Sql 获取城市名称不要以元音开头或以元音结尾

从电台查询不以元音开头或不以元音结尾的城市名称列表。您的结果不能包含重复项

输入格式

车站表描述如下：

我写了下面的查询，但效果很好。有什么建议吗

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从城市不象“^[aeiouAEIOU].[aeiouAEIOU]$”的车站选择不同的城市
假设您使用的是MySQL，下面是您要查找的内容

select distinct city
from station 
where not lower(substr(city,1,1)) in ('a','e','i','o','u')
      or not lower(substr(city,-1,1)) in ('a','e','i','o','u');

选择城市与车站不同的城市，如“^[^aeiouAEIOU].*.[^aeiouAEIOU]$”

select distinct city
from station 
where not lower(substr(city,1,1)) in ('a','e','i','o','u')
      or not lower(substr(city,-1,1)) in ('a','e','i','o','u');

脚注：和
此SQL查询将对您有所帮助。
如果您使用的是MS SQL，请遵循以下代码行：

SELECT DISTINCT CITY 
FROM STATION  
WHERE CITY NOT LIKE '[AEIOU]%' OR CITY NOT LIKE '%[aeiou]';

SELECT DISTINCT CITY 
FROM STATION 
WHERE (CITY NOT IN (SELECT DISTINCT CITY FROM STATION WHERE CITY LIKE '%a' OR CITY LIKE '%e' OR CITY LIKE '%i' OR CITY LIKE '%o' OR CITY LIKE '%u'))
OR 
(CITY NOT IN (SELECT CITY FROM STATION WHERE CITY LIKE 'A%' OR CITY LIKE 'E%' OR CITY LIKE 'I%' OR CITY LIKE 'O%' OR CITY LIKE 'U%'));

如果您使用的是MySQL，请遵循以下代码行：

SELECT DISTINCT CITY 
FROM STATION  
WHERE CITY NOT LIKE '[AEIOU]%' OR CITY NOT LIKE '%[aeiou]';

SELECT DISTINCT CITY 
FROM STATION 
WHERE (CITY NOT IN (SELECT DISTINCT CITY FROM STATION WHERE CITY LIKE '%a' OR CITY LIKE '%e' OR CITY LIKE '%i' OR CITY LIKE '%o' OR CITY LIKE '%u'))
OR 
(CITY NOT IN (SELECT CITY FROM STATION WHERE CITY LIKE 'A%' OR CITY LIKE 'E%' OR CITY LIKE 'I%' OR CITY LIKE 'O%' OR CITY LIKE 'U%'));

请尝试以下操作：

SELECT city 
FROM station 
WHERE left(city,1) not regexp 'a|e|i|o|u' or right(city,1) not regexp 'a|e|i|o|u' 
GROUP BY city

SELECT city 
FROM station 
WHERE left(city,1) not regexp 'a|e|i|o|u' or right(city,1) not regexp 'a|e|i|o|u' 
GROUP BY city

如果您使用的是
Oracle
，此SQL查询将对您有所帮助：

SELECT DISTINCT city
FROM station
WHERE regexp_like (city, '^[^aeiouAEIOU].*') 
OR regexp_like (city, '.*[^aeiouAEIOU]$');

如果您使用的是Oracle，另一个答案是：

SELECT DISTINCT (CITY) 
FROM STATION 
WHERE NOT regexp_like(lower(CITY),'^[aeiou].*[aeiou]$');

您必须将插入符号字符放在方括号内，这意味着没有任何元音。解释是我用过的吗

SELECT DISTINCT CITY FROM STATION
WHERE CITY NOT LIKE 'a%' 
AND CITY NOT LIKE 'e%'
AND CITY NOT LIKE 'i%'
AND CITY NOT LIKE 'o%'
AND CITY NOT LIKE 'u%'
AND CITY NOT LIKE '%a' 
AND CITY NOT LIKE '%e'
AND CITY NOT LIKE '%i'
AND CITY NOT LIKE '%o'
AND CITY NOT LIKE '%u'

试试这个：

select DISTINCT city from STATION 
where  (CITY NOT LIKE 'a%'
        AND CITY  NOT LIKE 'e%'
        AND CITY NOT LIKE 'i%' 
        AND CITY NOT LIKE 'o%'
        AND CITY NOT LIKE 'u%')
OR (CITY NOT LIKE '%a' AND
    CITY  NOT LIKE '%e' AND
    CITY NOT LIKE '%i' AND 
    CITY NOT LIKE '%o' AND 
    CITY NOT LIKE '%u')

SELECT DISTINCT CITY FROM STATION WHERE CITY NOT RLIKE '^[aeiouAEIOU]' 
AND
CITY NOT RLIKE '[aeiouAEIOU]$'

这对我有用

SELECT DISTINCT CITY FROM STATION WHERE NOT CITY RLIKE '^[AEIOUaeiou]' AND NOT CITY RLIKE '.*[AEIOUaeiou]$' 

