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

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 characters. The result cannot contain duplicates.

For this is I wrote a query like WHERE NAME LIKE 'a%' that had 25 conditions, each vowel for every other vowel, which is quite unwieldy. Is there a better way to do it?

You could use a regular expression:

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

in Microsoft SQL server you can achieve this from below query:

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

Or

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

Update --Added Oracle Query

> --Way 1 --It should work in all Oracle versions

SELECT DISTINCT CITY FROM STATION WHERE REGEXP_LIKE(LOWER(CITY), '^[aeiou]') and  REGEXP_LIKE(LOWER(CITY), '[aeiou]$');

> --Way 2 --it may fail in some versions of Oracle

SELECT DISTINCT CITY FROM STATION WHERE REGEXP_LIKE(LOWER(CITY), '^[aeiou].*[aeiou]');

> --Way 3 --it may fail in some versions of Oracle

SELECT DISTINCT CITY FROM STATION WHERE REGEXP_LIKE(CITY, '^[aeiou].*[aeiou]', 'i');

Use a regular expression.

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

^ and $ anchor the match to the beginning and end of the value.

In my test, this won't use an index on the name column, so it will need to perform a full scan, as would

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

I think to make it use an index you'd need to use a union of queries that each test the first letter.

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]$'

You can try one simple solution for MySQL:

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

You may try this

    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');

You could substring the first and last character and compare it with IN keyword,

WHERE SUBSTRING(NAME,1,1) IN (a,e,i,o,u) AND SUBSTRING(NAME,-1) IN (a,e,i,o,u) 

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'
)

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');

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]'

The below one worked for me in MySQL:

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');

Try the following:

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

I hope this will help

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') ;

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');

In MSSQL, this could be the way:

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

Try this for beginning with vowel

Oracle:

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

For MS access or MYSQL server

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

you can also do a hard code like this, where you are checking each and every case possible, it's easy to understand for beginners

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'

You can use LEFT() and RIGHT() functions. Left(CITY,1) will get the first character of CITY from left. Right(CITY,1) will get the first character of CITY from right (last character of CITY).

DISTINCT is used to remove duplicates. To make the comparison case-insensitive, we will use the LOWER() function.

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')

Try this below code,

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

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');

Both of the statements below work in 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;

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

Try the following:

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

For oracle :

SELECT DISTINCT city
FROM   station
WHERE REGEXP_LIKE(city, '^[aeiou].*[aeiou]$','i') ;

My simple solution

SELECT DISTINCT CITY
FROM STATION
WHERE CITY  LIKE '[a,e,i,o,u]%[a,e,i,o,u]';

Worked for me using simple left, right functions in MS SQL

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

SELECT DISTINCT CITY From STATION WHERE LOWER(SUBSTR(CITY,1,1)) IN ('a','e','i','o','u');

this will work in my sql

This works on oracle:

select distinct COLUMN_NAME
from TABLE_NAME
where 
    --convert first character and check for vowel
    lower(substr(COLUMN_NAME, 1, 1)) IN ('a', 'e', 'i', 'o', 'u')
    and
    --convert last character and check for vowel
    lower(substr(COLUMN_NAME, -1, 1)) IN ('a', 'e', 'i', 'o', 'u')
order by COLUMN_NAME asc;

If you are using Sequal server management studio (SSMS)

SELECT DISTINCT CITY FROM STATION WHERE CITY LIKE '[a,e,i,o,u]%[a,e,i,o,u]';

This works in MYSQL:

SELECT DISTINCT CITY
FROM STATION
WHERE CITY REGEXP '^[a,e,i,o,u]' AND 
CITY REGEXP '[a,e,i,o,u]$';

Try this as I tried and it worked for me.

SELECT DISTINCT CITY 
FROM STATION  
WHERE CITY REGEXP '^[aeiou]' AND CITY REGEXP '[aeiou]$';

FOR Query the list of CITY names from STATION which have vowels (i.e., a, e, i, o, and u) as both their first and last characters. Your result cannot contain duplicates.

SELECT MY SQL AS the Preferred Engine

SELECT DISTINCT(CITY) FROM STATION WHERE CITY REGEXP '^[AEIOUaeiou]' AND CITY REGEXP '[AEIOUaeiou]$' ;

Very simple answer.