This question already has answers here:
SQL select only rows with max value on a column [duplicate]
(27 answers)
Closed 4 years ago.
Here is my table:
+----+------------+-----------+---------------+
| id | product_id | price | date |
+----+------------+-----------+---------------+
| 1 | 4 | 2000 | 2019-02-10 |
| 2 | 5 | 1600 | 2019-02-11 |
| 3 | 4 | 850 | 2019-02-11 |
| 4 | 5 | 1500 | 2019-02-13 |
+----+------------+-----------+---------------+
I need to get a list of unique product ids that are the latest (newest, in other word, bigger date) ones. So this is the expected result:
+------------+-----------+---------------+
| product_id | price | date |
+------------+-----------+---------------+
| 4 | 850 | 2019-02-11 |
| 5 | 1500 | 2019-02-13 |
+------------+-----------+---------------+
Any idea how can I achieve that?
Here is my query:
SELECT id, product_id, price, MAX(date)
FROM tbl
GROUP BY product_id
-- ot selects the max `date` with random price like this:
+------------+-----------+---------------+
| product_id | price | date |
+------------+-----------+---------------+
| 4 | 2000 | 2019-02-11 |
| 5 | 1600 | 2019-02-13 |
+------------+-----------+---------------+
-- See? Prices are wrong
You could use a correlated subquery
select t1.* from table t1
where t1.date=( select max(date) from table t2
where t1.product_id=t2.product_id
)
Select *from
table1 t1
where (t1.product_id, t1.date) in
(select t2.product_id, max(t2.date)
from table1 t2
where t1.product_id = t2.product_id
)
Don't use a GROUP BY. Use a filter:
SELECT id, product_id, price, MAX(date)
FROM tbl
WHERE tbl.date = (SELECT MAX(t2.date)
FROM tbl t2
WHERE t2.product_id = tbl.product_id
);
With an index on (product_id, date), this is probably the fastest method.
If you can have duplicates on a given date, you can resolve them with:
SELECT id, product_id, price, MAX(date)
FROM tbl
WHERE tbl.id = (SELECT t2.id
FROM tbl t2
WHERE t2.product_id = tbl.product_id
ORDER BY t2.date DESC
LIMIT 1
);
My solution is with the analytic function first_value
SELECT distinct product_id,
first_value(price) over (partition by product_id order by date desc) last_price,
first_value(date) over (partition by product_id order by date desc) last_date
FROM tbl
Assuming that you are using a modern version of MySQL (8.0), you can use this:
select *
from (
SELECT id
, product_id
, price
, date
, row_number() over (partition by product_id order by date desc) rn
FROM tbl
) a
where rn = 1
Related
create table dt
(
id varchar(20),
user_id int,
name varchar(20),
td DATE,
amount float
);
INSERT INTO dt VALUES('blah',1, 'Rodeo', '2018-01-20', 10.12);
INSERT INTO dt VALUES('blahblah',1, 'Rodeo', '2019-01-01', 40.44);
INSERT INTO dt VALUES('sas',2, 'Janice', '2018-02-05', 18.18);
INSERT INTO dt VALUES('dsdcd',3, 'Sam', '2019-01-26', 16.13);
INSERT INTO dt VALUES('sdc',2, 'Janice', '2018-02-01', 12.19);
INSERT INTO dt VALUES('scsc',2, 'Janice', '2017-12-06', 5.10);
+----------+---------+--------+------------+--------+
| id | user_id | name | td | amount |
+----------+---------+--------+------------+--------+
| blah | 1 | Rodeo | 2018-01-20 | 10.12 |
| blahblah | 1 | Rodeo | 2019-01-01 | 40.44 |
| sas | 2 | Janice | 2018-02-05 | 18.18 |
| dsdcd | 3 | Sam | 2019-01-26 | 16.13 |
| sdc | 2 | Janice | 2018-02-01 | 12.19 |
| scsc | 2 | Janice | 2017-12-06 | 5.1 |
+----------+---------+--------+------------+--------+
For the above table how i can get this output. I can achieve this by windowing function but not sure how to do this by correlated subquery. Appreciate any help!
Output
Basically difference of users first transaction amount from their latest transaction amount. If the user has only one transaction then the difference is 0
User_id name amount
1 Rodeo 30.32 [40.44(latest trans) - 10.12 (min trans)]
3 Sam 0
2 Janice 13.08 [18.18 (latest trans) - 5.1 (min trans)]
With 2 subqueries to get the latest and earliest amounts:
select distinct t.user_id, t.name,
(select amount from dt
where user_id = t.user_id
order by td desc limit 1
)
-
(select amount from dt
where user_id = t.user_id
order by td limit 1
) amount
from dt t
See the demo.
