022369|小数点是本地化的，因此使用逗号作为小数点分隔符的区域性将遇到逗号输出。是|1022370|当然，这是一种令人沮丧的做法，即让数据层进行格式化(但在我的示例中没有其他层；用户实际上是在运行存储过程并将结果放入电子邮件:/+)|1022371|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.microsoft.com/en-us/dotnet/standard/base-types/custom-numeric-format-strings^0|0|1|6|0|B|2H|0|0|0|X|8|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@$D|14|E|15|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|16|8|@]|9|@$D|17|E|18|1|19]]|A|$]]|$1|J|3|K|5|L|7|1A|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1B|8|@$D|1C|E|1D|F|G]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1E|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1F|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

I was reluctant to cast to float because of the potential for more digits to be in my decimal than float can represent

<code>FORMAT</code> when used with a standard .net format string 'g8' returned the scientific notation in cases of very small decimals (eg 1e-08) which was also unsuitable

Using a custom format string (<a href="https://docs.microsoft.com/en-us/dotnet/standard/base-types/custom-numeric-format-strings" rel="noreferrer">https://docs.microsoft.com/en-us/dotnet/standard/base-types/custom-numeric-format-strings</a>) allowed me to achieve what I wanted:

<pre><code>DECLARE @n DECIMAL(9,6) =1.23;
SELECT @n
--&gt; 1.230000
SELECT FORMAT(@n, '0.######')
--&gt; 1.23
</code></pre>

If you want your number to have at least one trailing zero, so 2.0 does not become 2, use a format string like <code>0.0#####</code>

The decimal point is localised, so cultures that use a comma as decimal separator will encounter a comma output where the . is

Of course, this is the discouragable practice of having the data layer doing formatting (but in my case there is no other layer; the user is literally running a stored procedure and putting the result in an email :/ )

I have a column <code>DECIMAL(9,6)</code> i.e. it supports values like 999,123456.

But when I insert data like 123,4567 it becomes 123,456700

How to remove those zeros?

Remove trailing zeros from decimal in SQL Server

SQL Server

我有一个列DECIMAL(9,6)，即它支持像999,123456这样的值。但是当我插入像123,4567这样的数据时，它变成了123,456700如何删除这些零？

问在SQL Server中删除小数的尾随零
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