Question

I want to list (a sorted list) all my entries from an attribute called streetNames in my table/relation Customers. eg. I want to achieve the following order:

Street_1A
Street_1B
Street_2A
Street_2B
Street_12A
Street_12B

A simple order by streetNames will do a lexical comparision and then Street_12A and B will come before Street_2A/B, and that is not correct. Is it possible to solve this by pure SQL?

Answer 1

The reliable way to do it (reliable in terms of ‘to sort your data correctly’, not ‘to solve your general problem’) is to split the data into street name and house number and sort both of them on their own. But this requires knowing where the house number starts. And this is the tricky part – making the assumption best fits your data.

You should use something like the following to refactor your data and from now on store the house number in a separate field. All this string-juggling won’t perform too well when it comes to sorting large data sets.

Assuming it is the last thing in the street name, and it contains a number:

DECLARE @test TABLE (   street VARCHAR(100) )  INSERT INTO @test (street) VALUES('Street') INSERT INTO @test (street) VALUES('Street 1A') INSERT INTO @test (street) VALUES('Street1 12B') INSERT INTO @test (street) VALUES('Street 22A') INSERT INTO @test (street) VALUES('Street1 200B-8a') INSERT INTO @test (street) VALUES('') INSERT INTO @test (street) VALUES(NULL)  SELECT   street,   CASE      WHEN LEN(street) > 0 AND CHARINDEX(' ', REVERSE(street)) > 0     THEN CASE       WHEN RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1) LIKE '%[0-9]%'       THEN LEFT(street, LEN(street) - CHARINDEX(' ', REVERSE(street)))     END   END street_part,   CASE      WHEN LEN(street) > 0 AND CHARINDEX(' ', REVERSE(street)) > 0     THEN CASE        WHEN RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1) LIKE '%[0-9]%'       THEN RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1)     END   END house_part,   CASE      WHEN LEN(street) > 0 AND CHARINDEX(' ', REVERSE(street)) > 0     THEN CASE        WHEN RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1) LIKE '%[0-9]%'       THEN CASE         WHEN PATINDEX('%[a-z]%', LOWER(RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1))) > 0         THEN CONVERT(INT, LEFT(RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1), PATINDEX('%[^0-9]%', LOWER(RIGHT(street, CHARINDEX(' ', REVERSE(street)) - 1))) - 1))       END     END   END house_part_num FROM   @test  ORDER BY   street_part,   house_part_num,   house_part 

This assumes these conditions:

• a street address can have a house number
• a house number must be the last thing in a street address (no ‘525 Monroe Av.’)
• a house number should start with a digit to be sorted correctly
• a house number can be a range (‘200-205’), this would be sorted below 200
• a house number must not contain spaces or recognition fails (When you look at your data, you could apply something like REPLACE(street, ' - ', '-') to sanitize common patterns beforehand.)
• the whole thing is still an approximation that certainly deviates from what it would look like in a telephone book, for example