Or:
select t.user_id, t.name,
max(t.latest * t.amount) - max(t.earliest * t.amount) amount
from (
select d.user_id, d.name, d.amount,
d.td = g.earliestdate earliest, d.td = g.latestdate latest
from dt d inner join (
select user_id, min(td) earliestdate, max(td) latestdate
from dt
group by user_id
) g on d.user_id = g.user_id and d.td in (earliestdate, latestdate)
) t
group by t.user_id, t.name
See the demo.
Results:
| user_id | name | amount |
| ------- | ------ | ------ |
| 1 | Rodeo | 30.32 |
| 2 | Janice | 13.08 |
| 3 | Sam | 0 |
This is similar to SQL select only rows with max value on a column, but you need to do it twice: once for the earliest row, again for the latest row.
SELECT t1.user_id, t1.name, t1.amount - t2.amount ASA amount
FROM (
SELECT dt1.user_id, dt1.name, dt1.amount
FROM dt AS dt1
JOIN (
SELECT user_id, name, MAX(td) AS maxdate
FROM dt
GROUP BY user_id, name) AS dt2
ON dt1.user_id = dt2.user_id AND dt1.name = dt2.name AND dt1.td = dt2.maxdate
) AS t1
JOIN (
SELECT dt1.user_id, dt1.name, dt1.amount
FROM dt AS dt1
JOIN (
SELECT user_id, name, MIN(td) AS mindate
FROM dt
GROUP BY user_id, name) AS dt2
ON dt1.user_id = dt2.user_id AND dt1.name = dt2.name AND dt1.td = dt2.mindate
) AS t2
ON t1.user_id = t2.user_id AND t1.name = t2.name
Approach using Correlated Subquery:
Query
SELECT user_id,
name,
Round(Coalesce ((SELECT t1.amount
FROM dt t1
WHERE t1.user_id = dt.user_id
ORDER BY t1.td DESC
LIMIT 1) - (SELECT t2.amount
FROM dt t2
WHERE t2.user_id = dt.user_id
ORDER BY t2.td ASC
LIMIT 1), 0), 2) AS amount
FROM dt
GROUP BY user_id,
name;
| user_id | name | amount |
| ------- | ------ | ------ |
| 1 | Rodeo | 30.32 |
| 2 | Janice | 13.08 |
| 3 | Sam | 0 |
View on DB Fiddle
You can try this as well
Select t3.user_id, t3.name, max(t3.new_amount) FROM (
Select t1.user_id, t2.name, (t1.amount - t2.amount) as new_amount
FROM dt t1
INNER JOIN dt t2
ON t1.user_id=t2.user_id
Order by t1.user_id ASC, t1.td DESC, t2.user_id ASC, t2.td ASC
) as t3
group by t3.user_id,t3.name;
Demo
This question already has answers here:
How can I SELECT rows with MAX(Column value), PARTITION by another column in MYSQL?
(22 answers)
Closed 3 years ago.
I want to sort a table by name. These should not be alphabetic but the largest id. I have this table.
id name
---|-----|
1 | abc |
2 | abc |
3 | def |
4 | def |
5 | def |
6 | abc |
7 | abc |
8 | def |
That's what i need
id name
---|-----|
8 | def |
5 | def |
4 | def |
3 | def |
7 | abc |
6 | abc |
2 | abc |
1 | abc |
Does anyone have an idea?
This is really shouting for window-functions:
SELECT *
FROM your_table t
ORDER BY MAX(id) OVER (PARTITION BY name) DESC, id DESC
With your select from the comment under Barmar's answer:
SELECT *
FROM posts p, influencer i
WHERE p.i_name = i.i_name
ORDER BY MAX(p.p_id) OVER (PARTITION BY i.i_name) DESC, p.p_id DESC
Perhaps also have a look here: MySQL Manual
Join the table with a subquery that gets the largest ID for each name and order by that.
