See My Table and query here (on sqlfiddle) Myquery is fetching exact results that I need but it takes long time when data is more than 1000 rows. I need to make it Efficient
Select `id`, `uid`, case when left_leg > right_leg then left_leg
else right_leg end as Max_leg,`date` from
(
SELECT * FROM network t2 where id in (select id from
(select id,uid,`date` from (SELECT * FROM network order by uid,
`date` desc) t4 group by uid) t3) and
(left_leg>=500 or right_leg>=500))t1
Want to pick data from network againt latest dates for each uid
Wanted to pick data where left_leg >=500 or right_leg >=500
Wanted to pick only bigger of two legs (left or right)
Whole query might have some problems but Core issue is with this code
SELECT * FROM network t2 where id in (select id from
(select id,uid,`date` from (SELECT * FROM network order by uid,
`date` desc) t4 group by uid) t3)
I want to improve this query because it fetches esults so much slow when data grows.
Taking your description:
first find the max date for each uid
then find the associate ids
filter on the leg values
Example:
SELECT id, uid, GREATEST(left_leg, right_leg) as max_leg
FROM network
WHERE (uid, `date`) IN (SELECT uid, MAX(`date`)
FROM network
GROUP BY uid)
AND (left_leg > 500 or right_leg > 500)
Note that this means that if the latest date for a uid does not have some legs > 500, then the records won't show up in the result. If you want the latest records with legs > 500, the leg filter has to be moved in.
Add an index on (uid,date):
ALTER TABLE network ADD INDEX uid_date( uid, date )
and try something like:
SELECT n.id, n.uid, greatest( n.right_leg, n.left_leg ) as max_leg, n.date
FROM
( SELECT uid,max(date) as latest FROM network
WHERE right_leg>=500 OR left_leg >=500
GROUP BY uid
) ng
JOIN network n ON n.uid = ng.uid AND n.date = ng.latest
Related
I have a big table of messages with date and room columns. and 2 billion rows.
now i want keep only last 50 messages for every room and delete previous messages.
can i do it with a fast query ?
this question is unique , i didn't found any other question for delete rows over a grouped and ordered selection
You cannot do it in a fast query. You have a lot of data.
I would suggest creating a new table. You can then replace the data in your first table, if necessary.
Possibly the most efficient method to get the 50 rows -- assuming that date is unique for each room:
select t.*
from t
where t.date >= coalesce((select t2.date
from t t2
where t2.room = t.room
order by t2.date desc
limit 1
), t.date
);
For this to have any hope of performance you want an index on (room, date).
You can also try row_number() in MySQL 8+:
select . . . -- list the columns
from (select t.*, row_number() over (partition by room order by date desc) as seqnum
from t
) t
where seqnum <= 50;
Then you can replace the data by doing:
create table temp_t as
select . . . -- one of the select queries here;
truncate table t; -- this gets rid of all the data, so be careful
insert into t
select *
from temp_t;
Massive inserts are much more efficient than massive updates, because the old data does not need to be logged (nor the pages locked and other things).
You can use Rank() function to get top 50 results for each group ordered by date desc, so the last entries will be in top.
http://www.mysqltutorial.org/mysql-window-functions/mysql-rank-function/
Then you left join that subquery on your table on id ( or room and date, if those are unique and you don’t have id in your table)
The last step would be to filter all such result that have null in subquery and delete those.
The full code will look something like this:
DELETE T FROM YOURTABLE T
LEFT JOIN (
SELECT *,
RANK() OVER (PARTITION BY
ROOM
ORDER BY
[DATE] DESC
) DATE_RANK
) AS T2
ON T.[DATE] = T2.[DATE]
AND T.ROOM = T2.ROOM
AND T2.DATE_RANK<=50
WHERE T2.DATE IS NULL
I have the following table in MYSQL:
DATETIME dt
VARCHAR location
FLOAT temperature
This table contains many temperatures measured at different locations. Now I need a SQL query to get the last measurements done for each location.
I actually have no idea how to do that.
Note that while doing group by aggregate its better to have the column selected prior doing group by.
select
t1.location,
t2.t2_dt as dt,
t1.temperature
from table_name t1
join
(
select
location,
temperature,
max(dt) as t2_dt
from table_name
group by location
)t2
on t2.location = t1.location AND t1.dt = t2.t2_dt
group by t1.location
DEMO
This will do your job
select * from
(select location, temperature, dt from my_table) as tab
join
(select max(dt) as max_dt from my_table
group by location) as max_date_tab
on (tab.dt = max_date_tab.max_dt)
I have two tables, one holds user info (id, name, etc) and another table that holds user tickets and ticket status (ticket_id, user_id, ticket_status, etc).
I want to produce a list of ALL the users for example: ( SELECT * FROM user_table )
And for each user I need a count of their tickets for example:
(SELECT t1.user_id, COUNT(*) FROM user_tickets t1 WHERE t1.ticket_status = 15 GROUP BY t1.ticket_status, t1.user_id )
I can do this query to achieve what I’m looking for but it takes 5sec. to run the query on 50000 tickets, while each query running separately only takes fraction of a second.
