I have two tables like these:
Table "users":
user_id | source
----------------
1 | 2
2 | 2
3 | 3
4 | 0
Table "sources":
source_id | name
----------------
1 | "one"
2 | "two"
3 | "three"
4 | "four"
Now I need to SELECT (*) FROM source and additionally COUNT "users" that have this source, BUT if there is an additional filter(requests by PHP mysqli), then additionally sort "sources" table by its users count.
What is the best way to do so, and is it possible to do in one statement?
--------------Added editing----------
The first part(SELECT with count from another table) I'm doing this way:
SELECT
id, name
(select count(*) from users where source = sources.id) as sourceUsersCount
FROM sources
And now, how to order this list by users count in each source?
Please check the below query if this is what you need.
select s.*,a.c from sources s
left join
(select count(*) as c,source as src
from user u join sources s
on s.source_id = u.source group by u.source) a
on s.source_id = a.src;
Count the number of users:
SELECT sources.*, COUNT(users.user_id) FROM sources
LEFT JOIN users ON users.source_id = sources.source_id
GROUP BY sources.source_id;
I assume by filters you mean the WHERE clause:
SELECT sources.*, COUNT(users.user_id) FROM sources
LEFT JOIN users ON users.source_id = sources.source_id
WHERE sources.source_id = 2
GROUP BY sources.source_id;
And you can always attach an ORDER BY on the end for sorting:
SELECT sources.*, COUNT(users.user_id) FROM sources
LEFT JOIN users ON users.source_id = sources.source_id
GROUP BY sources.source_id
ORDER BY sources.source_id DESC;
Achieved it by doing so:
SELECT
sources.*,
count(users.source) as sourceUsersCount
FROM sources
LEFT JOIN users ON sources.id = users.source
//In case of additional filters
WHERE
id != 0 AND (name LIKE %?% OR id LIKE %?%)
//\\
GROUP BY sources.id
//In case of sorting by "users" count
ORDER BY sourceUsersCount ASC
//\\
Is it the best way, or maybe there are some faster variants?
I have a table ("lms_attendance") of users' check-in and out times that looks like this:
id user time io (enum)
1 9 1370931202 out
2 9 1370931664 out
3 6 1370932128 out
4 12 1370932128 out
5 12 1370933037 in
I'm trying to create a view of this table that would output only the most recent record per user id, while giving me the "in" or "out" value, so something like:
id user time io
2 9 1370931664 out
3 6 1370932128 out
5 12 1370933037 in
I'm pretty close so far, but I realized that views won't accept subquerys, which is making it a lot harder. The closest query I got was :
select
`lms_attendance`.`id` AS `id`,
`lms_attendance`.`user` AS `user`,
max(`lms_attendance`.`time`) AS `time`,
`lms_attendance`.`io` AS `io`
from `lms_attendance`
group by
`lms_attendance`.`user`,
`lms_attendance`.`io`
But what I get is :
id user time io
3 6 1370932128 out
1 9 1370931664 out
5 12 1370933037 in
4 12 1370932128 out
Which is close, but not perfect. I know that last group by shouldn't be there, but without it, it returns the most recent time, but not with it's relative IO value.
Any ideas?
Thanks!
Query:
SQLFIDDLEExample
SELECT t1.*
FROM lms_attendance t1
WHERE t1.time = (SELECT MAX(t2.time)
FROM lms_attendance t2
WHERE t2.user = t1.user)
Result:
| ID | USER | TIME | IO |
--------------------------------
| 2 | 9 | 1370931664 | out |
| 3 | 6 | 1370932128 | out |
| 5 | 12 | 1370933037 | in |
Note that if a user has multiple records with the same "maximum" time, the query above will return more than one record. If you only want 1 record per user, use the query below:
SQLFIDDLEExample
SELECT t1.*
FROM lms_attendance t1
WHERE t1.id = (SELECT t2.id
FROM lms_attendance t2
WHERE t2.user = t1.user
ORDER BY t2.id DESC
LIMIT 1)
No need to trying reinvent the wheel, as this is common greatest-n-per-group problem. Very nice solution is presented.
