Using ORDER BY and GROUP BY together - mysql

My table looks like this (and I'm using MySQL):
m_id | v_id | timestamp
------------------------
6 | 1 | 1333635317
34 | 1 | 1333635323
34 | 1 | 1333635336
6 | 1 | 1333635343
6 | 1 | 1333635349
My target is to take each m_id one time, and order by the highest timestamp.
The result should be:
m_id | v_id | timestamp
------------------------
6 | 1 | 1333635349
34 | 1 | 1333635336
And i wrote this query:
SELECT * FROM table GROUP BY m_id ORDER BY timestamp DESC
But, the results are:
m_id | v_id | timestamp
------------------------
34 | 1 | 1333635323
6 | 1 | 1333635317
I think it causes because it first does GROUP_BY and then ORDER the results.
Any ideas? Thank you.

One way to do this that correctly uses group by:
select l.*
from table l
inner join (
select
m_id, max(timestamp) as latest
from table
group by m_id
) r
on l.timestamp = r.latest and l.m_id = r.m_id
order by timestamp desc
How this works:
selects the latest timestamp for each distinct m_id in the subquery
only selects rows from table that match a row from the subquery (this operation -- where a join is performed, but no columns are selected from the second table, it's just used as a filter -- is known as a "semijoin" in case you were curious)
orders the rows

If you really don't care about which timestamp you'll get and your v_id is always the same for a given m_i you can do the following:
select m_id, v_id, max(timestamp) from table
group by m_id, v_id
order by max(timestamp) desc
Now, if the v_id changes for a given m_id then you should do the following
select t1.* from table t1
left join table t2 on t1.m_id = t2.m_id and t1.timestamp < t2.timestamp
where t2.timestamp is null
order by t1.timestamp desc

Here is the simplest solution
select m_id,v_id,max(timestamp) from table group by m_id;
Group by m_id but get max of timestamp for each m_id.

You can try this
SELECT tbl.* FROM (SELECT * FROM table ORDER BY timestamp DESC) as tbl
GROUP BY tbl.m_id

SQL>
SELECT interview.qtrcode QTR, interview.companyname "Company Name", interview.division Division
FROM interview
JOIN jobsdev.employer
ON (interview.companyname = employer.companyname AND employer.zipcode like '100%')
GROUP BY interview.qtrcode, interview.companyname, interview.division
ORDER BY interview.qtrcode;

I felt confused when I tried to understand the question and answers at first. I spent some time reading and I would like to make a summary.
The OP's example is a little bit misleading.
At first I didn't understand why the accepted answer is the accepted answer.. I thought that the OP's request could be simply fulfilled with
select m_id, v_id, max(timestamp) as max_time from table
group by m_id, v_id
order by max_time desc
Then I took a second look at the accepted answer. And I found that actually the OP wants to express that, for a sample table like:
m_id | v_id | timestamp
------------------------
6 | 1 | 11
34 | 2 | 12
34 | 3 | 13
6 | 4 | 14
6 | 5 | 15
he wants to select all columns based only on (group by)m_id and (order by)timestamp.
Then the above sql won't work. If you still don't get it, imagine you have more columns than m_id | v_id | timestamp, e.g m_id | v_id | timestamp| columnA | columnB |column C| .... With group by, you can only select those "group by" columns and aggreate functions in the result.
By far, you should have understood the accepted answer.
What's more, check row_number function introduced in MySQL 8.0:
https://www.mysqltutorial.org/mysql-window-functions/mysql-row_number-function/
Finding top N rows of every group
It does the simlar thing as the accepted answer.
Some answers are wrong. My MySQL gives me error.
select m_id,v_id,max(timestamp) from table group by m_id;
#abinash sahoo
SELECT m_id,v_id,MAX(TIMESTAMP) AS TIME
FROM table_name
GROUP BY m_id
#Vikas Garhwal
Error message:
[42000][1055] Expression #2 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'testdb.test_table.v_id' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by

Why make it so complicated? This worked.
SELECT m_id,v_id,MAX(TIMESTAMP) AS TIME
FROM table_name
GROUP BY m_id

Just you need to desc with asc. Write the query like below. It will return the values in ascending order.
SELECT * FROM table GROUP BY m_id ORDER BY m_id asc;

Related

How to fetch rows from which sum of a single integer/float column sums upto a certain value

I have a table. It has the following structure
goods_receiving_items
id
item_id
quantity
created_at
I am trying to fetch rows against which have the following conditions
Has one item_id
When the sum of the quantity column equals a certain value
So for example I have the following data
+----+---------+----------+------------+
| id | item_id | quantity | created_at |
+----+---------+----------+------------+
| 1 | 2 | 11 | 2019-10-10 |
| 2 | 3 | 110 | 2019-10-11 |
| 3 | 2 | 20 | 2019-11-09 |
| 4 | 2 | 5 | 2019-11-10 |
| 5 | 2 | 1 | 2019-11-11 |
+----+---------+----------+------------+
I have tried the following query:
SET #sum:= 0;
SELECT item_id, created_at, (#sum:= #sum + quantity) AS SUM, quantity
FROM goods_receiving_items
WHERE item_id = 2 AND #sum<= 6
ORDER BY created_at DESC
If I don't use ORDER BY, then the query will give me ID '1'. But if I use ORDER BY it will return all the rows with item_id = 2.
What should be returned are IDs '5' and '4' exclusively in this order
I can't seem to resolve this and ORDER BY is essential to my task.
Any help would be appreciated
You should use the order by on the resulting set
you could do this using a subquery
SET #sum:= 0;
select t.*
from t (
SELECT item_id
, created_at
, (#sum:= #sum + quantity) as sum
, quantity
FROM goods_receiving_items
WHERE item_id = 2 AND #sum<= 6
) t
ORDER BY created_at DESC
You should try an INNER JOIN with SELECT min(created_at) or SELECT max(created_at)
From MYSQL docs:
...the selection of values from each group cannot be influenced by
adding an ORDER BY clause. Sorting of the result set occurs after
values have been chosen, and ORDER BY does not affect which values the
server chooses.
The answers on the following might help in more detail: MYSQL GROUP BY and ORDER BY not working together as expected
After searching around, I have made up the following query
SELECT
t.id, t.quantity, t.created_at, t.sum
FROM
( SELECT
*,
#bal := #bal + quantity AS sum,
IF(#bal >= $search_number, #doneHere := #doneHere + 1 , #doneHere) AS whereToStop
FROM goods_receiving_items
CROSS JOIN (SELECT #bal := 0.0 , #doneHere := 0) var
WHERE item_id = $item_id
ORDER BY created_at DESC) AS t
WHERE t.whereToStop <= 1
ORDER BY t.created_at ASC
In the above query, $search_number is a variable that holds the value that has to be reached. $item_id is the item we are searching against.
This will return all rows for which the sum of the column quantity makes up the required sum. The sum will be made with rows in descending order by created_at and then will be rearranged in ascending order.
I was using this query to calculate the cost when a certain amount of items are being used in an inventory management system; so this might help someone else do the same. I took most of the query from another question here on StackOverflow

MYSQL selecting top 4 sums based on criteria

Hoping you will be able to help me with this MYSQL statement. I have a table like so:
|id |duration |start |
|1110460 |8.2 |20171211 |
|2221104 |8.9 |20171112 |
|1110460 |3.2 |20171113 |
|1110460 |4.4 |20171214 |
|3331938 |3.2 |20180115 |
|3331722 |5.4 |20171216 |
|1948212 |9.2 |20171217 |
|9219302 |3.2 |20171218 |
What I want to do is list the top 4 IDs by total duration for a given start month in descending order.
For example, for the top 4 IDs for 201712:
|id |duration |
|1110460 |12.6 |
|1948212 |9.2 |
|3331722 |5.4 |
|9219302 |3.2 |
Any help would be appreciated. This is what I have so far, but it has been returning the incorrect results:
SELECT id, sum(duration) FROM table WHERE
start LIKE '201712%' ORDER BY sum(duration) DESC LIMIT 4
You should use group by
SELECT id, sum(duration)
FROM table
WHERE start LIKE '201712%'
GROUP BY id
ORDER BY sum(duration) DESC LIMIT 4
To get 4 distinct maximum sums and their ids you could use following
SELECT a.id, a.sum_duration
FROM (SELECT id, sum(duration) sum_duration
FROM demo
WHERE `start` LIKE '201712%'
GROUP BY id
) a
JOIN (SELECT distinct sum(duration) max_durations
FROM demo
WHERE `start` LIKE '201712%'
GROUP BY id
ORDER BY sum(duration) DESC
LIMIT 4
) b on a.sum_duration = b.max_durations
The above version will return morethan 4 rows if 2 ids have same result of sum
DEMO

MySQL - select rows under an ID, group by column value that has the latest timestamp

Table:
----------------------------------------------------
ID | field_name | field_value | timestamp
----------------------------------------------------
2 | postcode | LS1 | 2016-11-09 16:45:15
2 | age | 34 | 2016-11-09 16:45:22
2 | job | Scientist | 2016-11-09 16:45:27
2 | age | 38 | 2016-11-09 16:46:40
7 | postcode | LS5 | 2016-11-09 16:47:05
7 | age | 24 | 2016-11-09 16:47:44
I wonder if anyone could give me a few pointers, based on the above data, I would like to query by ID 2, return a row for each unique field_name (if more than one row exists under the same id with the same field_name then just return the row with the latest timestamp).
I have managed to almost achieve this by grouping the field_name, which will return a list of unique rows but not necessarily the latest row.
SELECT * FROM fragment WHERE (id = :id) GROUP BY field_name
I would really be grateful for any pointers on what exactly I should do here, and how I could fit something along the lines of MAX(timestamp) in this query,
Many thanks!
Consider you first need a set of data for each ID, FieldName with the max time stamp. (generate that set) as an inline view (B below). Then, join this set (B) back to your base set allowing the inner join to eliminate the unwanted rows.
SELECT A.ID, A.field_name, A.field_value, A.timestamp
FROM Table A
INNER JOIN (SELECT ID, field_name, MAX(timestamp) TS
FROM table
GROUP BY ID, field_name) B
on A.ID = B.ID
and A.field_name = B.field_name
and A.timestamp = B.TS
Outside of MySQL this could be done using window/analytical functions as you would be able to assign a row number to each record and eliminate those > 1 something like....
SELECT B.*
FROM (SELECT A.ID
, A.field_name
, A.field_Vale
, A.timestamp
, Rownumber() over (Order by A.timestamp Desc) RN
FROM Table A ) B
WHERE B.RN = 1
or using a cross apply with a limit or top.
The Simpliest way to do:
SELECT *
FROM fragment fra1
WHERE (id = :id)
and timestamp = (select max(timestamp)
from fragment fra2
where fra2.id = fra1.id
and fra2.field_name = fra1.field_name)
GROUP BY field_name

Query to Segment Results Based on Equal Sets of Column Value

I'd like to construct a single query (or as few as possible) to group a data set. So given a number of buckets, I'd like to return results based on a specific column.
So given a column called score which is a double which contains:
90.00
91.00
94.00
96.00
98.00
99.00
I'd like to be able to use a GROUP BY clause with a function like:
SELECT MIN(score), MAX(score), SUM(score) FROM table GROUP BY BUCKETS(score, 3)
Ideally this would return 3 rows (grouping the results into 3 buckets with as close to equal count in each group as is possible):
90.00, 91.00, 181.00
94.00, 96.00, 190.00
98.00, 99.00, 197.00
Is there some function that would do this? I'd like to avoid returning all the rows and figuring out the bucket segments myself.
Dave
create table test (
id int not null auto_increment primary key,
val decimal(4,2)
) engine = myisam;
insert into test (val) values
(90.00),
(91.00),
(94.00),
(96.00),
(98.00),
(99.00);
select min(val) as lower,max(val) as higher,sum(val) as total from (
select id,val,#row:=#row+1 as row
from test,(select #row:=0) as r order by id
) as t
group by ceil(row/2)
+-------+--------+--------+
| lower | higher | total |
+-------+--------+--------+
| 90.00 | 91.00 | 181.00 |
| 94.00 | 96.00 | 190.00 |
| 98.00 | 99.00 | 197.00 |
+-------+--------+--------+
3 rows in set (0.00 sec)
Unluckily mysql doesn't have analytical function like rownum(), so you have to use some variable to emulate it. Once you do it, you can simply use ceil() function in order to group every tot rows as you like. Hope that it helps despite my english.
set #r = (select count(*) from test);
select min(val) as lower,max(val) as higher,sum(val) as total from (
select id,val,#row:=#row+1 as row
from test,(select #row:=0) as r order by id
) as t
group by ceil(row/ceil(#r/3))
or, with a single query
select min(val) as lower,max(val) as higher,sum(val) as total from (
select id,val,#row:=#row+1 as row,tot
from test,(select count(*) as tot from test) as t2,(select #row:=0) as r order by id
) as t
group by ceil(row/ceil(tot/3))

MySQL - Exclude rows from Select based on duplication of two columns

I am attempting to narrow results of an existing complex query based on conditional matches on multiple columns within the returned data set. I'll attempt to simplify the data as much as possible here.
Assume that the following table structure represents the data that my existing complex query has already selected (here ordered by date):
+----+-----------+------+------------+
| id | remote_id | type | date |
+----+-----------+------+------------+
| 1 | 1 | A | 2011-01-01 |
| 3 | 1 | A | 2011-01-07 |
| 5 | 1 | B | 2011-01-07 |
| 4 | 1 | A | 2011-05-01 |
+----+-----------+------+------------+
I need to select from that data set based on the following criteria:
If the pairing of remote_id and type is unique to the set, return the row always
If the pairing of remote_id and type is not unique to the set, take the following action:
Of the sets of rows for which the pairing of remote_id and type are not unique, return only the single row for which date is greatest and still less than or equal to now.
So, if today is 2011-01-10, I'd like the data set returned to be:
+----+-----------+------+------------+
| id | remote_id | type | date |
+----+-----------+------+------------+
| 3 | 1 | A | 2011-01-07 |
| 5 | 1 | B | 2011-01-07 |
+----+-----------+------+------------+
For some reason I'm having no luck wrapping my head around this one. I suspect the answer lies in good application of group by, but I just can't grasp it. Any help is greatly appreciated!
/* Rows with exactly one date - always return regardless of when date occurs */
SELECT id, remote_id, type, date
FROM YourTable
GROUP BY remote_id, type
HAVING COUNT(*) = 1
UNION
/* Rows with more than one date - