entityMap|blocks|key|8ddue|text|我不知道为什么你需要，或者在数学上任何其他公式都可以实现。但从技术上讲，这可以实现如下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|51h9m|DECLARE+@count+INT
SET+@count+=+1000

DECLARE+@result+INT

SET+@result+=+CASE+WHEN+@count%2510+=+0+THEN+@count+ELSE+CAST('1'%2BREPLICATE('0',LEN(@count))+AS+INT)+end

SELECT+@result|code-block|syntax|javascript|f29kq^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

I don't know why you require or mathematically any other formula can be implemented. but technically that can be achieved as follows

<pre><code>DECLARE @count INT
SET @count = 1000

DECLARE @result INT

SET @result = CASE WHEN @count%10 = 0 THEN @count ELSE CAST('1'+REPLICATE('0',LEN(@count)) AS INT) end

SELECT @result
</code></pre>

entityMap|blocks|key|aniev|text|这是一种相当粗糙的方式，但它确实有效。该查询主要查看您的数字的位数，假设您想要的下一个整数以1开头，然后将相关数量的0添加到它上面。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|al1op|注意:这只会查找下一个增量，而不会向下舍入。|3l7n3|同样，10的话你会得到100，1000的话你会得到10000。|ea3lc|declare+@number+int+=+98999;
declare+@len+int+=+len(@number);
declare+@stringtoreturn+nvarchar(200)='1';

declare+@runs+int+=+1;

while+@runs<=@len
++++begin
++++select+@stringtoreturn+=+@stringtoreturn+%2B+'0';
++++select+@runs=@runs%2B1;
++++end

select+@stringtoreturn;|code-block|syntax|javascript|esqin^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|N|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|O|9|@]|A|@]|B|$]]|$2|G|4|H|6|I|8|P|9|@]|A|@]|B|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

This is a fairly crude way of doing it, however it does work. The query basically looks at the number of digits your number has, assumes the next integer you want starts with a 1 and then adds the relevant number of 0's to it.

Note this only looks for the next increment and does not round down.

Also for 10 you will get 100 and for 1000 you will get 10000.

<pre><code>declare @number int = 98999;
declare @len int = len(@number);
declare @stringtoreturn nvarchar(200)='1';

declare @runs int = 1;

while @runs&lt;=@len
 begin
 select @stringtoreturn = @stringtoreturn + '0';
 select @runs=@runs+1;
 end

select @stringtoreturn;
</code></pre>

entityMap|blocks|key|5rv6n|text|这适用于正数(大于零的数字)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|auaim|select+power(10,+ceiling(+log10(the_number)+))+from+mytable;|code-block|syntax|javascript|ogs7|如果数字已经是10的幂(1，10，100，...)，则返回数字本身。|1ogcg^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]]|$2|H|4|I|6|7|8|M|9|@]|A|@]|B|$]]|$2|J|4|-4|6|7|8|N|9|@]|A|@]|B|$]]]]

This works for positive numbers (numbers greater than zero):

<pre><code>select power(10, ceiling( log10(the_number) )) from mytable;
</code></pre>

In case the number is already a power of ten (1, 10, 100, ...) , the number itself is returned.

entityMap|blocks|key|8gf3e|text|您可以仅使用算术运算来完成此操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bmckk|select+power(10,+floor(log(v.n+-+0.1,+10))+%2B+1)
from+(values+(1),+(10),+(8),+(9982),+(124))+v(n)|code-block|syntax|javascript|238ha^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

You can do this with just arithmetic operations:

<pre><code>select power(10, floor(log(v.n - 0.1, 10)) + 1)
from (values (1), (10), (8), (9982), (124)) v(n)
</code></pre>

I'm looking for a way to find next greater number starting by 1 and followed by zeros in Microsoft SQL. Numbers could vary in digits. ie:

Query: 9856, Result after procedure: 10000
Query: 98999, Result after procedure: 100000

Thanks.

EDIT: There is no chance of having negative numbers. This is a calculation for a energy meter. For example, numbers can go up to 99999 or 999999 or 9999999. When energy overcome that number, it will start again at 0. So I can't read what energy has been used in that period. To know it, I need to calculate the number as asked, then perform some basic maths.

There is no need for knowing what is going on on 10, 100, etc, because of the nature of the operation. It will only be used when the above escenario happend.

Find next greater number with zeros in sql number

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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 我正在寻找一种在Microsoft SQL中查找下一个更大的数字的方法，从1开始，然后是0。数字的位数可能会有所不同。即： 查询: 9856，过程后结果: 10000查询: 98999，过程后结果: 100000 谢谢。 编辑:没有机会有负数。这是对能量计的计算。例如，数字可以上升到99999、999999或9999...

问在sql number中查找下一个更大的带零的数字
EN

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问在sql number中查找下一个更大的带零的数字EN