I prefer the most simplistic solution (see SQLFiddle, updated Justin's) without subqueries (thus easy to use in views):
SELECT t1.*
FROM lms_attendance AS t1
LEFT OUTER JOIN lms_attendance AS t2
ON t1.user = t2.user
AND (t1.time < t2.time
OR (t1.time = t2.time AND t1.Id < t2.Id))
WHERE t2.user IS NULL
This also works in a case where there are two different records with the same greatest value within the same group - thanks to the trick with (t1.time = t2.time AND t1.Id < t2.Id). All I am doing here is to assure that in case when two records of the same user have same time only one is chosen. Doesn't actually matter if the criteria is Id or something else - basically any criteria that is guaranteed to be unique would make the job here.
Based in #TMS answer, I like it because there's no need for subqueries but I think ommiting the 'OR' part will be sufficient and much simpler to understand and read.
SELECT t1.*
FROM lms_attendance AS t1
LEFT JOIN lms_attendance AS t2
ON t1.user = t2.user
AND t1.time < t2.time
WHERE t2.user IS NULL
if you are not interested in rows with null times you can filter them in the WHERE clause:
SELECT t1.*
FROM lms_attendance AS t1
LEFT JOIN lms_attendance AS t2
ON t1.user = t2.user
AND t1.time < t2.time
WHERE t2.user IS NULL and t1.time IS NOT NULL
Already solved, but just for the record, another approach would be to create two views...
CREATE TABLE lms_attendance
(id int, user int, time int, io varchar(3));
CREATE VIEW latest_all AS
SELECT la.user, max(la.time) time
FROM lms_attendance la
GROUP BY la.user;
CREATE VIEW latest_io AS
SELECT la.*
FROM lms_attendance la
JOIN latest_all lall
ON lall.user = la.user
AND lall.time = la.time;
INSERT INTO lms_attendance
VALUES
(1, 9, 1370931202, 'out'),
(2, 9, 1370931664, 'out'),
(3, 6, 1370932128, 'out'),
(4, 12, 1370932128, 'out'),
(5, 12, 1370933037, 'in');
SELECT * FROM latest_io;
Click here to see it in action at SQL Fiddle
If your on MySQL 8.0 or higher you can use Window functions:
Query:
DBFiddleExample
SELECT DISTINCT
FIRST_VALUE(ID) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS ID,
FIRST_VALUE(USER) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS USER,
FIRST_VALUE(TIME) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS TIME,
FIRST_VALUE(IO) OVER (PARTITION BY lms_attendance.USER ORDER BY lms_attendance.TIME DESC) AS IO
FROM lms_attendance;
Result:
| ID | USER | TIME | IO |
--------------------------------
| 2 | 9 | 1370931664 | out |
| 3 | 6 | 1370932128 | out |
| 5 | 12 | 1370933037 | in |
The advantage I see over using the solution proposed by Justin is that it enables you to select the row with the most recent data per user (or per id, or per whatever) even from subqueries without the need for an intermediate view or table.
And in case your running a HANA it is also ~7 times faster :D
Ok, this might be either a hack or error-prone, but somehow this is working as well-
SELECT id, MAX(user) as user, MAX(time) as time, MAX(io) as io FROM lms_attendance GROUP BY id;
select b.* from
(select
`lms_attendance`.`user` AS `user`,
max(`lms_attendance`.`time`) AS `time`
from `lms_attendance`
group by
`lms_attendance`.`user`) a
join
(select *
from `lms_attendance` ) b
on a.user = b.user
and a.time = b.time
I have tried one solution which works for me
SELECT user, MAX(TIME) as time
FROM lms_attendance
GROUP by user
HAVING MAX(time)
I have a very large table and all of the other suggestions here were taking a very long time to execute. I came up with this hacky method that was much faster. The downside is, if the max(date) row has a duplicate date for that user, it will return both of them.