SELECT t1.user_id, COUNT(*)
FROM user_tickets t1
LEFT JOIN user_table t2 ON t1.user_id = t2.id
WHERE t2.group_id = 20 AND t1.status_id = 15
GROUP BY t1.status_id, user_id
Any idea how to write the query to get same performance as each separately?
An indexing where clause fixed the problem.
I have a mysql query. I need to get last value from columns Lat,Lng from my table but serial_number column needs to be distinct.
How to make such a query?
This is needed as I am using this coordinates to load it to Google map. So when the Google maps loads I need to have a marker on each last coordinates where vehicle is.
SELECT m.*
FROM (
SELECT DISTINCT serial_number
FROM mytable
) md
JOIN mytable m
ON m.id =
(
SELECT id
FROM mytable mi
WHERE mi.serial_number = md.serial_number
ORDER BY
mi.time DESC, mi.id DESC
LIMIT 1
)
Create an index on (serial_number, time, id) for this to work fast.
If you want to retrieve the last record for a certain serial_number, just use this:
SELECT *
FROM mytable
WHERE serial_number = :my_serial_number
ORDER BY
time DESC, id DESC
LIMIT 1
1#
Assuming that max ID will always give you last lat and lon, the query becomes quite simple -
SELECT t2.*
FROM table t2
where t2.id IN
(
SELECT max(t1.id)
FROM table t1
GROUP BY t1.serial_number
)
2#
If you need to consider time also, then you will need to do it this way. Here, in the inner query, max_time of each serial_number is obtained. Then this max_time and serial_number is joined with the outer table time and serial_number respectively, to get distinct records with last lat and lon.
SELECT *
FROM table t2,
(
SELECT max(t1.time) max_time, t1.serial_number
FROM table t1
GROUP BY t1.serial_number
) new_table
WHERE t2.time=new_table.max_time
AND t2.serial_number=new_table.serial_number
Try this
select distinct serial_number, *
from table t
inner join table t1 on t1.serial_number = t.serial_number and t1.id = (select max id from table t2 where t2.serial_number = t1.serial_number)
I have a table defined like this:
CREATE TABLE mytable (id INT NOT NULL AUTO_INCREMENT, PRIMARY KEY(id),
user_id INT REFERENCES user(id) ON UPDATE CASCASE ON DELETE RESTRICT,
amount REAL NOT NULL CHECK (amount > 0),
record_date DATE NOT NULL
);
CREATE UNIQUE INDEX idxu_mybl_key ON mytable (user_id, amount, record_date);
I want to write a query that will have two columns:
user_id
amount
There should be only ONE entry in the returned result set for a given user. Furthermore, the amount figure returned should be the last recoreded amount for the user (i.e. MAX(record_date).
The complication arises because weights are recorded on different dates for different users, so there is no single LAST record_date for all users.
How may I write (preferably an ANSI SQL) query to return the columns mentioned previously, but ensuring that its only the amount for the last recorded amount for the user that is returned?
As an aside, it is probably a good idea to return the 'record_date' column as well in the query, so that it is eas(ier) to verify that the query is working as required.
I am using MySQL as my backend db, but ideally the query should be db agnostic (i.e. ANSI SQL) if possible.
First you need the last record_date for each user:
select user_id, max(record_date) as last_record_date
from mytable
group by user_id
Now, you can join previous query with mytable itself to get amount for this record_date:
select
t1.user_id, last_record_date, amount
from
mytable t1
inner join
( select user_id, max(record_date) as last_record_date
from mytable
group by user_id
) t2
on t1.user_id = t2.user_id
and t1.record_date = t2.last_record_date
A problem appears becuase a user can have several rows for same last_record_date (with different amounts). Then you should get one of them, sample (getting the max of the different amounts):
select
t1.user_id, t1.record_date as last_record_date, max(t1.amount)
from
mytable t1
inner join
( select user_id, max(record_date) as last_record_date
from mytable
group by user_id
) t2
on t1.user_id = t2.user_id
and t1.record_date = t2.last_record_date
group by t1.user_id, t1.record_date
I do not now about MySQL but in general SQL you need a sub-query for that. You must join the query that calculates the greatest record_date with the original one that calculates the corresponding amount. Roughly like this:
SELECT B.*
FROM
(select user_id, max(record_date) max_date from mytable group by user_id) A
join
mytable B
on A.user_id = B.user_id and A.max_date = B.record_date
SELECT datatable.* FROM
mytable AS datatable
INNER JOIN (
SELECT user_id,max(record_date) AS max_record_date FROM mytable GROUP BS user_id
) AS selectortable ON
selectortable.user_id=datatable.user_id
AND
selectortable.max_record_date=datatable.record_date
in some SQLs you might need
SELECT MAX(user_id), ...
in the selectortable view instead of simply SELECT user_id,...
The definition of maximum: there is no larger(or: "more recent") value than this one. This naturally leads to a NOT EXISTS query, which should be available in any DBMS.
SELECT user_id, amount
FROM mytable mt
WHERE mt.user_id = $user
AND NOT EXISTS ( SELECT *
FROM mytable nx
WHERE nx.user_id = mt.user_id
AND nx.record_date > mt.record_date
)
;
BTW: your table definition allows more than one record to exist for a given {id,date}, but with different amounts. This query will return them all.