I have a table like below:
id | deptid | last_activities
1 | 1 | MMD PURCASHING
2 | 1 | JKO / REPORTING
3 | 2 | STAND BY CC
4 | 3 | JKO / REPORTING
My query:
SELECT id, deptid, last_activities FROM dept_activities GROUP BY deptid ORDER BY id DESC
And the result like below:
id | deptid | last_activities
4 | 3 | JKO / REPORTING
3 | 2 | STAND BY CC
1 | 1 | MMD PURCASHING
There i want result like this:
id | deptid | last_activities
4 | 3 | JKO / REPORTING
3 | 2 | STAND BY CC
2 | 1 | JKO / REPORTING
How could it be? How right query?
Please try this :
SELECT * FROM dept_activities AS da1
JOIN
(SELECT MAX(id) as max_id FROM dept_activities GROUP BY deptid DESC) AS da2
ON (da1.id = da2.max_id);
Using a sub query to get the max id for each deptid, and then joining that against the table gives you something like this:-
SELECT a.id, a.deptid, a.last_activities
FROM dept_activities a
INNER JOIN
(
SELECT MAX(id) AS MaxId, deptid
FROM dept_activities
GROUP BY deptid
) Sub1
ON a.deptid = Sub1.deptid
AND a.id = Sub1.MaxId
ORDER BY a.id DESC
Another possible solution using GROUP_CONCAT, although not keen on this (and will fail if last_activity contains the delimiter you use for GROUP_CONCAT).
SELECT MAX(id), deptid, SUBSTRING_INDEX(GROUP_CONCAT(last_activities ORDER BY id DESC), ',', 1)
FROM dept_activities
GROUP BY a.deptid
Related
There are a lot of questions dealing with max values but I can't find any that relate to this issue.
ID | Company | Result
----------------------
1 | 1 | A
2 | 1 | C
3 | 1 | B <--
4 | 2 | C
5 | 2 | B
6 | 2 | A <!--
7 | 3 | C
8 | 3 | A
9 | 3 | B <--
I need to output the Companies whose last Result (based on ID) was "B".
To further complicate the issue, the $query will be used this:
select * from table where Company in ($query)
Any ideas? Thanks!
On MySQL 8+, here is a query you may try using analytic functions:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY Company ORDER BY ID DESC) rn
FROM yourTable
)
SELECT ID, Company, Result
FROM cte
WHERE rn = 1 AND Result = 'B';
Demo
On earlier versions of MySQL, we can try joining to a subquery which finds the most recent record for each company:
SELECT t1.*
FROM yourTable t1
INNER JOIN
(
SELECT Company, MAX(ID) AS MAX_ID
FROM yourTable
GROUP BY Company
) t2
ON t1.Company = t2.Company AND
t1.ID = t2.MAX_ID
WHERE
t1.Result = 'B';
Demo
Please see the picture for ERROR SCREENSHOT
Table: Candidate
+-----+---------+
| id | Name |
+-----+---------+
| 1 | A |
| 2 | B |
| 3 | C |
| 4 | D |
| 5 | E |
+-----+---------+
Table: Vote
+-----+--------------+
| id | CandidateId |
+-----+--------------+
| 1 | 2 |
| 2 | 4 |
| 3 | 3 |
| 4 | 2 |
| 5 | 5 |
+-----+--------------+
id is the auto-increment primary key, CandidateId is the id appeared in Candidate table.
Write a sql to find the name of the winning candidate, the above example will return the winner B.
+------+
| Name |
+------+
| B |
+------+
Notes:
You may assume there is no tie, in other words there will be at most one winning candidate.
Why this code can't work? Just try to use without limit
SELECT c.Name AS Name
FROM Candidate AS c
JOIN
(SELECT r.CandidateId AS can, MAX(r.Total_vote) AS big
FROM (SELECT CandidateId, COUNT(id) AS Total_vote
FROM Vote
GROUP BY CandidateId) AS r) AS v
ON c.id = v.can;
In your query, here: SELECT r.CandidateId AS can, MAX(r.Total_vote) AS big
you use MAX aggregate function, without group by, which is not correct SQL.
Try:
SELECT Candidate.* FROM Candidate
JOIN (
SELECT CandidateId, COUNT(id) AS Total_vote
FROM Vote
GROUP BY CandidateId
ORDER BY COUNT(id) DESC LIMIT 1
) v
ON Candidate.id = v.CandidateId
This is a join/group by query with order by:
select c.name
from candidate c join
vote v
on v.candidateid = c.id
group by c.id, c.name
order by count(*) desc
limit 1;
SELECT c.Name AS Name
FROM Candidate AS c JOIN (SELECT r.CandidateId AS can
FROM
(SELECT CandidateId, COUNT(id) AS Total_vote
FROM Vote
GROUP BY CandidateId) AS r
WHERE r.Total_vote = (SELECT MAX(r.Total_vote) FROM (SELECT
CandidateId, COUNT(id) AS Total_vote
FROM Vote
GROUP BY CandidateId) r)) AS v
ON c.id = v.can;
This is updated code
My code has two errors. The first one is "use of an aggregate like Max requires a Group By clause if there are any non-aggregated columns in the select list", but not sure why my previous code still can run and show no error. Maybe the system add the group by function automatically when it run.
The second one is that max can't be used with Group by in this format.
Sorry to confuse you about my title. I am building an auction system and I am having a difficulty in getting the user's winning item.
Example I have a table like this:
the columns are:
id, product_id, user_id, status, is_winner, info, bidding_price, bidding_date
here's my sql fiddle:
http://sqlfiddle.com/#!9/7097d/1
I want to get every user's item that they already win. So I need to identify if they are the last who bid in that item.
I need to filter it using a user_id.
