Question

我想查询表CITY中的STATION(id, city, longitude, latitude)名称列表，这些名称的元音是他们的第一个和最后一个字符。结果不能包含重复项。

为此，我写了一个像WHERE NAME LIKE 'a%'这样的查询，它有25个条件，每个元音都用于其他每个元音，这是非常笨拙的。有没有更好的方法呢？

Answer 1

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Answer 2

在 Microsoft SQL服务器中，您可以从以下查询中实现此目的：

SELECT distinct City FROM STATION WHERE City LIKE '[AEIOU]%[AEIOU]'

或

SELECT distinct City FROM STATION WHERE City LIKE '[A,E,I,O,U]%[A,E,I,O,U]'

Answer 3

使用正则表达式。

WHERE name REGEXP '^[aeiou].*[aeiou]$'

^和$将匹配锚定到值的开头和结尾。

在我的测试中，这不会在name列上使用索引，因此需要执行完整扫描，

WHERE name LIKE 'a%a' OR name LIKE 'a%e' ...

我认为要使用一个索引，你需要使用每个测试第一个字母的查询联合。

SELECT * FROM table
WHERE name LIKE 'a%' AND name REGEXP '[aeiou]$'
UNION
SELECT * FROM table
WHERE name LIKE 'e%' AND name REGEXP '[aeiou]$'
UNION
SELECT * FROM table
WHERE name LIKE 'i%' AND name REGEXP '[aeiou]$'
UNION
SELECT * FROM table
WHERE name LIKE 'o%' AND name REGEXP '[aeiou]$'
UNION
SELECT * FROM table
WHERE name LIKE 'u%' AND name REGEXP '[aeiou]$'

Answer 4

你可以试试这个

    select city
    from station where SUBSTRING(city,1,1) in ('A','E','I','O','U') and 
    SUBSTRING(city,-1,1) in ('A','E','I','O','U');

Answer 5

您可以尝试一个简单的MySQL解决方案：

SELECT DISTINCT city FROM station WHERE city REGEXP "^[aeiou].*";

Answer 6

您可以对第一个和最后一个字符进行子字符串，并将其与IN关键字

进行比较

SELECT DISTINCT city
FROM   station
WHERE  city RLIKE '^[aeiouAEIOU].*[aeiouAEIOU]$'

Answer 7

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%'
) AND (CITY LIKE '%a' 
    OR CITY LIKE '%e'
    OR CITY LIKE '%i'
    OR CITY LIKE '%o'
    OR CITY LIKE '%u'
)

Answer 8

以下查询适用于Orale DB：

select distinct(city) from station where upper(substr(city, 1,1)) in ('A','E','I','O','U') and upper(substr(city, length(city),1)) in ('A','E','I','O','U');

Answer 9

尝试以下方法：

select distinct city 
from station 
where city like '%[aeuio]'and city like '[aeuio]%' Order by City;

Answer 10

下面的一个在MySQL中对我有用：

SELECT DISTINCT CITY FROM STATION WHERE SUBSTR(CITY,1,1) IN ('A','E','I','O','U') AND SUBSTR(CITY,-1,1) in ('A','E','I','O','U');

Answer 11

对于MS访问或MYSQL服务器

SELECT city FROM station
WHERE City LIKE '[aeiou]%'and City LIKE '%[aeiou]';

Answer 12

您还可以编写这样的硬代码，在每种情况下都进行检查，对于初学者来说很容易理解

SELECT DISTINCT CITY 
FROM STATION
WHERE CITY LIKE 'A%A' OR CITY LIKE 'E%E' OR CITY LIKE 'I%I' OR CITY LIKE 'O%O' OR
CITY LIKE 'U%U' OR CITY LIKE 'A%E' OR CITY LIKE 'A%I' OR CITY LIKE 'A%O' OR 
CITY LIKE 'A%U' OR CITY LIKE 'E%A' OR CITY LIKE 'E%I' OR CITY LIKE 'E%O' OR 
CITY LIKE 'E%U' OR CITY LIKE 'I%A' OR CITY LIKE 'I%E' OR CITY LIKE 'I%O' OR 
CITY LIKE 'I%U' OR CITY LIKE 'O%A' OR CITY LIKE 'O%E' OR CITY LIKE 'O%I' OR 
CITY LIKE 'O%U' OR CITY LIKE 'U%A' OR CITY LIKE 'U%E' OR CITY LIKE 'U%I' OR 
CITY LIKE 'U%O'