SELECT t1.*
FROM YourTable AS t1
JOIN (
SELECT name, MAX(id) AS maxid
FROM YourTable
GROUP BY name
) AS t2 ON t1.name = t2.name
ORDER BY maxid DESC, id DESC
If you have two tables, you can still join them with the subquery.
select p1.*, i.*
FROM posts AS p1
JOIN influencers AS i ON p1.i_name = i.i_name
JOIN (
SELECT i_name, MAX(p_id) AS maxid
FROM posts
GROUP BY i_name
) AS p2 ON p1.i_name = p2.i_name
ORDER BY p2.maxid DESC, p1.p_id DESC
I have a table like this:
+----+---------+------------+
| id | price | date |
+----+---------+------------+
| 1 | 340 | 2018-09-02 |
| 2 | 325 | 2018-09-05 |
| 3 | 358 | 2018-09-08 |
+----+---------+------------+
And I need to make a view which has a row for every day. Something like this:
+----+---------+------------+
| id | price | date |
+----+---------+------------+
| 1 | 340 | 2018-09-02 |
| 1 | 340 | 2018-09-03 |
| 1 | 340 | 2018-09-04 |
| 2 | 325 | 2018-09-05 |
| 2 | 325 | 2018-09-06 |
| 2 | 325 | 2018-09-07 |
| 3 | 358 | 2018-09-08 |
+----+---------+------------+
I can do that using PHP with a loop (foreach) and making a temp variable which holds the previous price til there is a new date.
But I need to make a view ... So I should do that using pure-SQL .. Any idea how can I do that?
You could use a recursive CTE to generate the records in the "gaps". To avoid that an infinite gap after the last date is "filled", first get the maximum date in the source data and make sure not to bypass that date in the recursion.
I have called your table tbl:
with recursive cte as (
select id,
price,
date,
(select max(date) date from tbl) mx
from tbl
union all
select cte.id,
cte.price,
date_add(cte.date, interval 1 day),
cte.mx
from cte
left join tbl
on tbl.date = date_add(cte.date, interval 1 day)
where tbl.id is null
and cte.date <> cte.mx
)
select id,
price,
date
from cte
order by 3;
demo with mysql 8
Here is an approach which should work without analytic functions. This answer uses a calendar table join approach. The first CTE below is the base table on which the rest of the query is based. We use a correlated subquery to find the most recent date earlier than the current date in the CTE which has a non NULL price. This is the basis for finding out what the id and price values should be for those dates coming in from the calendar table which do not appear in the original data set.
WITH cte AS (
SELECT cal.date, t.price, t.id
FROM
(
SELECT '2018-09-02' AS date UNION ALL
SELECT '2018-09-03' UNION ALL
SELECT '2018-09-04' UNION ALL
SELECT '2018-09-05' UNION ALL
SELECT '2018-09-06' UNION ALL
SELECT '2018-09-07' UNION ALL
SELECT '2018-09-08'
) cal
LEFT JOIN yourTable t
ON cal.date = t.date
),
cte2 AS (
SELECT
t1.date,
t1.price,
t1.id,
(SELECT MAX(t2.date) FROM cte t2
WHERE t2.date <= t1.date AND t2.price IS NOT NULL) AS nearest_date
FROM cte t1
)
SELECT
(SELECT t2.id FROM yourTable t2 WHERE t2.date = t1.nearest_date) id,
(SELECT t2.price FROM yourTable t2 WHERE t2.date = t1.nearest_date) price,
t1.date
FROM cte2 t1
ORDER BY
t1.date;
Demo
Note: To make this work on MySQL versions earlier than 8+, you would need to inline the CTEs above. It would result in verbose code, but, it should still work.
Since you are using MariaDB, it is rather trivial:
MariaDB [test]> SELECT '2019-01-01' + INTERVAL seq-1 DAY FROM seq_1_to_31;
+-----------------------------------+
| '2019-01-01' + INTERVAL seq-1 DAY |
+-----------------------------------+
| 2019-01-01 |
| 2019-01-02 |
| 2019-01-03 |
| 2019-01-04 |
| 2019-01-05 |
| 2019-01-06 |
(etc)
There are variations on this wherein you generate a large range of dates, but then use a WHERE to chop to what you need. And use LEFT JOIN with the sequence 'derived table' on the 'left'.
Use something like the above as a derived table in your query.
I need to join together 2 SQL statements and both of those statements work on their own. But I don't know how to combine both into 1 SQL statement.
I have two tables in 1st statement, TR120 and TR1201.
The SQL is this:
select
PRODUCT, PRICE, QUANTITY, INVOICE.DATE
from
TR1201
left join
(select
DATE, ID as INVOICE_ID, INVOICE
from TR120) as INVOICE on INVOICE.INVOICE_ID = ID
where
INVOICE.DATE >= '2016-06-01' and INVOICE.DATE <= '2016-06-30'
This returns a list of all the products I sold, with price, quantity and date of sales in a specific time frame from 01-06-16 till 30-06-16.
Now I need to find out the latest price that I bought product for in different two tables TR100 and TR1001 based on the product and date of sale from the 1st SQL statement.
select
PRODUCT, PRICE, SUP.DATE
from
TR1001
left join
(select
DATE, ID as SUP_ID, SUP_INVOICE
from TR100) as SUP on SUP.SUP_ID = ID
This returns a list of all the products that I have bought with a price and a date. I only need last record from this query based on product and date of purchased.
