SQL update column3 after group by column1 and compare value in column2 - mysql

Question

If I have a table T that look like this: where id is the unique auto-increment primary key. Difference column is default to 0. I want to UPDATE only the difference of largestId - secondLargestId in each id_str group while the rest remains unchanged.
id_str id Value Difference
2380 1 21.01 0
2380 3 22.04 0
2380 5 22.65 0
2380 8 23.11 0
2380 10 35.21 0
20100 2 37.07 0
20100 4 38.17 0
20100 6 38.97 0
20103 7 57.98 0
20103 9 60.83 0
The result I want is:
id_str id Value Difference
2380 1 21.01 0
2380 3 22.04 0
2380 5 22.65 0
2380 8 23.11 0
2380 10 35.21 12.1
20100 2 37.07 0
20100 4 38.17 0
20100 6 38.97 0.8
20103 7 57.98 0
20103 9 60.83 2.85
How can I write the query?

Answer 1

This should do the trick in MySQL.
CREATE TABLE SomeTable
( id_str VARCHAR(10),
id INTEGER,
value_ DECIMAL(7,5),
difference DECIMAL(7,5)
);
INSERT INTO SomeTable VALUES(2380,1,21.01,0);
INSERT INTO SomeTable VALUES(2380,3,22.04,0);
INSERT INTO SomeTable VALUES(2380,5,22.65,0);
INSERT INTO SomeTable VALUES(2380,8,23.11,0);
INSERT INTO SomeTable VALUES(2380,10,35.21,0);
INSERT INTO SomeTable VALUES(20100,2,37.07,0);
INSERT INTO SomeTable VALUES(20100,4,38.17,0);
INSERT INTO SomeTable VALUES(20100,6,38.97,0);
INSERT INTO SomeTable VALUES(20103,7,57.98,0);
INSERT INTO SomeTable VALUES(20103,9,60.83,0);
UPDATE SomeTable,
(SELECT T1.id AS id_updt,
T1.value_ - T2.value_ AS diff_updt
FROM (SELECT id_str,
id,
value_,
(
CASE id_str
WHEN #curStr THEN #curRow := #curRow + 1
ELSE #curRow := 1
AND #curStr := id_str
END
) AS rnk
FROM SomeTable,
(SELECT #curRow := 0, #curStr := '') r
ORDER
BY id_str DESC,
id DESC
) AS T1
INNER
JOIN (SELECT id_str,
id,
value_,
(
CASE id_str
WHEN #curStr THEN #curRow := #curRow + 1
ELSE #curRow := 1
AND #curStr := id_str
END
) AS rnk
FROM SomeTable,
(SELECT #curRow := 0, #curStr := '') r
ORDER
BY id_str DESC,
id DESC
) AS T2
ON T1.id_str = T2.id_str
AND T1.rnk = 1
AND T2.rnk = 2
) AS UPDT
SET SomeTable.difference = UPDT.diff_updt
WHERE SomeTable.id = UPDT.id_updt;
Deprecated solution - This will work for a DBMS that supports the rank function.
UPDATE SomeTable
FROM ( SELECT RNK1.id AS id_updt,
RNK1.value_ - RNK2.value_ AS diff_updt
FROM (SELECT id_str,
RANK() OVER
( PARTITION BY id_str
ORDER BY id DESC
) AS id_rnk
FROM SomeTable
) AS RNK1
INNER
JOIN (SELECT id_str,
RANK() OVER
( PARTITION BY id_str
ORDER BY id DESC
) - 1 AS id_rnk_decrement
FROM SomeTable
) AS RNK2
ON RNK1.id_str = RNK2.id_str
AND RNK1.id_rnk = RNK2.id_rnk_decrement
WHERE RNK1.id_rnk = 1
) AS UPDT
SET SomeTable.difference_ = UPDT.diff_updt
WHERE SomeTable.id = UPDT.id_updt;

Answer 2

You can find the two greatest ids per group with the following query:
select t1.id_str, max(t1.id) as id1, (
select max(t2.id)
from mytable t2
where t2.id_str = t1.id_str
and t2.id < max(t1.id)
) as id2
from mytable t1
group by t1.id_str;
Result:
| id_str | id1 | id2 |
|--------|-----|-----|
| 2380 | 10 | 8 |
| 20100 | 6 | 4 |
| 20103 | 9 | 7 |
Use it as subquery in your update statement:
update mytable u
join (
select t1.id_str, max(t1.id) as id1, (
select max(t2.id)
from mytable t2
where t2.id_str = t1.id_str
and t2.id < max(t1.id)
) as id2
from mytable t1
group by t1.id_str
) t on t.id1 = u.id
join mytable t1 on t1.id = t.id1
join mytable t2 on t2.id = t.id2
set u.Difference = t1.Value - t2.Value;
The table will now contain:
| id_str | id | Value | Difference |
|--------|----|-------|------------|
| 2380 | 1 | 21.01 | 0 |
| 2380 | 3 | 22.04 | 0 |
| 2380 | 5 | 22.65 | 0 |
| 2380 | 8 | 23.11 | 0 |
| 2380 | 10 | 35.21 | 12.1 |
| 20100 | 2 | 37.07 | 0 |
| 20100 | 4 | 38.17 | 0 |
| 20100 | 6 | 38.97 | 0.8 |
| 20103 | 7 | 57.98 | 0 |
| 20103 | 9 | 60.83 | 2.85 |
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