entityMap|blocks|key|c82q0|text|SELECT+DISTINCT+CITY
FROM+STATION
WHERE+MOD(ID,+2)+=+0|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|bfsbt|unstyled^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|G|9|@]|A|@]|B|$C|D]]|$2|E|4|-4|6|F|8|H|9|@]|A|@]|B|$]]]]

<pre><code>SELECT DISTINCT CITY
FROM STATION
WHERE MOD(ID, 2) = 0
</code></pre>


entityMap|blocks|key|k6m1|text|您可以尝试此查询：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|dt5vp|SELECT+DISTINCT+CITY+FROM+STATION+WHERE+(ID+%25+2)+=+0+|code-block|syntax|javascript|dh3fc^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>You can try this query:</p>

<pre><code>SELECT DISTINCT CITY FROM STATION WHERE (ID % 2) = 0 
</code></pre>


entityMap|blocks|key|dlmsu|text|对于DB2:-从工作站选择不同的城市|type|unstyled|depth|inlineStyleRanges|entityRanges|data|9v6ki|其中MOD(ID，2)+=+0；|69jhs|对于Oracle:-从工作站选择不同的城市|5eaq0|其中MOD(ID，2)+=+0；|4nbe2|对于MySQL:-+SELECT+DISTINCT+CITY+FROM+STATION+WHERE+MOD(ID，2)+=+0；--或者--+SELECT+DISTINCT+CITY+FROM+STATION|3q2d7|其中ID%252=0；|3444e|对于MSSQL:-从站点选择不同的城市|5c442|其中ID%252=0；|dbsik^0|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|R|9|@]|A|@]|B|$]]|$2|C|4|D|6|7|8|S|9|@]|A|@]|B|$]]|$2|E|4|F|6|7|8|T|9|@]|A|@]|B|$]]|$2|G|4|H|6|7|8|U|9|@]|A|@]|B|$]]|$2|I|4|J|6|7|8|V|9|@]|A|@]|B|$]]|$2|K|4|L|6|7|8|W|9|@]|A|@]|B|$]]|$2|M|4|N|6|7|8|X|9|@]|A|@]|B|$]]|$2|O|4|P|6|7|8|Y|9|@]|A|@]|B|$]]|$2|Q|4|-4|6|7|8|Z|9|@]|A|@]|B|$]]]]

<p>For DB2:-
SELECT DISTINCT CITY 
FROM STATION</p>

<h2>WHERE MOD(ID, 2) = 0;</h2>

<p>For Oracle:-
SELECT DISTINCT CITY 
FROM STATION</p>

<h2>WHERE MOD(ID, 2) = 0;</h2>

<p>For MySQL:-
SELECT DISTINCT CITY 
FROM STATION
WHERE MOD(ID, 2) = 0;
-- Or --
SELECT DISTINCT CITY 
FROM STATION</p>

<h2>WHERE ID%2=0;</h2>

<p>For MSSQL:-
SELECT DISTINCT CITY 
FROM STATION</p>

<h2>WHERE ID%2=0;</h2>


entityMap|blocks|key|9tqfi|text|SELECT+distinct+CITY+FROM+STATION
WHERE+MOD(ID,2)=0;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|susk|unstyled^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|G|9|@]|A|@]|B|$C|D]]|$2|E|4|-4|6|F|8|H|9|@]|A|@]|B|$]]]]

<pre><code>SELECT distinct CITY FROM STATION
WHERE MOD(ID,2)=0;
</code></pre>


entityMap|blocks|key|7dei4|text|MYSQL：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1qalg|SELECT+DISTINCT+CITY+FROM+STATION+WHERE+(ID+%25+2)=0;|code-block|syntax|javascript|1lb0i|ORACLE：|a9o6j|SELECT+DISTINCT+CITY+FROM+STATION+WHERE+MOD(ID,+2)=0;|ehas7^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|N|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|O|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

<p>MYSQL: </p>

<pre><code>SELECT DISTINCT CITY FROM STATION WHERE (ID % 2)=0;
</code></pre>

<p>ORACLE: </p>

<pre><code>SELECT DISTINCT CITY FROM STATION WHERE MOD(ID, 2)=0;
</code></pre>


entityMap|blocks|key|j4sh|text|从MOD(ID，2)=0的站点中选择唯一的城市；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|260n4^0|0^^$0|$]|1|@$2|3|4|5|6|7|8|D|9|@]|A|@]|B|$]]|$2|C|4|-4|6|7|8|E|9|@]|A|@]|B|$]]]]

<p>SELECT UNIQUE CITY FROM STATION WHERE MOD(ID,2)=0;</p>


entityMap|blocks|key|7s0k9|text|SELECT+DISTINCT+CITY+FROM+STATION+WHERE+(ID+%25+2)+=+0;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|c40mj|Distinct表示不允许重复，(Id+%25+2)告诉SQL将Id除以2并返回余数。因为我们只寻找偶数值，所以我们将其设置为"=+0"，如2%252=0、4%252=0等。如果我们寻找奇数Ids，我们将使用"=+1“，因为当奇数除以2时总是有1的余数。希望这会有帮助！|unstyled|4j6gn^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|J|9|@]|A|@]|B|$]]|$2|H|4|-4|6|G|8|K|9|@]|A|@]|B|$]]]]