TR120
ID | INVOICE | DATE
1 | 000001 |2016-06-05
2 | 000002 |2016-06-15
3 | 000003 |2016-06-25
TR1201
ID | PRODUCT | PRICE A | QUANTITY
1 | A | 2,00 | 5
2 | A | 2,00 | 2
3 | A | 2,00 | 1
TR100
ID | SUP_INVOICE | DATE
1 | 160001 | 2016-05-30
2 | 160002 | 2016-06-16
TR1001
ID | PRODUCT | PRICE B
1 | A | 0,5
2 | A | 0,7
The result I am trying to get is this:
PRODUCT | PRICE A (tr1201) | QUANTITY | DATE (tr100) | PRICE B (tr1001)
A | 2 | 5 | 2016-05-30 | 0,5
A | 2 | 2 | 2016-05-15 | 0,5
A | 2 | 1 | 2016-05-16 | 0,7
That is all I want to do :(
Have you tried first_value?
FIRST_VALUE ( [scalar_expression ] )
OVER ( [ partition_by_clause ] order_by_clause [ rows_range_clause ] )
it works like this:
select distinct id,
first_value(price) over (partition by id (,sup) order by date DESC (latest, ASC for oldest)) as last_price
from table;
Documentation can be found here: https://msdn.microsoft.com/en-us/library/hh213018.aspx
I don't have your tables so cannot test and therefore am providing advice only.
I think what you need is an Outer apply like this instead of joins
select
T1.Product
, T1.Price
, T2.DATE -- Alias this
, T2.Price -- Alias this
, T3.DATE -- Alias this
, T3.Price -- Alias this
from T1
OUTER APPLY (
select top 1
Date
,Price
from table2
WHERE ID = T1.Id AND product = T1.Product-- plus any other joins
ORDER BY Date desc
) as T2
OUTER APPLY (
select top 1
Date
,Price
from table3
WHERE ID = T1.Id AND product = T1.Product-- plus any other joins
ORDER BY Date desc
) as T3
From MySQL - Get row number on select
I know how to get the row number / rank using this mysql query:
SELECT #rn:=#rn+1 AS rank, itemID
FROM (
SELECT itemID
FROM orders
ORDER BY somecriteria DESC
) t1, (SELECT #rn:=0) t2;
The result returns something like this:
+--------+------+
| rank | itemID |
+--------+------+
| 1 | 265 |
| 2 | 135 |
| 3 | 36 |
| 4 | 145 |
| 5 | 123 |
| 6 | 342 |
| 7 | 111 |
+--------+------+
My question is: How can I get the result in 1 simple SINGLE QUERY that returns items having lower rank than itemID of 145, i.e.:
+--------+------+
| rank | itemID |
+--------+------+
| 5 | 123 |
| 6 | 345 |
| 7 | 111 |
+--------+------+
Oracle sql query is also welcomed. Thanks.
An Oracle solution (not sure if it meets your criteria of "one simple single query"):
WITH t AS
(SELECT item_id, row_number() OVER (ORDER BY some_criteria DESC) rn
FROM orders)
SELECT t2.rn, t2.item_id
FROM t t1 JOIN t t2 ON (t2.rn > t1.rn)
WHERE t1.item_id = 145;
My assumption is no repeating values of item_id.
Attempting to put this in MySQL terms, perhaps something like this might work:
SELECT t2.rank, t2.itemID
FROM (SELECT #rn:=#rn+1 AS rank, itemID
FROM (SELECT itemID
FROM orders
ORDER BY somecriteria DESC), (SELECT #rn:=0)) t1 INNER JOIN
(SELECT #rn:=#rn+1 AS rank, itemID
FROM (SELECT itemID
FROM orders
ORDER BY somecriteria DESC), (SELECT #rn:=0)) t2 ON t2.rank > t1.rank
WHERE t1.itemID = 145;
Disclaimer: I don't work with MySQL much, and it's untested. The Oracle piece works.
SELECT #rn:=#rn+1 AS rank, itemID
FROM (
SELECT itemID
FROM orders
ORDER BY somecriteria DESC
) t1, (SELECT #rn:=0) t2
where rank >
(
select rank from
(
SELECT #rn:=#rn+1 AS rank, itemID
FROM
(
SELECT itemID
FROM orders
ORDER BY somecriteria DESC
) t1, (SELECT #rn:=0) t2
) x where itemID = 145
) y