Split address column in streetname and number using SELECT statement - mysql

Question

I have a mysql table with an address column.
Now I need to SELECT the streetname and number separately.
Address
Wallstreet 20
New Yorkavenue 30
New London Street 40
Needs to be:
Street: Number:
Wallstreet 20
New Yorkavenue 30
New London Street 40
Any ideas?
Thanks in advance!

Answer 1

If you assume that the number is the final "word" and separated by a space:
select replace(address, substring_index(address, ' ', -1), '') as street,
substring_index(address, ' ', -1) as number
I happen to think that those two assumptions are very big assumptions, meaning that this might not work on all your rows.

Answer 2

For Mysql probably you could create a MYSQL SUBSTRING_INDEX to separate the fields if the numbers are only in the address number and the address has no numbers.
Example
SELECT
REPLACE(address, SUBSTRING_INDEX(address, ' ', -1), '') as ADDRESS,
SUBSTRING_INDEX(address, ' ', -1) as NUMBER
FROM
ADDRESSES
it's not a really good method in performance and probably clould be done with other ways but if the schema is allways like the example it could works
Also probably is better in performance to do it on client side in the language that fetch the data.

Answer 3

You could use some string functions, like SUBSTRING_INDEX and LEFT.
Getting the Number is easy:
SELECT
SUBSTRING_INDEX(Street, ' ', -1)
(yes, it's not actually a number, but I suppose it's the last part of the string after the last space, it can also be a string as 20/C).
Getting the street name is a little more tricky:
SELECT
LEFT(Street,
CHAR_LENGTH(Street)
-CHAR_LENGTH(SUBSTRING_INDEX(Street, ' ', -1))
-1
) AS street_name,
SUBSTRING_INDEX(Street, ' ', -1) AS street_number
FROM
tablename

