mysql split string [duplicate] - mysql

Question

I have a column that has comma separated data:
1,2,3
3,2,1
4,5,6
5,5,5
I'm trying to run a search that would query each value of the CSV string individually.
0<first<5 and 1<second<3 and 2<third<4
I get that I could return all queries and split it myself and compare it myself. I'm curious if there is a way to do this so MySQL does that processing work.
Thanks!

Answer 1

Use
substring_index(`column`,',',1) ==> first value
substring_index(substring_index(`column`,',',-2),',',1)=> second value
substring_index(substring_index(`column`,',',-1),',',1)=> third value
in your where clause.
SELECT * FROM `table`
WHERE
substring_index(`column`,',',1)<0
AND
substring_index(`column`,',',1)>5

Answer 2

It seems to work:
substring_index ( substring_index ( context,',',1 ), ',', -1)
substring_index ( substring_index ( context,',',2 ), ',', -1)
substring_index ( substring_index ( context,',',3 ), ',', -1)
substring_index ( substring_index ( context,',',4 ), ',', -1)
it means 1st value, 2nd, 3rd, etc.
Explanation:
The inner substring_index returns the first n values that are comma separated. So if your original string is "34,7,23,89", substring_index( context,',', 3) returns "34,7,23".
The outer substring_index takes the value returned by the inner substring_index and the -1 allows you to take the last value. So you get "23" from the "34,7,23".
Instead of -1 if you specify -2, you'll get "7,23", because it took the last two values.
Example:
select * from MyTable where substring_index(substring_index(prices,',',1),',',-1)=3382;
Here, prices is the name of a column in MyTable.

Answer 3

Usually substring_index does what you want:
mysql> select substring_index("foo#gmail.com","#",-1);
+-----------------------------------------+
| substring_index("foo#gmail.com","#",-1) |
+-----------------------------------------+
| gmail.com |
+-----------------------------------------+
1 row in set (0.00 sec)

Answer 4

You may get what you want by using the MySQL REGEXP or LIKE.
See the MySQL Docs on Pattern Matching

Answer 5

As an addendum to this, I've strings of the form:
Some words 303
where I'd like to split off the numerical part from the tail of the string.
This seems to point to a possible solution:
http://lists.mysql.com/mysql/222421
The problem however, is that you only get the answer "yes, it matches", and not the start index of the regexp match.

Answer 6

Here is another variant I posted on related question. The REGEX check to see if you are out of bounds is useful, so for a table column you would put it in the where clause.
SET #Array = 'one,two,three,four';
SET #ArrayIndex = 2;
SELECT CASE
WHEN #Array REGEXP CONCAT('((,).*){',#ArrayIndex,'}')
THEN SUBSTRING_INDEX(SUBSTRING_INDEX(#Array,',',#ArrayIndex+1),',',-1)
ELSE NULL
END AS Result;
SUBSTRING_INDEX(string, delim, n) returns the first n
SUBSTRING_INDEX(string, delim, -1) returns the last only
REGEXP '((delim).*){n}' checks if there are n delimiters (i.e. you are in bounds)

Answer 7

Building on #Oleksiy's answer, here is one that can work with strings of variable segment lengths (within reasonable limits), for example comma-separated addresses:
SELECT substring_index ( substring_index ( address,',',1 ), ',', -1) AS address_line_1,
IF(address_parts > 1, substring_index ( substring_index ( address,',',2 ), ',', -1), '') AS address_line_2,
IF(address_parts > 2, substring_index ( substring_index ( address,',',3 ), ',', -1), '') AS address_line_3,
IF(address_parts > 3, substring_index ( substring_index ( address,',',4 ), ',', -1), '') AS address_line_4,
IF(address_parts > 4, substring_index ( substring_index ( address,',',5 ), ',', -1), '') AS address_line_5
FROM (
SELECT address, LENGTH(address) - LENGTH(REPLACE(address, ',', '')) AS address_parts
FROM mytable
) AS addresses

Answer 8

It's working..
SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(
SUBSTRING_INDEX(SUBSTRING_INDEX(SUBSTRING_INDEX(col,'1', 1), '2', 1), '3', 1), '4', 1), '5', 1), '6', 1)
, '7', 1), '8', 1), '9', 1), '0', 1) as new_col
FROM table_name group by new_col;