If I do a query like this:
SELECT MAX(product_id) AS product_id FROM auction_product_bidding
WHERE user_id = 3;
it will get only the product_id that is 12 and the product_id of 9 did not get. Product ID 9 is also that last bid of the user_id 3.
Can you help me? I hope you got my point. Thanks. Sorry if my question a little bit confusing.
According to your question, seems 11 is also what you want, try this query:
SELECT apd.product_id
FROM auction_product_bidding apd
JOIN (
SELECT MAX(bidding_date) AS bidding_date, product_id
FROM auction_product_bidding
GROUP BY product_id
) t
ON apd.product_id = t.product_id
AND apd.bidding_date = t.bidding_date
WHERE apd.user_id = 3;
Check Demo Here
select id,product_id,user_id,status,is_winner,info,bidding_price,bidding_date,rank
from
( SELECT apb.*,
greatest(#rank:=if(product_id=#prodGrp,#rank+1,1),-1) as rank,
#prodGrp:=product_id as dummy
FROM auction_product_bidding apb
cross join (select #prodGrp:=-1,#rank:=0) xParams
order by product_id,bidding_date DESC
) xDerived
where user_id=3 and rank=1;
That user won 9,11,12
+----+------------+---------+--------+-----------+------+---------------+---------------------+------+
| id | product_id | user_id | status | is_winner | info | bidding_price | bidding_date | rank |
+----+------------+---------+--------+-----------+------+---------------+---------------------+------+
| 60 | 9 | 3 | | 0 | | 75000.00 | 2016-08-02 16:31:23 | 1 |
| 59 | 11 | 3 | | 0 | | 15000.00 | 2016-08-02 12:04:16 | 1 |
| 68 | 12 | 3 | | 0 | | 18000.00 | 2016-08-10 09:20:01 | 1 |
+----+------------+---------+--------+-----------+------+---------------+---------------------+------+
SELECT product_id FROM auction_product_bidding where bidding_price= any
(select max(bidding_price) from auction_product_bidding group by product_id)
and user_id='3';
select * from
(select product_id,user_id,max(bidding_price) from
(select * from auction_product_bidding order by bidding_price desc) a
group by product_id) b
where user_id=3;
Answer:
product_id user_id max(bidding_price)
9 3 75000
11 3 15000
12 3 18000
An idea could be to sort the table desc by date and select every distinct row by product_id and customer_id. Something like
SELECT DISTINCT prod_id, user_id FROM (
SELECT * FROM auction_product_bidding ORDER BY date DESC
)
You want everything that bids last in 3, is it right ?
I have a table like this:
+----+---------+------------+
| id | conn_id | read_date |
+----+---------+------------+
| 1 | 1 | 2010-02-21 |
| 2 | 1 | 2011-02-21 |
| 3 | 2 | 2011-02-21 |
| 4 | 2 | 2013-02-21 |
| 5 | 2 | 2014-02-21 |
+----+---------+------------+
I want the second highest read_date for particular 'conn_id's i.e. I want a group by on conn_id. Please help me figure this out.
Here's a solution for a particular conn_id :
select max (read_date) from my_table
where conn_id=1
and read_date<(
select max (read_date) from my_table
where conn_id=1
)
If you want to get it for all conn_id using group by, do this:
select t.conn_id, (select max(i.read_date) from my_table i
where i.conn_id=t.conn_id and i.read_date<max(t.read_date))
from my_table t group by conn_id;
Following answer should work in MSSQL :
select id,conn_id,read_date from (
select *,ROW_NUMBER() over(Partition by conn_id order by read_date desc) as RN
from my_table
)
where RN =2
There is an intresting article on use of rank functions in MySQL here : ROW_NUMBER() in MySQL
If your table design as ID - date matching (ie a big id always a big date), you can group by id, otherwise do the following:
$sql_max = '(select conn_id, max(read_date) max_date from tab group by 1) as tab_max';
$sql_max2 = "(select tab.conn_id,max(tab.read_date) max_date2 from tab, $sql_max
where tab.conn_id = tab_max.conn_id and tab.read_date < tab_max.max_date
group by 1) as tab_max2";
$sql = "select tab.* from tab, $sql_max2
where tab.conn_id = tab_max2.conn_id and tab.read_date = tab_max2.max_date2";
I have a table of the following structure:
ID | COMPANY_ID | VERSION | TEXT
---------------------------------
1 | 1 | 1 | hello
2 | 1 | 2 | world
3 | 2 | 1 | foo
is there a way to get the most recent version of records only, i.e. I would want to have as a result set the IDs 2 and 3?
I'm sure there are better ways, but I tend to use this kind of query:
SELECT *
FROM
(SELECT * FROM test ORDER BY VERSION DESC) AS my_table
GROUP BY COMPANY_ID
Produces this result set:
ID | COMPANY_ID | VERSION | TEXT
---------------------------------
2 | 1 | 2 | world
3 | 2 | 1 | foo
Try this:
SELECT *
FROM (
SELECT company_id, MAX(version) maxVersion
FROM table
GROUP BY company_id ) as val
JOIN table t ON (val.company_id = t.company_id AND t.version = val.maxversion)
If your IDs are ordered (newer version iff higher id):
SELECT t.*, a.maxversion
FROM (
SELECT MAX(id) maxid, MAX(version) maxversion
FROM table
GROUP BY company_id
) a
INNER JOIN table t
ON a.maxid = t.id
However, if your IDs are not properly ordered, you need to use the following query:
SELECT t.*
FROM (
SELECT company_id, MAX(version) maxversion
FROM table
GROUP BY company_id
) v
INNER JOIN table t
ON v.company_id = t.company_id
AND v.maxversion = t.version
(assuming there's an UNIQUE constraint/index on (company_id, version))