MySQL query to categorised address based on values using regular Expression - mysql

Question

I have a table which has one column of addresses. I have some 10 - 11 places name.
When i query that table using 'Select * ...', i want to create a new column which matches the values with address fields and store that values into new column of exist else 'Not Found'.
The table has address column as below. I want to extract areas from it such as BTM Layot, Wilson Garden
When i do the select query, the output should be that address field and one more field which will give me the abstract location area from address field. And if any value does not matches the address field then it shoud display as 'Area Nt Specified'

Answer 1

Consider a cross join query (query with no joins but a list of tables in FROM clause) between the larger table of addresses (t1) and smaller table of your 10-11 places (t2) holding BTM Layot, Wilson Garden... values. This will be scalable instead of manually entering/editing places in an IN clause.
Then use a LIKE expression in a WHERE clause to match the places which are a part of the larger address string. However, to return all original address values with matched places use the LEFT JOIN...NOT NULL query with cross join as derived table (sub).
SELECT `maintable`.`address`, IFNULL(sub.`place`, 'Area Nt Specified') As matchplaces
FROM `maintable`
LEFT JOIN
(SELECT t1.ID, t1.address, t2.place
FROM `maintable` As t1,
(SELECT `place` FROM `placestable`) As t2
WHERE t1.address LIKE Concat('%',t2.place,'%')) As sub
ON `maintable`.ID = sub.ID
WHERE `maintable`.ID IS NOT NULL;
If really need to use regular expression, replace the LIKE expression in derived table with below:
WHERE t1.address regexp t2.place

Answer 2

If you have a list of know places, then you can do:
select (case when address regexp '(, BTM Layout|, Bapuji Nagar|, Adugodi)$'
then substring_index(address, ', ', -1)
else 'Not Found'
end)
You can expand the regular expression to include as many places as you like.
Or alternatively, you don't really need a regular expression:
select (case when substring_index(address, ', ', -1) in ('BTM Layout', 'Bapuji Nagar', 'Adugodi', . . .)
then substring_index(address, ', ', -1)
else 'Not Found'
end)