I need to select all branches from branches table grouped by store_id order by distance
In other words i need to get the nearest branch of each store
select distinct `branch_id`, `store_id`, `branch_name`, distance
from `store_branches`
groupBy store_id having MIN(distance)
orderBy `distance` asc
MIN(distance) couse the query to return an empty result
I used having because orderBy is not doing the job ,because groupBy is applied before orderBy
With a correlated subquery:
select sb.`branch_id`, sb.`store_id`, sb.`branch_name`, sb.distance
from `store_branches` sb
where sb.`distance` = (select min(distance) from `store_branches` where `store_id` = sb.`store_id`)
or with NOT EXISTS:
select sb.`branch_id`, sb.`store_id`, sb.`branch_name`, sb.distance
from `store_branches` sb
where not exists (
select 1 from `store_branches`
where `store_id` = sb.`store_id` and `distance` < sb.`distance`
)
For MySql 8.0 you could use RANK() window function:
select sb.`branch_id`, sb.`store_id`, sb.`branch_name`, sb.distance
from (
select *, rank() over (partition by `store_id` order by `distance`) rnk
from `store_branches`
) sb
where sb.rnk = 1
Related
I'm attempting to create an SQL query that retrieves the total_cost for every row in a table. Alongside that, I also need to collect the most dominant value for both columnA and columnB, with their respective values.
For example, with the following table contents:
cost
columnA
columnB
target
250
Foo
Bar
XYZ
200
Foo
Bar
XYZ
150
Bar
Bar
ABC
250
Foo
Bar
ABC
The result would need to be:
total_cost
columnA_dominant
columnB_dominant
columnA_value
columnB_value
850
Foo
Bar
250
400
Now I can get as far as calculating the total cost - that's no issue. I can also get the most dominant value for columnA using this answer. But after this, I'm not sure how to also get the dominant value for columnB and the values too.
This is my current SQL:
SELECT
SUM(`cost`) AS `total_cost`,
COUNT(`columnA`) AS `columnA_dominant`
FROM `table`
GROUP BY `columnA_dominant`
ORDER BY `columnA_dominant` DESC
WHERE `target` = "ABC"
UPDATE: Thanks to #Barmar for the idea of using a subquery, I managed to get the dominant values for columnA and columnB:
SELECT
-- Retrieve total cost.
SUM(`cost`) AS `total_cost`,
-- Get dominant values.
(
SELECT `columnA`
FROM `table`
GROUP BY `columnA`
ORDER BY COUNT(*) DESC
LIMIT 1
) AS `columnA_dominant`,
(
SELECT `columnB`
FROM `table`
GROUP BY `columnB`
ORDER BY COUNT(*) DESC
LIMIT 1
) AS `columnB_dominant`
FROM `table`
WHERE `target` = "XYZ"
However, I'm still having issues figuring out how to calculate the respective values.
You might get close, if we want to get percentage values we can try to add COUNT(*) at subquery to get max count by columnA and columnB then do division by total count
SELECT
SUM(cost),
(
SELECT tt.columnA
FROM T tt
GROUP BY tt.columnA
ORDER BY COUNT(*) DESC
LIMIT 1
) AS columnA_dominant,
(
SELECT tt.columnB
FROM T tt
GROUP BY tt.columnB
ORDER BY COUNT(*) DESC
LIMIT 1
) AS columnB_dominant,
(
SELECT COUNT(*)
FROM T tt
GROUP BY tt.columnA
ORDER BY COUNT(*) DESC
LIMIT 1
) / COUNT(*) AS columnA_percentage,
(
SELECT COUNT(*)
FROM T tt
GROUP BY tt.columnB
ORDER BY COUNT(*) DESC
LIMIT 1
) / COUNT(*) AS columnB_percentage
FROM T t1
If your MySQL version supports the window function, there is another way which reduce table scan might get better performance than a correlated subquery
SELECT SUM(cost) OVER(),
FIRST_VALUE(columnA) OVER (ORDER BY counter1 DESC) columnA_dominant,
FIRST_VALUE(columnB) OVER (ORDER BY counter2 DESC) columnB_dominant,
