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ID       RoomNo

1     4, 5, 6, 7, 8
2     4, 5, 6
3     6, 7, 8
4     5, 6
5     4, 5, 6, 7
6     4, 15, 6, 7


'RoomNo' field type is varchar.

When I use
SELECT * FROM table WHERE RoomNo LIKE '%5%' ;

am getting value but thats not proper because i need value contain 5 but it will get 15 also in id - 6

and also i tried to use this query
SELECT * FROM table WHERE FIND_IN_SET(7, RoomNo);


but it not showing result
Posted

1 solution

The problem is your data. %5% literally means <zero or more characters>5<zero or more characters>.
You could edit the first query to exclude results that are not valid such as;
SELECT * FROM table WHERE RoomNo LIKE '%5%' and RoomNo NOT LIKE '%15%'

But this is inefficient and would require you to enter every other value that you do not want but still includes a 5. The number of exclusions is determined by the number of possible characters, if you are talking about numbers only then you would have 20 - see below;
05, 15, 25, 35, 45, 55, 65, 75, 85, 95
50, 51, 52, 53, 54, 55, 56, 57, 58, 59
If you also had the possibility of characters a-z then you would have 52 exceptions.

Additionally my understanding of the function FIND_IN_SET does not work the way you are using it. It returns the position in a string, not the position in a table - refer to the below link for documentation.
http://www.w3resource.com/mysql/string-functions/mysql-find_in_set-function.php[^]
   

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