Count rows where value in row is also in previous row - mysql

Question

I want to get a count where the contents of a value in one row is also in the previous row.
Row | Item1 | Item2 | Item 3 |
1 | Dog | Cat | Rat
2 | Bird | Cat | Horse
3 | Horse | Dog | Rat
4 | Bird | Cat | Horse
5 | Horse | Bird | Cat
Row 2 would increase the count of Cat because Cat is in row 1 and 2
Row 3 would increase the count of Horse because Horse is also in Row 2
Row 4 would increase the count of Horse because Horse is also in Row 3
Row 5 would increase the count of Horse AND Cat because both of those appear in row 4.
There can be a max of 100 items or SKU's and I can index on any or all fields. At any given time there's probably between 1000 and 2000 rows.
I can't even wrap my head around where to begin with this query other than "SELECT * FROM table WHERE"

Answer 1

First, create table with all available unique values of SKU:
CREATE TABLE results(
id VARCHAR(255) NOT NULL PRIMARY KEY
);
-- All fields should be listed here one-by-one.
INSERT IGNORE INTO results (select Item1 from example);
INSERT IGNORE INTO results (select Item2 from example);
INSERT IGNORE INTO results (select Item3 from example);
Previous row could be obtained by left join primary table again with itself, i.e. LEFT JOIN example AS previous ON previous.id + 1 = example.id.
After that we've to check that each unique result exists in example table within current row and in previous row and finally get this:
SELECT
r.*,
SUM(
CASE WHEN r.id IN (
prv.Item1, prv.Item2, prv.Item3 -- All fields should be listed here.
) THEN 1 ELSE 0 END
) AS total
FROM
results AS r
LEFT JOIN
example AS cur ON r.id IN (
cur.Item1, cur.Item2, cur.Item3 -- All fields should be listed here.
)
LEFT JOIN
example AS prv ON prv.id + 1 = cur.id
GROUP BY
r.id
ORDER BY
cur.id
;
See working example http://www.sqlfiddle.com/#!9/7ebd85/1/0

Answer 2

This can be done with window functions (available in MySQL 8.0).
An option is to unpivot the resultset, and then use lag() to check the previous record. Assuming that ids are always increasing by 1, you can do:
select
item,
sum(case when id = lag_id + 1 then 1 else 0 end) cnt_consecutive
from (
select
t.*,
lag(id) over(partition by item order by id) lag_id
from (
select id, item1 item from mytable
union all select id, item2 from mytable
union all select id, item3 from mytable
) t
) t
group by item
order by item
If you don't have an incremented column, you can generate one with dense_rank():
select
item,
sum(case when new_id = lag_new_id + 1 then 1 else 0 end) cnt_consecutive
from (
select
t.*,
lag(new_id) over(partition by item order by new_id) lag_new_id
from (
select
t.*,
dense_rank() over(order by id) new_id
from (
select id, item1 item from mytable
union all select id, item2 from mytable
union all select id, item3 from mytable
) t
) t
) t
group by item
order by item
In this DB Fiddle, both queries return:
item | cnt_consecutive
:---- | --------------:
Bird | 1
Cat | 2
Dog | 0
Horse | 3
Rat | 0

Answer 3

I see #frost-nzcr4 suggestion is very good and I was doing my own version quite similar to that yesterday. However, the approach I'm doing is a bit different because I didn't create a table specifically to store the unique value. Instead, I was doing similarly like #GMB UNION sub-query and it end up to be something like this :
SELECT B.row, A.allitem,
SUM(CASE WHEN A.allitem IN (C.Item1, C.Item2, C.Item3) THEN 1
ELSE 0 END) AS total
FROM
-- this sub-query will be dynamic and UNION will eliminate any duplicate
(SELECT item1 AS allitem FROM mytable UNION
SELECT item2 FROM mytable UNION
SELECT item3 FROM mytable) AS A
LEFT JOIN mytable AS B ON A.allitem IN (B.Item1, B.Item2, B.Item3)
LEFT JOIN mytable AS C ON C.row + 1 = B.row
GROUP BY A.allitem
ORDER BY B.row;
Fiddle here : https://www.db-fiddle.com/f/bUUEsaeyPpAMfR2bK1VpBb/2
As you can see this is exactly similar query to frost's suggestion with only minor modification. In the sub-query allitem value will be updated as long as there are new values inserted so you don't need to keep inserting new unique data into a separate table.
Also, this query would normally get this is incompatible with sql_mode=only_full_group_by error on MySQL v5.7 above unless you remove the sql_mode.