FIRST_VALUE(counter1) OVER (ORDER BY counter1 DESC) / COUNT(*) OVER() columnA_percentage,
FIRST_VALUE(counter2) OVER (ORDER BY counter2 DESC) / COUNT(*) OVER() columnB_percentage
FROM (
SELECT *,
COUNT(*) OVER (PARTITION BY columnA) counter1,
COUNT(*) OVER (PARTITION BY columnB) counter2
FROM T
) t1
LIMIT 1
sqlfiddle
try this query
select sum(cost) as total_cost,p.columnA,q.columnB,p.columnA_percentage,q.columnB_percentage
from get_common,(
select top 1 columnA,columnA_percentage
from(
select columnA,count(columnA) as count_columnA,cast(count(columnA) as float)/(select count(columnA) from get_common) as columnA_percentage
from get_common
group by columnA)s
order by count_columnA desc
)p,
(select top 1 columnB,columnB_percentage
from (
select columnB,count(columnB) as count_columnB, cast(count(columnB) as float)/(select count(columnB) from get_common) as columnB_percentage
from get_common
group by columnB) t
order by count_columnB desc)q
group by p.columnA,q.columnB,p.columnA_percentage,q.columnB_percentage
so if you want to get the percent and dominant value you must make their own query like this
select top 1 columnA,columnA_percentage
from(
select columnA,count(columnA) as count_columnA,cast(count(columnA) as float)/(select count(columnA) from get_common) as columnA_percentage
from get_common
group by columnA)s
order by count_columnA desc
then you can join with the sum query to get all value you want
hope this can help you
My database table name is ledgers and fields are id, item_id, date, ...other fields
I Want the last record from (groupBy item_id order by date ASC). from each group.
I tried below query
select
`id`,
`item_id`,
`date`,
`opening_quantity`,
`closing_quantity`,
`item_rate`,
`item_value`,
`previous_rate`
from `ledgers`
where date(`date`) >= ? and date(`date`) <= ?
group by `item_id`
order by `date` desc
Can you guys please help.
You can filter with a correlated subquery:
select t.*
from `ledgers` t
where
date(t.`date`) >= ?
and date(t.`date`) <= ?
and t.`date` = (
select max(t1.`date`)
from `ledgers` t1
where t1.`item_id` = t.`item_id`
)
For performance, consider an index on (item_id, date).
Another option is to use rank() (available in MySQ 8.0 only):
select *
from (
select
t.*,
rank() over(partition by `item_id` order by `date` desc) rn
from `ledgers` t
where date(t.`date`) >= ? and date(t.`date`) <= ?
) t
where rn = 1
SELECT max(sum(`orderquantity`)), `medicinename`
FROM `orerdetails`
WHERE `OID`=
(
SELECT `OrderID`
FROM `order`
where `VID` = 5 AND `OrerResponse` = 1
)
GROUP BY `medicinename`
i want to get the max of the result(sum of the order quantity) but it gives error any soultion to solve this
You don't need Max() here. Instead sort your recordset by that Sum('orderquantity') descending, and take the first record returned:
SELECT sum(`orderquantity`) as sumoforderqty, `medicinename`
FROM `orerdetails`
WHERE `OID`=
(
SELECT `OrderID`
FROM `order`
where `VID` = 5 AND `OrerResponse` = 1
)
GROUP BY `medicinename`
ORDER BY sumoforderqty DESC
LIMIT 1
Ok, so I have the following query:
SELECT MIN(`date`), `player_name`
FROM `player_playtime`
GROUP BY `player_name`
I then need to use this result inside the following query:
SELECT DATE(`date`) , COUNT(DISTINCT `player_name`)
FROM `player_playtime /*Use previous query result here*/`
GROUP BY DATE( `date`) DESC LIMIT 60
How would I go about doing this?
You just need to write the first query as a subquery (derived table), inside parentheses, pick an alias for it (t below) and alias the columns as well.