Answer 4

I'm may be some what late, but issues are still the same nower days :-)
The question was, how to split/select a streetname (Straßenname) and the number (Hausnummer) out of a adress string?
Therefor I created a function, to implement German DIN 5008:
DROP FUNCTION IF EXISTS HAUSNUMMER_DIN_5008;
delimiter //
CREATE DEFINER=`vlw`#`%` FUNCTION `HAUSNUMMER_DIN_5008`(
oldStreet VARCHAR(255), formating BOOL
) RETURNS varchar(16) CHARSET utf8
BEGIN
SET #oldString := oldStreet;
SET #newString := "";
tokenLoop: LOOP
END LOOP tokenLoop;
-- are there no figures at the beginning of street name?
IF NOT #oldString REGEXP '^[1-9]' THEN
-- must be a word, to jump over
SET #splitPoint := LOCATE(" ", #oldString);
SET #oldString := SUBSTRING(#oldString, #splitPoint+1);
ELSE
-- Okay, we found the first figure
-- Are there any chars inside the string including "."
IF #oldString REGEXP '[a-z,A-Z,.,--,/," "]' THEN
-- Are there any char directly behind a figure
IF #oldString REGEXP '[0-9][a-z,A-Z,.,--,/," "]' THEN
-- now we have to check step by step
SET #i := 1;
tokenPos: LOOP
-- jump over the first figures
IF NOT SUBSTRING(#oldString, #i, 1) REGEXP '[0-9]' THEN
-- this is the first non figure
IF formating THEN
IF SUBSTRING(#oldString, #i, 1) REGEXP '[a-z,A-Z]' THEN
-- If a char is directly written after figures, then add a blank between
SET #oldString := CONCAT(SUBSTRING(#oldString,1,#i-1)," ",SUBSTRING(#oldString,#i));
LEAVE tokenPos;
ELSE
IF SUBSTRING(#oldString, #i, 1) = "." THEN
-- this must be part of the street name, so we will loop some what
LEAVE tokenPos;
END IF;
-- SET #newString := concat(">xx>",REPLACE(#oldString," ",""));
SET #newString := REPLACE(#oldString," ","");
LEAVE tokenLoop;
END IF;
LEAVE tokenPos;
ELSE
IF SUBSTRING(#oldString, #i, 1) = "." THEN
-- this must be part of the street name, so we will loop some what
LEAVE tokenPos;
ELSE
-- the street number is found
SET #newString := #oldString;
LEAVE tokenLoop;
END IF;
END IF;
END IF;
SET #i := #i+1;
END LOOP tokenPos;
END IF;
SET #splitPoint := LOCATE(" ", #oldString);
IF SUBSTRING(#oldString, #splitPoint+1) REGEXP '[1-9]' THEN
-- we have to split one more word
SET #oldString := SUBSTRING(#oldString, #splitPoint+1);
ELSE
SET #newString := #oldString;
LEAVE tokenLoop;
END IF;
ELSE
IF formating AND #oldString REGEXP '[//][1-9]' THEN
SET #i := LOCATE(#oldString,"//")+4;
SET #oldString := CONCAT(SUBSTRING(#oldString,1,#i)," ",SUBSTRING(#oldString,#i+1));
ELSEIF formating THEN
SET #oldString := REPLACE(#oldString," ","");
END IF;
SET #newString := #oldString;
LEAVE tokenLoop;
END IF;
END IF;
RETURN #newString;
END //
delimiter ;
There is an additional BOOLEAN parameter, to reformat the number part of the address against DIN 5008. But the reformating part isn't finaly done yet.
Now we can test it with some examples:
select HAUSNUMMER_DIN_5008("Mörikestr. 28/3",TRUE);
select HAUSNUMMER_DIN_5008("Nettelbeckstraße 6 a",TRUE);
select HAUSNUMMER_DIN_5008("Auf dem Brande 19a",TRUE); ==>> "19 a"
select HAUSNUMMER_DIN_5008("Auf dem Brande 19a",FALSE); ==>> "19a"
select HAUSNUMMER_DIN_5008("Anger 1-3",TRUE);
select HAUSNUMMER_DIN_5008("Straße des 17. Juni 12-16",TRUE);
select HAUSNUMMER_DIN_5008("L11 2",TRUE); -- z.B in Mannheim
select HAUSNUMMER_DIN_5008("111. 2",TRUE);
select HAUSNUMMER_DIN_5008("Züricher Straße 17// 28",,TRUE);
-- Some special formating tests
SELECT HAUSNUMMER_DIN_5008("Mörikestr. 28 / 3",TRUE);
SELECT HAUSNUMMER_DIN_5008("8 - 6",TRUE);
SELECT HAUSNUMMER_DIN_5008("Straße des 17. Juni 12 - 16",TRUE);
SELECT HAUSNUMMER_DIN_5008("Straße des 17. Juni 12- 16",TRUE);
SELECT HAUSNUMMER_DIN_5008("Straße des 17. Juni 12 -16",FALSE);
-- Next one is not DIN 5008, but was a loop issue inside function
SELECT HAUSNUMMER_DIN_5008("8 /App.6",FALSE);
SELECT HAUSNUMMER_DIN_5008("8 /App.6",TRUE);
SELECT HAUSNUMMER_DIN_5008("8/App.6",TRUE);
If you need only the street name, you have to use HAUSNUMMER_DIN_5008() with formating FALSE, otherwise you can't find the numberpart within your adress.
SET #address := "Auf dem Brande 19a";
SELECT SUBSTRING(#address, 1, LOCATE(HAUSNUMMER_DIN_5008(#address, FALSE),#address)-1);
SET #address := "Straße des 17. Juni 12-16";
SELECT SUBSTRING(#address, 1, LOCATE(HAUSNUMMER_DIN_5008(#address, FALSE),#address)-1);
That are my 5 cent
Christian Eickhoff

Answer 5

ERROR 1064 (42000) at line 5: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'END LOOP tokenLoop;
IF NOT #oldString REGEXP '^[1-9]' THEN
SET #sp' at line 9

Answer 6

For everyone coming from Google: As you can't trust humans, I chose this way:
SELECT
REGEXP_SUBSTR(address, '[a-z"äöüß"\-\. ]+') AS street ,
REGEXP_SUBSTR(address, '[0-9]+.*') AS number,
FROM
ADDRESSES
Because this will also work on:
Streetname 5
Streetname 5B
Streetname 5 B
Streetn.5B
Street-Name5
Streetname 5 (+trailing Space)
Street name5