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 ;
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
I would like to get values without the smallest and the biggest ones, so without entry with 2 and 29 in column NumberOfRepeating.
My query is:
SELECT Note, COUNT(*) as 'NumberOfRepeating'
WHERE COUNT(*) <> MAX(COUNT(*))AND COUNT(*) <> MIN(COUNT(*))
FROM Note GROUP BY Note;
SELECT Note, COUNT(*) as 'NumberOfRepeating'
FROM Notes
GROUP BY Note
HAVING count(*) <
(
SELECT max(t.maxi)
FROM (select
Note, COUNT(Note) maxi FROM Notes
GROUP BY Note
) as t
)
AND
count(*) >
(
SELECT min(t.min)
FROM (select
Note, COUNT(Note) min FROM Notes
GROUP BY Note
) as t
)
try this code.
One method would use order by and limit, twice:
select t.*
from (select t.*
from t
order by NumberOfRepeating asc
limit 99999999 offset 1
) t
order by NumberOfRepeating desc
limit 99999999 offset 1;
Try this code,
Select * from Note where NumberOfRepeating < (select MAX(NumberOfRepeating) from Note ) AND NumberOfRepeating > (select MIN(NumberOfRepeating) from Note );
Here in the code, as in your table Note is the name of the table, and NumberOfRepeating is the column name, as in your table.
Try this. It should work
SELECT *
FROM ( SELECT Note, COUNT(*) as 'NumberOfRepeating'
FROM Notes
GROUP BY Note
ORDER BY NumberOfRepeating DESC
LIMIT 1, 2147483647
) T1
ORDER BY T1.NumberOfRepeating
LIMIT 1, 2147483647
This query returns all items where the difference between the timestamps is less than 180 seconds.
The problem is, after all this is done, I need to then limit the results to the one most recent entry per facebook_id.
I tried using GROUP BY facebook_id, but it doesn't work because if I GROUP BY facebook_id before ORDER BY 'time', it picks the older entry instead of the newer entry which is not what I want.
Is there any way to GROUP BY after ORDER BY?
SELECT facebook_id,
TIMESTAMPDIFF(SECOND, `time`, '$mytime') AS `timediff`
FROM `table`
WHERE `facebook_id` != $fbid
HAVING `timediff` <= '180'
ORDER BY `time` DESC
Thanks for your help!
Note: I did try the suggested solutions to this question but had no success. GROUP BY after ORDER BY
You can use a self join by calculating maximum value of time column and join with 2 conditions one with facebook_id and second to match the time from table to the max_time of second query which will return recent entry against each facebook_id
SELECT t.*,
TIMESTAMPDIFF(SECOND, `time`, '$mytime') AS `timediff`
FROM `table` t
JOIN (
SELECT facebook_id,MAX(`time`) max_time FROM `table` GROUP BY facebook_id
) t1
ON(t.facebook_id= t1.facebook_id AND t.`time` = t1.max_time)
WHERE t.`facebook_id` != $fbid
HAVING `timediff` <= '180'
ORDER BY t.`time` DESC
SELECT fid, timediff
FROM (
SELECT facebook_id as fid,
TIMESTAMPDIFF(SECOND, `time`, '$mytime') AS timediff
FROM `table`
WHERE `facebook_id` != $fbid
HAVING `timediff` <= '180'
ORDER BY `time` DESC
) entries
GROUP BY entries.fid
Please let me know if you have any questions!
I'm using an union statement in mysql but i've some problems sorting the results. The ORDER statement doesn't works at all, the results comes out always sorted by the id field.
Here an example query:
SELECT a.* FROM ( ( select * from ticket_ticket AS t1 WHERE ticket_active=1 ORDER BY t1.ticket_date_last_modified DESC )
UNION ( select * from ticket_ticket AS t2 WHERE ticket_active=0 ORDER BY t2.ticket_date_last_modified DESC, t2.ticket_status_id DESC ) )
AS a LIMIT 0,20;
I want to order the results of the first SELECT by last_modified time, and the second SELECT by time and status. But the ORDER statement get just skipped. The results always come out ordered by the ticket_id ( the PRIMARY KEY ).
What's wrong in this query ?
Thanks!
Ok, i've fixed it writing the query this way:
SELECT a.*
FROM
(SELECT *
FROM ticket_ticket
WHERE ticket_active=1
ORDER BY ticket_date_last_modified DESC) AS a
UNION ALL
SELECT b.*
FROM
(SELECT *
FROM ticket_ticket
WHERE ticket_active=0
ORDER BY ticket_date_last_modified DESC, ticket_status_id DESC) AS b LIMIT 0,
20;
You are using a UNION query that will return distinct values, and the order of the returned rows is not guaranteed.
But you don't need an union query for this:
select *
from ticket_ticket AS t1
ORDER BY
ticket_active!=1,
ticket_date_last_modified DESC,
ticket_status_id DESC
LIMIT 0,20;