The DISTINCT can also be safely removed as the internal GROUP BY makes it redundant:
SELECT DATE(`date`) AS `date` , COUNT(`player_name`) AS `player_count`
FROM (
SELECT MIN(`date`) AS `date`, `player_name`
FROM `player_playtime`
GROUP BY `player_name`
) AS t
GROUP BY DATE( `date`) DESC LIMIT 60 ;
Since the COUNT is now obvious that is only counting rows of the derived table, you can replace it with COUNT(*) and further simplify the query:
SELECT t.date , COUNT(*) AS player_count
FROM (
SELECT DATE(MIN(`date`)) AS date
FROM player_playtime
GROUP BY player_name
) AS t
GROUP BY t.date DESC LIMIT 60 ;
Let's say I was looking for the second most highest record.
Sample Table:
CREATE TABLE `my_table` (
`id` int(2) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`value` int(10),
PRIMARY KEY (`id`)
);
INSERT INTO `my_table` (`id`, `name`, `value`) VALUES (NULL, 'foo', '200'), (NULL, 'bar', '100'), (NULL, 'baz', '0'), (NULL, 'quux', '300');
The second highest value is foo. How many ways can you get this result?
The obvious example is:
SELECT name FROM my_table ORDER BY value DESC LIMIT 1 OFFSET 1;
Can you think of other examples?
I was trying this one, but LIMIT & IN/ALL/ANY/SOME subquery is not supported.
SELECT name FROM my_table WHERE value IN (
SELECT MIN(value) FROM my_table ORDER BY value DESC LIMIT 1
) LIMIT 1;
Eduardo's solution in standard SQL
select *
from (
select id,
name,
value,
row_number() over (order by value) as rn
from my_table t
) t
where rn = 1 -- can pick any row using this
This works on any modern DBMS except MySQL. This solution is usually faster than solutions using sub-selects. It also can easily return the 2nd, 3rd, ... row (again this is achievable with Eduardo's solution as well).
It can also be adjusted to count by groups (adding a partition by) so the "greatest-n-per-group" problem can be solved with the same pattern.
Here is a SQLFiddle to play around with: http://sqlfiddle.com/#!12/286d0/1
This only works for exactly the second highest:
SELECT * FROM my_table two
WHERE EXISTS (
SELECT * FROM my_table one
WHERE one.value > two.value
AND NOT EXISTS (
SELECT * FROM my_table zero
WHERE zero.value > one.value
)
)
LIMIT 1
;
This one emulates a window function rank() for platforms that don't have them. It can also be adapted for ranks <> 2 by altering one constant:
SELECT one.*
-- , 1+COALESCE(agg.rnk,0) AS rnk
FROM my_table one
LEFT JOIN (
SELECT one.id , COUNT(*) AS rnk
FROM my_table one
JOIN my_table cnt ON cnt.value > one.value
GROUP BY one.id
) agg ON agg.id = one.id
WHERE agg.rnk=1 -- the aggregate starts counting at zero
;
Both solutions need functional self-joins (I don't know if mysql allows them, IIRC it only disallows them if the table is the target for updates or deletes)
The below one does not need window functions, but uses a recursive query to enumerate the rankings:
WITH RECURSIVE agg AS (
SELECT one.id
, one.value
, 1 AS rnk
FROM my_table one
WHERE NOT EXISTS (
SELECT * FROM my_table zero
WHERE zero.value > one.value
)
UNION ALL
SELECT two.id
, two.value
, agg.rnk+1 AS rnk
FROM my_table two
JOIN agg ON two.value < agg.value
WHERE NOT EXISTS (
SELECT * FROM my_table nx
WHERE nx.value > two.value
AND nx.value < agg.value
)
)
SELECT * FROM agg
WHERE rnk = 2
;
(the recursive query will not work in mysql, obviously)
You can use inline initialization like this:
select * from (
select id,
name,
value,
#curRank := #curRank + 1 AS rank
from my_table t, (SELECT #curRank := 0) r
order by value desc
) tb
where tb.rank = 2
SELECT name
FROM my_table
WHERE value < (SELECT max(value) FROM my_table)
ORDER BY value DESC
LIMIT 1
SELECT name
FROM my_table
WHERE value = (
SELECT min(r.value)
FROM (
SELECT name, value
FROM my_table
ORDER BY value DESC
LIMIT 2
) r
)
LIMIT 1