blocks|key|1189988|text|这几乎就是解决问题的方法：Adding+Leading+Zeros+To+Integer+Values|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1189989|因此，为了保存链接，查询如下所示，其中#Numbers是表，Num是列：|style|CODE|1189990|+++SELECT+RIGHT('000000000'+%2B+CONVERT(VARCHAR(8),Num),+8)+FROM+#Numbers|code-block|syntax|javascript|1189991|对于负值或正值|1189992|declare+@v+varchar(6)
select+@v+=+-5

SELECT+case++when+@v+<+0+
then+'-'+else+''+end+%2B+RIGHT('00000'+%2B+replace(@v,'-',''),+5)+|1189993|entityMap|0|LINK|mutability|MUTABLE|url|http://wiki.lessthandot.com/index.php/Adding_Leading_Zeros_To_Integer_Values^0|D|12|0|0|J|8|U|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|6|7|12|8|@$A|13|B|14|F|G]|$A|15|B|16|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|17|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|C|$]]|$1|O|3|P|5|J|7|19|8|@]|9|@]|C|$K|L]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

That is pretty much the way: <a href="http://wiki.lessthandot.com/index.php/Adding_Leading_Zeros_To_Integer_Values" rel="noreferrer">Adding Leading Zeros To Integer Values</a>

So, to save following the link, the query looks like this, where <code>#Numbers</code> is the table and <code>Num</code> is the column:

<pre><code> SELECT RIGHT('000000000' + CONVERT(VARCHAR(8),Num), 8) FROM #Numbers
</code></pre>

for negative or positive values

<pre><code>declare @v varchar(6)
select @v = -5

SELECT case when @v &lt; 0 
then '-' else '' end + RIGHT('00000' + replace(@v,'-',''), 5) 
</code></pre>

blocks|key|3867683|text|我会这样做。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3867684|SELECT+RIGHT(REPLICATE('0',+5)+%2B+CAST(Field2+AS+VARCHAR(5),5)+|code-block|syntax|javascript|3867685|不一定都是“更容易”，或者更高效，但读起来更好。可以进行优化以消除对“正确”的需求|3867686|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 would do it like this.

<pre><code>SELECT RIGHT(REPLICATE('0', 5) + CAST(Field2 AS VARCHAR(5),5) 
</code></pre>

Not necessarily all that "Easier", or more efficient, but better to read. Could be optimized to remove the need for "RIGHT"

blocks|key|4710818|text|如果你负担得起/想要在你的数据库中有一个函数，你可以使用类似这样的东西：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4710819|CREATE+FUNCTION+LEFTPAD
+++++++++++(@SourceString+VARCHAR(MAX),
++++++++++++@FinalLength++INT,
++++++++++++@PadChar++++++CHAR(1))+
RETURNS+VARCHAR(MAX)
AS
BEGIN
++RETURN
++++(SELECT+Replicate(@PadChar,+@FinalLength+-+Len(@SourceString))+%2B+@SourceString)
END|code-block|syntax|javascript|4710820|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 can afford/want to have a function in your database you could use something like:

<pre><code>CREATE FUNCTION LEFTPAD
 (@SourceString VARCHAR(MAX),
 @FinalLength INT,
 @PadChar CHAR(1)) 
RETURNS VARCHAR(MAX)
AS
BEGIN
 RETURN
 (SELECT Replicate(@PadChar, @FinalLength - Len(@SourceString)) + @SourceString)
END
</code></pre>

blocks|key|3867758|text|另一种方式(不进行强制转换或转换)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3867759|SELECT+RIGHT(REPLACE(STR(@NUM),'+','0'),5)|code-block|syntax|javascript|3867760|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|$]]

Another way (without CAST or CONVERT):

<pre><code>SELECT RIGHT(REPLACE(STR(@NUM),' ','0'),5)
</code></pre>

blocks|key|1182840|text|如果您希望通过添加不同数量的零在最终结果中获得一致数量的字符串总数，这里有一点修改(针对vsql)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1182841|SELECT+
++CONCAT(
++++REPEAT('0',+9-length(TO_CHAR(var1))),+
++++CAST(var1+AS+VARCHAR(9))
++)+as+var1|code-block|syntax|javascript|1182842|您可以根据需要将9替换为任何数字！|1182843|BRD|1182844|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

If you want to get a consistent number of total strings in the final result by adding different number of zeros, here is a little bit modification (for vsql)

<pre><code>SELECT 
 CONCAT(
 REPEAT('0', 9-length(TO_CHAR(var1))), 
 CAST(var1 AS VARCHAR(9))
 ) as var1
</code></pre>

You can replace 9 by any number for your need!

BRD

I need to return two fields from a database concatenated as 'field1-field2'. The second field is an int, but needs to be returned as a fixed length of 5 with leading 0's. The method i'm using is:

<pre><code>SELECT Field1 + '-' + RIGHT('0000' + CAST(Field2 AS varchar),5) FROM ...
</code></pre>

Is there a more efficient way to do this?

Most efficient method for adding leading 0's to an int in sql

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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我需要从数据库返回两个连接为'field1-field2‘的字段。第二个字段是int，但需要以固定长度5和前导0返回。我使用的方法是：SELECT Field1 + '-' + RIGHT('0000' + CAST(Field2 AS varchar),5) FROM ...有没有更有效的方法来做到这一点？

问在sql中将前导0添加到int的最有效方法
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问在sql中将前导0添加到int的最有效方法EN