First record combined with GROUP BY - mysql

Question

Let's say I got a table "values" which contains the fields
id (int)
name (varchar)
value (float)
timestamp (int)
Now I want to to calculate the highest lowest and first value (timestamp based) for each name on the entire values table.
Is this possible to be achieved in one single performant query? I stumbled upon the 'first_value' function, but that one doesn't seem to work. I tried the following query, using joins, but also without success.
SELECT
a.name,
b.value as open,
MIN(a.value) as low,
MAX(a.value) as high
FROM values a
LEFT JOIN values b
ON a.name = b.name AND b.id = MIN(a.id)
GROUP BY a.name;
Isn't there some sort of function which would make something similar as this possible?
SELECT
name,
FIRST_VALUE(value) as open,
MIN(value) as low,
MAX(value) as high
FROM values
GROUP BY name
ORDER BY timestamp ASC;
Example data
id name value timestamp
1 USD 3 16540
2 EUR 5 16540
3 GBP 4 16540
4 EUR 2 16600
5 USD 4 16600
6 GBP 5 16600
7 USD 6 16660
8 EUR 7 16660
9 GBP 6 16660
10 USD 5 16720
11 EUR 5 16720
12 GBP 7 16720
13 EUR 8 16780
14 USD 7 16780
15 GBP 8 16780
Example output
name open low high
USD 3 3 7
EUR 5 2 8
GBP 4 4 8
I'm using MySQL-client version: 5.6.39
A tie should not be possible, if it does, I don't care which value gets picked.

Answer 1

If you are running MySQL 8.0, this can be quite easily solved with window functions:
select name, value open, low, high
from (
select
name,
value,
min(value) over(partition by name) low,
max(value) over(partition by name) high,
row_number() over(partition by name order by timestamp) rn
from mytable
) x
where rn = 1
Demo on DB Fiddle:
| name | open | low | high |
| ---- | ---- | --- | ---- |
| EUR | 5 | 2 | 8 |
| GBP | 4 | 4 | 8 |
| USD | 3 | 3 | 7 |
In earlier versions, you could:
use a correlated subquery to filter on the first record for each name
join the table with an aggregate query that computes the min and max of each name
Query:
select
t.name,
t.value open,
t0.low,
t0.high
from
mytable t
inner join (
select name, min(value) low, max(value) high from mytable group by name
) t0 on t0.name = t.name
where t.timestamp = (
select min(t1.timestamp) from mytable t1 where t1.name = t.name
);
Demo on MySQL 5.6 DB Fiddle: same results as above
This could also be achieved using inline subqueries (which may actually perform better):
select
t.name,
t.value open,
(select min(value) from mytable t1 where t1.name = t.name) low,
(select max(value) from mytable t1 where t1.name = t.name) high
from
mytable t
where timestamp = (
select min(t1.timestamp) from mytable t1 where t1.name = t.name
)
Demo on MySQL 5.6 DB Fiddle

Answer 2

in one single performant query
Do it logically and let the DBMS worry about performance. If that isn't fast enough, check your indexes.
The value associated with the first timestamp requires a join. You can find the first timestamp easily enough. Getting a value from a row associated with a given row: that's what joins are for.
So, we have:
SELECT
name,
value as open,
v1.low
v1.high
FROM values as v join (
select name,
min(timestamp) as timestamp,
min(value) as low,
max(value) as high
FROM values
GROUP BY name
) as v1
on v.name = v1.name and v.timestamp = v1.timestamp

Answer 3

This solution seems to have the best performance.
SELECT
name,
CAST(SUBSTRING_INDEX(GROUP_CONCAT(CAST(value AS CHAR) ORDER BY TIMESTAMP ASC), ',', 1) AS DECIMAL(10, 6)) AS open,
MIN(value) AS low,
MAX(value) AS high
FROM mytable
GROUP BY name
ORDER BY name ASC