match a number using regex for comma separated number - mysql

Question

I have a string that contains number with separated by comma like below.
15,22,20,26,33,445,40,44,22,225,115,2
I want to know if a number say 15 is in that string or not.The problem is that 15 and 115 both are a match.Same for other number say 2, for this case 20 , 25, and 225 are match.For both cases only it should return if there is 15 or 2 in the string.I tried using like keyword but it's not working. It also return the rows with 115 or 20, 225, 222 whille matching 15 and 2 respectively. Can anyone suggest a regex pattern?
Update
I have a query like below where I was using like keyword, but I was getting wrong result for above reason.
SELECT DISTINCT A.id,A.title,A.title_hi,A.cId,B.id as cid1,A.report_type ,A.icon_img_url, A.created_at , A.news_date
FROM tfs_report_news A, tfs_commodity_master B
WHERE (',' + RTRIM(A.cId) + ',') LIKE ('%,' + B.id + ',%')
AND A.ccId = B.ccId AND A.`report_type`= "M"
AND A.isDeleted=0 AND A.isActive=1 AND B.isDeleted=0
AND B.status=1
AND A.news_date= (SELECT MAX(T.news_date)
FROM tfs_report_news T WHERE (',' + RTRIM(T.cId) + ',')
LIKE ('%,' + B.id + ',%'))
ORDER BY created_at desc, id desc limit 100;
Here tfs_report_news has the string 15,22,20,26,33,445,40,44,22,225,115,2 as column name cId and individual cId like 15 or 2 is id of tfs_commodity_master

Answer 1

In MySQL, what you asked for is the purpose of string function find_in_set():
Returns a value in the range of 1 to N if the string str is in the string list strlist consisting of N substrings. A string list is a string composed of substrings separated by , characters [...] Returns 0 if str is not in strlist or if strlist is the empty string. Returns NULL if either argument is NULL.
So to check if a value is present in the list, you can just do:
find_in_set('15', '15,22,20,26,33,445,40,44,22,225,115,2') > 0
Side note: here is a recommended reading.

Answer 2

Use FIND_IN_SET:
SELECT
CASE WHEN FIND_IN_SET('15', csv) > 0 THEN 'yes' ELSE 'no' END AS result
FROM yourTable;
Another option would be to use LIKE:
SELECT
CASE WHEN CONCAT(',', csv, ',') LIKE '%,15,%' THEN 'yes' ELSE 'no' END AS result
FROM yourTable;
Finally, you could also use REGEXP here:
SELECT
CASE WHEN csv REGEXP '[[:<:]]15[[:>:]]' THEN 'yes' ELSE 'no' END AS result
FROM yourTable;