I have a table storing the scores (with the date) of players they did at each game.
Example:
john 154 10/02/2014
mat 178 09/02/2014
eric 270 08/02/2014
mat 410 07/02/2014
john 155 06/02/2014
In this example I want "eric 270 08/02/2014" because thins is the oldest of the most recents.
Which request must I do to retrieve that ?
As I understand it, you request the oldest entry among the set containing the most recent one of each user.
In such a case, you can deal with your problem using a subquery given the last date for each user, then used in the main query to select and sort only the most recent entry of each user.
SELECT scores.*
FROM scores
INNER JOIN
(
SELECT max(date) last, name
FROM scores
GROUP BY name
) last_temp_table
ON scores.name = last_temp_table.name
AND scores.date = last_temp_table.last
ORDER BY scores.date ASC LIMIT 1;
More info in different SO threads such as MySQL order by before group by
The question as worded doesn't make much sense unless you define most recent.
If I assume that you have some criteria like: "Give the oldest event that happened within the last 3 days" then that is a simple matter of ordering and limiting across a date range.
select * from events where ts >= CURDATE() - 3
order by ts asc
limit 1
Related
I already asked this question earlier but forgot a few (important) details or got them wrong.
My table in MySQL 8.0.29 looks like this
UserID
Appointment
Description
Bob
2022-06-01
Cleaning
Bob
2022-06-03
Toothache
John
2022-06-02
Braces
I'm trying to get the latest appointment for every person sorted by oldest first.
The query should return
UserID
Appointment
Description
John
2022-06-02
Braces
Bob
2022-06-03
Toothache
Using one of the previous answers I get
SELECT Name, Appointment, Description
FROM (
SELECT Name, Appointment, Description, ROW_NUMBER() OVER(PARTITION BY Name ORDER BY Appointment DESC) rn) t1
WHERE rn = 1
The problem is the database currently has 3 million rows and it'll continue to grow so this query ends up being pretty slow.
My plan is to consume the data in chunks so I'd prefer the query having "pagination". Something like a LIMIT 0, 5000 to get 5000 records at a time.
I'm open to even re-architecting the database if it comes to that.
For now i've resorted to creating a new table that just keeps the latest appointment for each user.
You are halfway there. Use that query as a 'derived table' instead of making it permanent:
SELECT b.*
FROM ( SELECT user_id, MAX(appointment) AS last_date)
FROM tbl
GROUP BY user_id ) AS x
JOIN tbl AS b ON b.user_id = x.user_id
AND b.appointment = x.last_date
And be sure to have INDEX(user_id, appointment)
I would be interested to see if this and the "OVER" approach both give the same results and which is faster.
I am stuck with a select I have to do, I have a data base where a new claim file is registered in the table called “claims”, in this table every file is registered as follows :
Sorry, i have attached above a print screen with how the tables look, i don't know why are as bellow when i post it.
ClaimFileNumber || Vehicle number || ……. || OpeningDate
1 abc 20170302
2 bcd 20170302
3 efg 20170301
4 hij 20170301
I need a select which can help me to find out how many claim files are open on each day from when this year started until now, ordered by top 5 days for each month like for example, on the month of May we have: 20170506 - 300 claims, 20170511 – 295 claims, 20170509 – 200 claims etc.
Or it is ok a select which can give me the number of claims opened per day and order them desc.
The problem is that the date stored in table OpeningDate it is stored as numeric and not as date, this is the tricky part at least for me.
I cannot use a select like “select count (OpeningDate) from claim where openingdate = 20170302” for each day because there are more than 200 days from when the year have started.
Thank you in advance for your help.
This should do it:
SELECT OpeningDate, COUNT(OpeningDate)
FROM claim
WHERE LEFT(OpeningDate, 4) = '2017'
GROUP BY OpeningDate
ORDER BY OpeningDate ASC, COUNT(OpeningDate) DESC
You need group by:
select OpeningDate,count(1) from your_table group by OpeningDate
For top 5, you need order by and limit
select OpeningDate,count(1)
from your_table
group by OpeningDateorder
order by 2 desc
limit 5
I have a data set like this:
User Date Status
Eric 1/1/2015 4
Eric 2/1/2015 2
Eric 3/1/2015 4
Mike 1/1/2015 4
Mike 2/1/2015 4
Mike 3/1/2015 2
I'm trying to write a query in which I will retrieve users whose MOST RECENT transaction status is a 4. If it's not a 4 I don't want to see that user in the results. This dataset could have 2 potential results, one for Eric and one for Mike. However, Mike's most recent transaction was not a 4, therefore:
The return result would be:
User Date Status
Eric 3/1/2015 4
As this record is the only record for Eric that has a 4 as his latest transaction date.