Answer 13

尝试以下代码，

SELECT DISTINCT CITY
FROM   STATIOn
WHERE  city RLIKE '^[aeiouAEIOU].*.[aeiouAEIOU]$'

Answer 14

对于oracle：

-Xlog-implicits

Answer 15

我希望这会有所帮助

select distinct city from station where lower(substring(city,1,1)) in ('a','e','i','o','u') and lower(substring(city,length(city),length(city))) in ('a','e','i','o','u') ;

Answer 16

在 MS SQL 中使用简单的左、右函数为我工作

select city from station where left(city,1) in ('a','e','i','o','u') and right(city,1) in ('a','e','i','o','u')

Answer 17

select distinct(city) from STATION 
where lower(substr(city, -1)) in ('a','e','i','o','u') 
      and lower(substr(city, 1,1)) in ('a','e','i','o','u');

Answer 18

我的简单解决方案:::

选择不同的城市

从站

在哪个城市喜欢'[a，e，i，o，u]％[a，e，i，o，u]';

Answer 19

您可以使用以下正则表达式 并反转结果 ：

^[^aeiou]|[^aeiou]$

即使输入包含单个字符，此方法也有效。它应该可以在不同的正则表达式引擎中使用。

MySQL

SELECT city
FROM (
    SELECT 'xx' AS city UNION
    SELECT 'ax'         UNION
    SELECT 'xa'         UNION
    SELECT 'aa'         UNION
    SELECT 'x'          UNION
    SELECT 'a'
) AS station
WHERE NOT city REGEXP '^[^aeiou]|[^aeiou]$'

PostgreSQL

WHERE NOT city ~ '^[^aeiou]|[^aeiou]$'

Oracle

WHERE NOT REGEXP_LIKE(city, '^[^aeiou]|[^aeiou]$')`

SQL Server

不支持正则表达式。将LIKE子句与方括号一起使用：

WHERE city LIKE '[aeiou]%' AND city LIKE '%[aeiou]'

Answer 20

您可以使用LEFT()和RIGHT()函数。 Left(CITY,1)将从左侧获得CITY的第一个字符。 Right(CITY,1)将从右开始获得CITY的第一个字符（CITY的最后一个字符）。

DISTINCT用于删除重复项。为了使比较不区分大小写，我们将使用LOWER()函数。

SELECT DISTINCT CITY
FROM STATION
WHERE LOWER(LEFT(CITY,1)) IN ('a', 'e', 'i', 'o', 'u') AND
      LOWER(RIGHT(CITY,1)) IN ('a', 'e', 'i', 'o', 'u')

Answer 21

以下两个语句都适用于Microsoft SQL SERVER

SELECT DISTINCT
    city
FROM
    station
WHERE
    SUBSTRING(lower(CITY), 1, 1) IN ('a', 'e', 'i', 'o', 'u')
    AND SUBSTRING(lower(CITY), LEN(CITY), 1) IN ('a', 'e', 'i', 'o', 'u');

SELECT DISTINCT
    City
FROM
    Station
WHERE
    City LIKE '[A, E, O, U, I]%[A, E, O, U, I]'
ORDER BY
    City;

Answer 22

在Oracle中：

SELECT DISTINCT city 
FROM station 
WHERE SUBSTR(lower(CITY),1,1) IN ('a','e','i','o','u') AND SUBSTR(lower(CITY),-1) IN ('a','e','i','o','u');

Answer 23

尝试以下方法：

select distinct city from station where city REGEXP '^[aeiou]' and city REGEXP '[aeiou]$';

Answer 24

以元音

开头尝试此操作

甲骨文：

select distinct *field* from *tablename* where SUBSTR(*sort field*,1,1) IN('A','E','I','O','U') Order by *Sort Field*;

Answer 25

在MSSQL中，这可能是这样的：

transpiled

Answer 26

SELECT DISTINCT city
FROM   station
WHERE  city RLIKE '^[^aeiouAEIOU]'OR city RLIKE'[^aeiouAEIOU]$'

Answer 27

Select distinct(city) from station where city RLIKE '^[aeiouAEIOU]'

非常简单的答案。

用于检查名称是否以元音开头和结尾的SQL查询

27 个答案: