blocks|key|361363|text|您可以将列转换为int|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|361364|select+*+
from+A+test1+
inner+join+B+test2+
++++++on+cast(test2.tag_num+as+int)+=+cast(test1.tag_num+as+int)|code-block|syntax|javascript|361365|或者首先使用int作为表中的数据类型。|361366|entityMap^0|8|3|0|0|6|3|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|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

You can cast the columns to <code>int</code>

<pre><code>select * 
from A test1 
inner join B test2 
 on cast(test2.tag_num as int) = cast(test1.tag_num as int)
</code></pre>

Or better use <code>int</code> as data type in the table in the first place.

blocks|key|359104|text|您应该将字符串cast到int并对其进行连接|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|359105|SELECT+*
++++FROM+A+AS+test1+INNER+JOIN+B+AS+test2+
++++ON+CAST(test2.tag_num+AS+INT)+=+CAST(test1.tag_num+AS+INT)|code-block|syntax|javascript|359106|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/ms187928.aspx^0|7|4|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

you should <a href="https://msdn.microsoft.com/en-us/library/ms187928.aspx" rel="nofollow">cast</a> the string to int and join on them

<pre><code>SELECT *
 FROM A AS test1 INNER JOIN B AS test2 
 ON CAST(test2.tag_num AS INT) = CAST(test1.tag_num AS INT)
</code></pre>

blocks|key|3307180|text|您可以保留数据类型为NVARCHAR，并使用SUBSTRING删除以下内容中的前导零：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3307181|select+*+from+A+test1+
inner+join+B+test2+on+test1.tag_num+=
SUBSTRING(test2.tag_num,+PATINDEX('%25[%5E0]%25',+test2.tag_num%2B'.'),+LEN(test2.tag_num))|code-block|syntax|javascript|3307182|entityMap^0|A|8|M|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can keep datatype of <code>NVARCHAR</code> and use <code>SUBSTRING</code> to remove leading zero in following:

<pre><code>select * from A test1 
inner join B test2 on test1.tag_num =
SUBSTRING(test2.tag_num, PATINDEX('%[^0]%', test2.tag_num+'.'), LEN(test2.tag_num))
</code></pre>

blocks|key|361403|text|如果您希望使用索引，则希望避免在两端进行类型转换。一种方法是使用like|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|361404|select+*
from+A+inner+join
+++++B
+++++on+b.tag_num+like+'%25'+%2B+a.tag_num;|code-block|syntax|javascript|361405|然而，这太笼统了，因为通配符将匹配太多的东西。另一种方法是使用in|361406|select+*
from+A+inner+join
+++++B
+++++on+b.tag_num+in+(a.tag_num,+'0'+%2B+a.tag_num,+'00'+%2B+a.tag_num)|361407|这有一些希望，因为在b.tag_num.++Getting+a+sargable+expression+fora.tag_num`上使用索引更具挑战性。|361408|entityMap^0|W|4|0|0|V|2|0|0|A|19|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

You want to avoid doing a type conversion on both sides, if you want any hope of using an index. One method is to use <code>like</code>:

<pre><code>select *
from A inner join
 B
 on b.tag_num like '%' + a.tag_num;
</code></pre>

However, this is too general, because the wildcards will match too many things. Another approach would be to use <code>in</code>:

<pre><code>select *
from A inner join
 B
 on b.tag_num in (a.tag_num, '0' + a.tag_num, '00' + a.tag_num)
</code></pre>

This has some hope of using an index on <code>b.tag_num. Getting a sargable expression for</code>a.tag_num` is more challenging.

blocks|key|3307212|text|这只是造型的一个简化版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3307213|select+*+from+A+test1+
inner+join+B+test2+on+1*test2.tag_num+=+test1.tag_num|code-block|syntax|javascript|3307214|乘以1会自动将结果转换为int。|offset|length|style|CODE|3307215|entityMap^0|0|0|C|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Just a shorten version of casting:

<pre><code>select * from A test1 
inner join B test2 on 1*test2.tag_num = test1.tag_num
</code></pre>

Multiplying by 1 will automatically cast result to <code>int</code>.

The tables are:

<pre><code>Table A: tag_num (nvarchar(6)) Table B: tag_num (nvarchar(6))
------------------------------ ------------------------------
883 00883
</code></pre>

The query:

<pre><code>select * from A test1 
inner join B test2 on test2.tag_num = test1.tag_num
</code></pre>

return 0 rows due to the leading zeros in Table <code>B</code>.

How can I return the result

<pre><code>A.tag_num | B.tag_num
883 | 00883
</code></pre>

even though Table <code>B</code> column contain leading zeros.

Which data type should I use?

Remove leading zero in join

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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这些表包括：Table A: tag_num (nvarchar(6))       Table B: tag_num (nvarchar(6))------------------------------       ------------------------------883                 ...

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