SQL Like statement with regular expressions

Question

My table contains some columns with ;-separated numbers like this :

1;2;43;22;20;12

and so on. It's also possible there's only 1 number in this column like 110 , or 2 numbers like this 110;143

I want to select the rows that contain a certain number in this column. The number is in a variable $search_var. Let's say I need to search for the number 1 in my column. If I use a select with like statement like so :

"SELECT * FROM Table WHERE ids LIKE '%".$search_var."%'"

I get all results containing '1' and not only '1', so I also get 11, 14, 110, 1999 etc. I think I need to use some sort of regex-statement but I'm lost here... Who can help ?

Answer 1

You might not need regex for this

Set @YourNUmber := 110;

SELECT * 
FROM Table 
WHERE ';' + ids + ';' LIKE '%;'+ @yourNumber + ';%'

This guarantees there are always ; surrounding all the numbers.

This is formatted for SQL Server. The variable syntax and wildcards might be different if you are using something else.

EDIT:

Thanks @FélixGagnon-Grenier for the suggestions. I think either of these two will work. See here for a SQL Fiddle example

SELECT * 
FROM T 
WHERE concat(';',ids,';') LIKE concat('%;', @YourNumber , ';%');


SELECT * 
FROM T 
WHERE LOCATE(concat(';', @YourNumber , ';'),concat(';',ids,';'))>0

Answer 2

Try this solution if you're using SQL Server. This searches for the number where adjcent characters are not numbers:

SELECT * FROM Table WHERE ids LIKE '%[^0-9]".$search_var."[^0-9]%'