<pre><code>SELECT DISTINCT CITY FROM STATION WHERE (ID % 2) = 0;
</code></pre>

<p>Distinct for not allowing Duplicates, (Id % 2) tells SQL to divide the Id by 2 and return the remainder. Since we're looking for only even values we set this to "= 0", as 2%2=0, 4%2=0, etc. If we were looking for odd-numbered Ids, we'd use "= 1" instead, since there's always a remainder of 1 when dividing odd numbers by 2. Hope this helps!</p>


entityMap|blocks|key|peba|text|在MySql中，您可以执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|51u2t|SELECT+DISTINCT+CITY+FROM+STATION+WHERE+(ID+%25+2)=0+ORDER+BY+CITY+;|code-block|syntax|javascript|e9ni^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>in MySql, you can do :</p>

<pre><code>SELECT DISTINCT CITY FROM STATION WHERE (ID % 2)=0 ORDER BY CITY ;
</code></pre>


entityMap|blocks|key|4blh5|text|如果问题想要“查询仅具有偶数ID号的车站的城市名称列表”，那么答案将是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3dss5|select+CITY+from+station+where+ID+%25+2+=+0;|code-block|syntax|javascript|fusn8|但由于进一步的标准，如“但必须从你的答案中排除重复项”。你的答案应该是|6gj3|select+DISTINCT+CITY+from+station+where+ID+%25+2+=+0;|7vcca^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|N|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|O|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|P|9|@]|A|@]|B|$F|G]]|$2|L|4|-4|6|7|8|Q|9|@]|A|@]|B|$]]]]

<p>If the question would like "Query a list of CITY names from STATION with even ID numbers only" then the answer would be</p>

<pre><code>select CITY from station where ID % 2 = 0;
</code></pre>

<p>But due the further criteria like "But must exclude duplicates from your answer." your answer should be </p>

<pre><code>select DISTINCT CITY from station where ID % 2 = 0;
</code></pre>


entityMap|blocks|key|aqlda|text|我建议另一种解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bs7no|++++SELECT+DISTINCT+CITY
++++FROM+STATION
++++WHERE+ID+LIKE+'%250'+OR+ID+LIKE+'%252'+OR+ID+LIKE+'%254'+OR+ID+LIKE+'%256'+OR+ID+LIKE+'%258';|code-block|syntax|javascript|ak7t3^0|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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>I suggest another solution:</p>

<pre class="lang-sql prettyprint-override"><code>    SELECT DISTINCT CITY
    FROM STATION
    WHERE ID LIKE '%0' OR ID LIKE '%2' OR ID LIKE '%4' OR ID LIKE '%6' OR ID LIKE '%8';
</code></pre>


entityMap|blocks|key|9km0m|text|在MSSQL中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5i4db|SELECT+CITY
FROM+STATION
WHERE+(ID+%25+2)+=+0
GROUP+BY+CITY+|code-block|syntax|javascript|5fbrk^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>In MSSQL:</p>
<pre><code>SELECT CITY
FROM STATION
WHERE (ID % 2) = 0
GROUP BY CITY 
</code></pre>


entityMap|blocks|key|2ret8|text|在MYSQL查询中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|era2e|++++select+distinct+city+from+station+where+(ID+%25+2)=0;|code-block|syntax|javascript|9lean^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]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

<p>in MYSQL query</p>

<pre><code>    select distinct city from station where (ID % 2)=0;
</code></pre>


<p>I'm trying to solve the Hackerrank problem <a href="https://www.hackerrank.com/challenges/weather-observation-station-3?h_r=next-challenge&amp;h_v=zen" rel="noreferrer">Weather Observation Station 3</a> which is stated below:</p>

<p><a href="https://i.stack.imgur.com/QeMt2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/QeMt2.png" alt="enter image description here"></a></p>

<p>I came up with the following solution:</p>

<pre><code>SELECT CITY
FROM STATION
WHERE MOD(ID, 2) = 0
</code></pre>

<p>but when I try to run it I get the compiler message "Wrong answer". I don't see what's wrong with this query?</p>


How to select columns with an even ID number 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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 我正在尝试解决Hackerrank问题Weather Observation Station 3，如下所述： ​ ? ​ 我想出了以下解决方案： SELECT CITYFROM STATIONWHERE MOD(ID, 2) = 0 但是当我尝试运行它时，我得到了编译器消息“错误的答案”。我看不出这个查询有什么问题？ 

问如何在SQL中选择ID为偶数的列？
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问如何在SQL中选择ID为偶数的列？EN

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问如何在SQL中选择ID为偶数的列？
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