Oracle解决方案：

选择与所处车站不同的城市
不象REGEXP_（上（城市），“^[AEIOU]”或
不象REGEXP_（上（城市），“[AEIOU]$”；
正确回答这个问题

SELECT DISTINCT CITY FROM STATION
WHERE NOT CITY RLIKE '^[AEIOUaeiou]' AND NOT CITY RLIKE '.*[AEIOUaeiou]$';

mysql查询：
从城市非REGEXP'^[aeiouAEIOU].[aeiouAEIOU]$所在的车站选择不同的城市
以下查询应该执行您想要的操作。它成功地为我工作

select distinct city
from station 
where not lower(substr(city,1,1)) in ('a','e','i','o','u')
      or not lower(substr(city,-1,1)) in ('a','e','i','o','u');

SELECT DISTINCT CITY 
FROM STATION
WHERE CITY NOT RLIKE '^[aeiouAEIOU]' OR CITY NOT RLIKE '[aeiouAEIOU]$'

这对我使用MYSQL很有效。我还按城市对结果进行分组，这是不需要的…只是更漂亮：

select distinct CITY
from STATION
where CITY NOT RLIKE '^[aeiouAEIOU]'
  OR CITY NOT RLIKE '[AEIOUaeiou]$'
GROUP BY CITY;

您可以在MySQL上尝试以下方法：

SELECT DISTINCT(CITY) 
FROM STATION 
WHERE CITY NOT REGEXP '^[aeiou]' 
    AND CITY NOT REGEXP '[aeiou]$';

此查询将为所问问题提供所需的输出。
这将按照您希望的方式工作：

这就是你要找的bro。
假设你在黑客排名上看到了这个问题，并且给出了表站模式的屏幕截图，也说明了同样的问题。这是我的答案，它在MYSQL中对我有效

SELECT distinct city FROM station 
WHERE City not LIKE '[aeiou]%' and CITY NOT LIKE '%[aeiou]'

select DISTINCT(city) from STATION where CITY REGEXP '^[^aeiou]' or CITY REGEXP '.*[^aeiou]$';


这是MS SQL SERVER所需的解决方案

从城市不象“[aeiouAEIOU]”和城市不象“[aeiouAEIOU]”的站点中选择不同的城市
这个怎么样：

SELECT CITY from STATION WHERE LOWER(CITY) NOT REGEXP '^[aeiou].*[aeiou]$' GROUP BY CITY

试试这个：

select DISTINCT city from STATION 
where  (CITY NOT LIKE 'a%'
        AND CITY  NOT LIKE 'e%'
        AND CITY NOT LIKE 'i%' 
        AND CITY NOT LIKE 'o%'
        AND CITY NOT LIKE 'u%')
OR (CITY NOT LIKE '%a' AND
    CITY  NOT LIKE '%e' AND
    CITY NOT LIKE '%i' AND 
    CITY NOT LIKE '%o' AND 
    CITY NOT LIKE '%u')

SELECT DISTINCT CITY FROM STATION WHERE CITY NOT RLIKE '^[aeiouAEIOU]' 
AND
CITY NOT RLIKE '[aeiouAEIOU]$'

此代码适用于MS SQL SERVER。
适用于MySQL
从城市注册表“^[^aeiou]|[^aeiou]$”所在的车站选择不同的城市
这是哪个数据库？大多数数据库不支持正则表达式。使用数据库支持的函数。它以什么方式不起作用？您使用的是什么？“SQL”只是一种查询语言，而不是特定数据库产品的名称。请为您正在使用的数据库产品添加一个标记，
postgresql
，
oracle
，
sqlserver
，
db2
，…@AriSingh:实际上大多数DBMS都支持正则表达式。但是由于没有真正的SQL标准（至少对于真正的POSIX正则表达式来说），每个DBMS产品的语法都非常不同，“或者/或者”听起来像是允许以元音开头和结尾？您假设OP使用的是
MySQL
。当问题被半心半意地贴上标签时回答，只会让其他提出类似问题的人感到困惑。同意！编辑了我的答案，尽管我只是在OPs答案中修改了查询。尽管MySQL中不同数据库的RLIKE不同，
RLIKE
和
REGEXP
是等效的。@AnandG MS SQL Server中会有什么？请解释为什么这是一个有效的答案并格式化代码。请，如果您为您的解决方案添加一些解释文本，答案将更加完整。仅包含代码的答案不被视为completeAvoid，使用语句的所有大写字母！欢迎来到SO。您应该添加一些解释，并格式化您的代码（在本例中使用三个反勾号和sql语法提示）。欢迎使用stackoverflow！在提供答案时，请添加上下文来解释代码中发生了什么。仅为代码的答案不完整。在MySQL中工作。关于“^[aeiou].[aeiou]$”，如何理解此处的逗号和星号？我看到这是一个流行的答案，但现在确定如何理解。*1。没有逗号，请检查。2. 
*
表示字符串应该从（aeiou）中的任何一个开始，并且可以对该字母进行零次或多次迭代。
select distinct city
from station 
where not lower(substr(city,1,1)) in ('a','e','i','o','u')
      or not lower(substr(city,-1,1)) in ('a','e','i','o','u');

SELECT DISTINCT city
FROM   station
WHERE  city REGEXP '^[^aeiouAEIOU]|[^aeiouAEIOU]$'

SELECT DISTINCT CITY 
FROM STATION 
WHERE 
  SUBSTRING(CITY,1,1) NOT IN ('A','E','I','O','U') 
  OR 
  SUBSTRING(CITY,1,1) NOT IN ('a','e','i','o','u')