Here's what I've tried so far:
SELECT
user, MAX(date) as dates, status
FROM
orders
GROUP BY
status,
user
This would get me to a unqiue record for every user for every status type. This would be a subquery, and the parent query would look like:
SELECT
user, dates, status
WHERE
status = 4
GROUP BY
user
However, this is clearly flawed as I don't want status = 4 records IF their most recent record is not a 4. I only want status = 4 when the latest date is a 4. Any thoughts?
SELECT user, date
, actualOrders.status
FROM (
SELECT user, MAX(date) as date
FROM orders
GROUP BY user) AS lastOrderDates
INNER JOIN orders AS actualOrders USING (user, date)
WHERE actualOrders.status = 4
;
-- Since USING is being used, there is not a need to specify source of the
-- user and date fields in the SELECT clause; however, if an ON clause was
-- used instead, either table could be used as the source of those fields.
Also, you may want to rethink the field names used if it is not too late and user and date are both found here.
SELECT user, date, status FROM
(
SELECT user, MAX(date) as date, status FROM orders GROUP BY user
)
WHERE status = 4
The easiest way is to include your order table a second time in a subquery in your from clause in order to retrieve the last date for each user. Then you can add a where clause to match the most recent date per user, and finally filter on the status.
select orders.*
from orders,
(
select ord_user, max(ord_date) ord_date
from orders
group by ord_user
) latestdate
where orders.ord_status = 4
and orders.ord_user = latestdate.ord_user
and orders.ord_date = latestdate.ord_date
Another option is to use the over partition clause:
Oracle SQL query: Retrieve latest values per group based on time
Regards,
I am trying to query a dataset from a single table, which contains quiz answers/entries from multiple users. I want to pull out the highest scoring entry from each individual user.
My data looks like the following:
ID TP_ID quiz_id name num_questions correct incorrect percent created_at
1 10154312970149546 1 Joe 3 2 1 67 2015-09-20 22:47:10
2 10154312970149546 1 Joe 3 3 0 100 2015-09-21 20:15:20
3 125564674465289 1 Test User 3 1 2 33 2015-09-23 08:07:18
4 10153627558393996 1 Bob 3 3 0 100 2015-09-23 11:27:02
My query looks like the following:
SELECT * FROM `entries`
WHERE `TP_ID` IN('10153627558393996', '10154312970149546')
GROUP BY `TP_ID`
ORDER BY `correct` DESC
In my mind, what that should do is get the two users from the IN clause, order them by the number of correct answers and then group them together, so I should be left with the 2 highest scores from those two users.
In reality it's giving me two results, but the one from Joe gives me the lower of the two values (2), with Bob first with a score of 3. Swapping to ASC ordering keeps the scores the same but places Joe first.
So, how could I achieve what I need?
You're after the groupwise maximum, which can be obtained by joining the grouped results back to the table:
SELECT * FROM entries NATURAL JOIN (
SELECT TP_ID, MAX(correct) correct
FROM entries
WHERE TP_ID IN ('10153627558393996', '10154312970149546')
GROUP BY TP_ID
) t
Of course, if a user has multiple records with the maximal score, it will return all of them; should you only want some subset, you'll need to express the logic for determining which.
MySql is quite lax when it comes to group-by-clauses - but as a rule of thumb you should try to follow the rule that other DBMSs enforce:
In a group-by-query each column should either be part of the group-by-clause or contain a column-function.
For your query I would suggest:
SELECT `TP_ID`,`name`,max(`correct`) FROM `entries`
WHERE `TP_ID` IN('10153627558393996', '10154312970149546')
GROUP BY `TP_ID`,`name`
Since your table seems quite denormalized the group by name-par could be omitted, but it might be necessary in other cases.
ORDER BY is only used to specify in which order the results are returned but does nothing about what results are returned - so you need to apply the max()-function to get the highest number of right answers.
I've got a table where the columns that matter look like this:
username
source
description
My goal is to get the 10 most recent records where a user/source combination is unique. From the following data:
1 katie facebook loved it!
2 katie facebook it could have been better.
3 tom twitter less then 140
4 katie twitter Wowzers!
The query should return records 2,3 and 4 (assume higher IDs are more recent - the actual table uses a timestamp column).
My current solution 'works' but requires 1 select to generate the 10 records, then 1 select to get the proper description per row (so 11 selects to generate 10 records) ... I have to imagine there's a better way to go. That solution is:
SELECT max(id) as MAX_ID, username, source, topic
FROM events
GROUP BY source, username
ORDER BY MAX_ID desc;
It returns the proper ids, but the wrong descriptions so I can then select the proper descriptions by the record ID.
Untested, but you should be able to handle this with a join:
SELECT
fullEvent.id,
fullEvent.username,
fullEvent.source,
fullEvent.topic
FROM
events fullEvent JOIN
(
SELECT max(id) as MAX_ID, username, source
FROM events
GROUP BY source, username
) maxEvent ON maxEvent.MAX_ID = fullEvent.id
ORDER BY fullEvent.id desc;