SELECT * FROM mb_web.devices_log WHERE CONCAT(dtime, '-', user_id) in (
SELECT concat(max(dtime), '-', user_id) FROM mb_web.devices_log GROUP BY user_id
)
select result from (
select vorsteuerid as result, count(*) as anzahl from kreditorenrechnung where kundeid = 7148
group by vorsteuerid
) a order by anzahl desc limit 0,1
I have done same thing like below
SELECT t1.*
FROM lms_attendance t1
WHERE t1.id in (SELECT max(t2.id) as id
FROM lms_attendance t2
group BY t2.user)
This will also reduce memory utilization.
Thanks.
Possibly you can do group by user and then order by time desc. Something like as below
SELECT * FROM lms_attendance group by user order by time desc;
Try this query:
select id,user, max(time), io
FROM lms_attendance group by user;
This worked for me:
SELECT user, time FROM
(
SELECT user, time FROM lms_attendance --where clause
) AS T
WHERE (SELECT COUNT(0) FROM table WHERE user = T.user AND time > T.time) = 0
ORDER BY user ASC, time DESC
I have the table:
id | date_submitted
1 | 01/01/2017
1 | 01/02/2017
2 | 01/03/2017
2 | 01/04/2017
I'm looking for the correct SQL to select each row, limited to one row per id that has the latest value in date_submitted.
So the SQL should return for the above table:
id | date_submitted
1 | 01/02/2017
2 | 01/04/2017
The query needs to select everything in the row, too.
Thanks for your help.
You can find max date for each id in subquery and join it with the original table to get all the rows with all the columns (assuming there are more columns apart from id and date_submitted) like this:
select t.*
from your_table t
inner join (
select id, max(date_submitted) date_submitted
from your_table
group by id
) t2 on t.id = t2.id
and t.date_submitted = t2.date_submitted;
Note that this query will return multiple rows for an id in case there are multiple rows with date_submitted equals to max date_submitted for that id. If you really want only one row per id, then the solution will be a bit different.
If you just need id and max date use:
select id, max(date_submitted) date_submitted
from your_table
group by id
I'm trying to sort the resutls of a SELECT statement using a custom order like so:
SELECT * FROM table ORDER BY FIELD(id,4,5,6) LIMIT 6
I was expecting to have returned rows with ids: 4,5,6,1,2,3 but instead I'm getting 1,2,3,7,8,9. What am I doing wrong?
As a side note: Prior to running this query, I'm pulling this sort order from the database using a different SELECT with a GROUP_CONCAT function like so:
SELECT group_concat(clickID ORDER BY count DESC separator ',') from table2 WHERE searchphrase='$searchphrase'
This results in the 4,5,6 which is then used in the main query. Is there a faster way to write this all in one statement?
Try it this way
SELECT *
FROM table1
ORDER BY FIELD(id, 4,5,6) > 0 DESC, id
LIMIT 6
Output:
| ID |
|----|
| 4 |
| 5 |
| 6 |
| 1 |
| 2 |
| 3 |
Here is SQLFiddle demo
There is no need of the FIELD function. That will only make things slow.
You just need to properly use the ORDER BY:
SELECT * FROM table
ORDER BY id IN (4,5,6) DESC, id
LIMIT 6
here's how to do it all in one query
SELECT DISTINCT t1.*
FROM table t1
LEFT JOIN table2 ON t1.id = t2.clickID AND t2.searchphrase='$searchphrase'
ORDER BY t2.clickID IS NULL ASC, t1.id ASC
When the LEFT JOIN finds no match, it sets the fields in t2 to NULL in the returned row. This orders by this nullness.
So lets say i have 2 tables.
table users:
id | name | date
table weight_tracking;
id | user_id | previous_weight | current_weight | date
weight_tracking table is being updated daily with user current weight and previous_weight.
I am trying to display all users ordering by previous_weight - current_weight (so by the difference in weight - who ever lost the most weight will show up first)
Can that be done with 1 call ?
SELECT users.*, weight_tracking.*
FROM users left join weight_tracking on users.id = weight_tracking.user_id
ORDER BY previous_weight - current_weight desc
if a user can have more than one row in weight_tracking table, you could use something like this:
SELECT users.*
FROM users left join weight_tracking on users.id = weight_tracking.user_id
GROUP BY users.id
ORDER BY max(previous_weight) - min(current_weight) desc