blocks|key|3015351|text|如果要支持五位数字，只需将代码行更改为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3015352|SET+@STOCK_LEDGER_NUMBER+=+'SLL'+%2B+'-'+%2B+RIGHT(('00000'%2B+CAST((@LASTNUMBER+%2B+1)+AS+VARCHAR(5))),+5)|code-block|syntax|javascript|3015353|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|$]]

If you want to support five digits numbers, just change your line of code to:

<pre><code>SET @STOCK_LEDGER_NUMBER = 'SLL' + '-' + RIGHT(('00000'+ CAST((@LASTNUMBER + 1) AS VARCHAR(5))), 5)
</code></pre>

blocks|key|796052|text|试试看..。它不需要任何修改就能处理任何正整数.|type|unstyled|depth|inlineStyleRanges|entityRanges|data|796053|DECLARE+
++++@last_number+INT,
++++@digit_count+INT,
++++@rep_start+INT,
++++@stock_ledger_number+VARCHAR(14);

SELECT+
++++@last_number+=+10000,
++++@digit_count+=+FLOOR(LOG10(@last_number+%2B+1)+%2B+1),
++++@rep_start+=+(ABS(4+-+@digit_count)+%2B+(4+-+@digit_count))+/+2,
++++@stock_ledger_number+=+STUFF('SLL-0000',+5+%2B+@rep_start,+@digit_count,+CONVERT(VARCHAR(10),+@last_number+%2B+1))

SELECT+last_number+=+@last_number,+stock_ledger_number+=+@stock_ledger_number;|code-block|syntax|javascript|796054|结果..。|796055|last_number+stock_ledger_number
-----------+-------------------
10000+++++++SLL-10001|796056|还有其他一些例子..。|796057|last_number+stock_ledger_number
-----------+-------------------
1+++++++++++SLL-0002

last_number+stock_ledger_number
-----------+-------------------
55++++++++++SLL-0056

last_number+stock_ledger_number
-----------+-------------------
332+++++++++SLL-0333

last_number+stock_ledger_number
-----------+-------------------
2018++++++++SLL-2019

last_number+stock_ledger_number
-----------+-------------------
12345+++++++SLL-12346

last_number+stock_ledger_number
-----------+-------------------
1234567890++SLL-1234567891

last_number+stock_ledger_number
-----------+-------------------
2147483646++SLL-2147483647|796058|当然，您不局限于使用标量变量一次执行这些操作。你可以在相当大的范围内.|796059|-与上述标量变量版本相同的逻辑，但适用于在一次执行中生成多个值。|796060|DECLARE+@last_number+INT+=+9900;

WITH++++
++++cte_Tally+AS+(
++++++++SELECT+TOP+(15000)
++++++++++++n+=+ROW_NUMBER()+OVER+(ORDER+BY+o1.object_id)+%2B+@last_number
++++++++FROM
++++++++++++sys.objects+o1
++++++++++++CROSS+JOIN+sys.objects+o2
++++++++)
SELECT+
++++STOCK_LEDGER_NUMBER+=+CONVERT(VARCHAR(14),+STUFF('SLL-0000',+5+%2B+rs.rep_start,+dc.digit_count,+CONVERT(VARCHAR(10),+t.n)))
FROM
++++cte_Tally+t
++++CROSS+APPLY+(+VALUES+(+CONVERT(INT,+FLOOR(LOG10(t.n)+%2B+1)))+)+dc+(digit_count)
++++CROSS+APPLY+(+VALUES+((ABS(4+-+dc.digit_count)+%2B+(4+-+dc.digit_count))+/+2)+)+rs+(rep_start);
GO+|796061|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Give this a shot... It'll handle any positive INT value without any modification...

<pre><code>DECLARE 
 @last_number INT,
 @digit_count INT,
 @rep_start INT,
 @stock_ledger_number VARCHAR(14);

SELECT 
 @last_number = 10000,
 @digit_count = FLOOR(LOG10(@last_number + 1) + 1),
 @rep_start = (ABS(4 - @digit_count) + (4 - @digit_count)) / 2,
 @stock_ledger_number = STUFF('SLL-0000', 5 + @rep_start, @digit_count, CONVERT(VARCHAR(10), @last_number + 1))

SELECT last_number = @last_number, stock_ledger_number = @stock_ledger_number;
</code></pre>

Results...

<pre><code>last_number stock_ledger_number
----------- -------------------
10000 SLL-10001
</code></pre>

And a few other examples...

<pre><code>last_number stock_ledger_number
----------- -------------------
1 SLL-0002

last_number stock_ledger_number
----------- -------------------
55 SLL-0056

last_number stock_ledger_number
----------- -------------------
332 SLL-0333

last_number stock_ledger_number
----------- -------------------
2018 SLL-2019

last_number stock_ledger_number
----------- -------------------
12345 SLL-12346

last_number stock_ledger_number
----------- -------------------
1234567890 SLL-1234567891

last_number stock_ledger_number
----------- -------------------
2147483646 SLL-2147483647
</code></pre>

<hr>

Of course you aren't restricted to doing these one at a time with scalar variables. You can crank them out in fairly large batches...

-- same logic as the scalar variable version above but adapted to generate multiple values in a single execution.

<pre><code>DECLARE @last_number INT = 9900;

WITH 
 cte_Tally AS (
 SELECT TOP (15000)
 n = ROW_NUMBER() OVER (ORDER BY o1.object_id) + @last_number
 FROM
 sys.objects o1
 CROSS JOIN sys.objects o2
 )
SELECT 
 STOCK_LEDGER_NUMBER = CONVERT(VARCHAR(14), STUFF('SLL-0000', 5 + rs.rep_start, dc.digit_count, CONVERT(VARCHAR(10), t.n)))
FROM
 cte_Tally t
 CROSS APPLY ( VALUES ( CONVERT(INT, FLOOR(LOG10(t.n) + 1))) ) dc (digit_count)
 CROSS APPLY ( VALUES ((ABS(4 - dc.digit_count) + (4 - dc.digit_count)) / 2) ) rs (rep_start);
GO 
</code></pre>

I'm using a SQL Server stored procedure to increase a customized number. I have an autonumber column and it assigns to <code>@LASTNUMBER</code> variable:

<pre><code>SET @STOCK_LEDGER_NUMBER = 'SLL' + '-' + RIGHT(('0000'+ CAST((@LASTNUMBER + 1)AS VARCHAR)),4)
</code></pre>

<ul>
<li>First output : 'SLL-0001'</li>
<li>Last output I can return : 'SLL-9999'</li>
</ul>

With this code, I can increase my number only to a maximum of 9999. 

When autonumber is set to 10000, this code returns 'SLL-0000'. But I need to show it as 'SLL-10000' and next number should be 'SLL-10001'

Please help me

How to increase a number with a format that i'm giving in SQL Server stored procedure

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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我正在使用Server存储过程来增加自定义的数字。我有一个自动编号列，它为@LASTNUMBER变量赋值：SET @STOCK_LEDGER_NUMBER = 'SLL' + '-' + RIGHT(('0000'+ CAST((@LASTNUMBER + 1)AS VARCHAR)),4)第一个输出：'SLL-000...

问如何使用我在Server存储过程中提供的格式来增加数字
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问如何使用我在Server存储过程中提供的格式来增加数字EN