blocks|key|1702377|text|如果1和2不在那里怎么办？如果它总是跳到下一个id，你应该选择其中的第一个按id排序的项目。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1702378|select+top+1+amt+from+table+order+by+id;++|code-block|syntax|javascript|1702379|这方面的语法可能有点偏差...我目前正在dbs之间跳转...|1702380|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|$]]

what if 1 and 2 aren't there? if it's always going to drop through to the next id you should select the first item in there ordered by id.

<pre><code>select top 1 amt from table order by id; 
</code></pre>

Syntax for this might be off a bit...I'm jumping between dbs currently...

blocks|key|698705|text|select+Amt++from+theTable+where+ID+in++(select+min(ID)+++from+theTable++where+(ID+<+3)+)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|698706|entityMap^0|0|2G|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>
select Amt 
from theTable
where ID in
 (select min(ID) 
 from theTable
 where (ID &lt; 3)
)
</code>

blocks|key|2765196|text|基于你的问题，也许有更好的方法...在不知道更多的情况下，这是我的答案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2765197|MSSQL：|2765198|select+top+1+*+from+myTable+where+ID+>=+1+order+by+ID|code-block|syntax|javascript|2765199|MySQL：|2765200|select+*+from+myTable+where+ID+>=+1+order+by+ID+limit+1|2765201|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Based on your question, there may be a better way... Without knowing more, this is my answer:

MSSQL:

<pre><code>select top 1 * from myTable where ID &gt;= 1 order by ID
</code></pre>

MySQL:

<pre><code>select * from myTable where ID &gt;= 1 order by ID limit 1
</code></pre>

blocks|key|4368710|text|这是一个特定于Oracle的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4368711|目前还不清楚你想要的解决方案有多普遍。|4368712|对于硬编码值1和2：|4368713|select+amt+from+(
++select+amt+from+table+where+id+between+1+and+2+order+by+id+asc
)+where+rownum+=1|code-block|syntax|javascript|4368714|将1和2替换为n和n%2B1，以泛化任何起始值n。|offset|length|style|CODE|4368715|如果您真的希望找到比您传入的ID更大的现有最低ID，则删除上限2，无论它有多高。|4368716|entityMap^0|0|0|0|0|7|1|9|3|L|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]|$M|11|N|12|O|P]|$M|13|N|14|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|T|$]]

This is an Oracle-specific answer.

It's not clear how general a solution you want.

For hardcoded values 1 and 2:

<pre><code>select amt from (
 select amt from table where id between 1 and 2 order by id asc
) where rownum =1
</code></pre>

Replace 1 and 2 with <code>n</code> and <code>n+1</code> to generalize for any starting value <code>n</code>.

Remove the upper bound of 2 if you really want to find the lowest existing ID greater than the one you passed in, regardless of how high it is.

blocks|key|1698914|text|您可能希望使用EXISTS+stament并创建一个函数http://www.sqlteam.com/article/using-exists|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1698915|entityMap|0|LINK|mutability|MUTABLE|url|http://www.sqlteam.com/article/using-exists^0|S|17|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

you probably want to use EXISTS stament and create a function
<a href="http://www.sqlteam.com/article/using-exists" rel="nofollow">http://www.sqlteam.com/article/using-exists</a>

blocks|key|4368765|text|如果数据库是SQL+Server，则可以使用案例状态来实现这一点|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4368766|http://msdn.microsoft.com/en-us/library/ms181765.aspx|offset|length|4368767|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|1H|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

If database is SQL Server you can achieve this with a CASE State

<a href="http://msdn.microsoft.com/en-us/library/ms181765.aspx" rel="nofollow">http://msdn.microsoft.com/en-us/library/ms181765.aspx</a>

blocks|key|4368800|text|看起来您想要ID最小的记录|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4368801|针对MySQL、SQLite、PostgreSQL的使用限制|4368802|select+amt+from+tbl+order+by+id+limit+1;|code-block|syntax|javascript|4368803|对于SQL+Server，Sybase使用top|4368804|select+top+1+amt+from+tbl+order+by+id;|4368805|对于Oracle，使用rownum|4368806|SELECT+*+FROM+(select+amt+from+tbl+order+by+id)+X+WHERE+ROWNUM=1;|4368807|对于ANSI+SQL92方法|4368808|SELECT+AMT+FROM+TBL+WHERE+ID+=+(SELECT+MIN(ID)+FROM+TBL)|4368809|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

It looks like you want the record with the lowest ID

For MySQL, SQLite, PostgreSQL use limit

<pre><code>select amt from tbl order by id limit 1;
</code></pre>

For SQL Server, Sybase use top

<pre><code>select top 1 amt from tbl order by id;
</code></pre>

For Oracle use rownum

<pre><code>SELECT * FROM (select amt from tbl order by id) X WHERE ROWNUM=1;
</code></pre>

For an ANSI SQL92 approach

<pre><code>SELECT AMT FROM TBL WHERE ID = (SELECT MIN(ID) FROM TBL)
</code></pre>

I have a table like below,

<pre><code>ID | Amt
-------
1 | $30
2 | $2
3 | $50
</code></pre>

I want to do a select on the amount based on the ID and the thing is that ID=1 is not guarantee to be there. So, I want to check if ID=1 exist then I select the amount correspond to the ID=1 else I want to select the amount where ID=2.

So something like,

<pre><code>Select (If ID= 1 then amt else amt(for ID2) from Table 
</code></pre>

How can I achieve this Sql select statement?

Thanks.
.

SQL select on condition

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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我有一个类似下面的表格，ID | Amt-------1  | $302  | $23  | $50我想根据ID在数量上做一个选择，问题是ID=1不能保证在那里。因此，我想检查ID=1是否存在，然后选择与ID=1对应的数量，否则我想选择ID=2所在的数量。所以像这样，Select (If ID= 1 then amt ...

问SQL select on condition
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