SQL query to check if a name begins and ends with a vowel

Question

I want to query the list of CITY names from the table STATION(id, city, longitude, latitude) which have vowels as both their first and last charact

Answer 1

You can use the following regular expression and invert the result:

^[^aeiou]|[^aeiou]$

This works even if the input consists of a single character. It should work across different regex engines.

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

No regular expression support. Use LIKE clause with square brackets:

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

Answer 2

In Oracle:

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 3

Try the following:

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

Answer 4

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 5

Try the following:

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

Answer 6

The below query will do for 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');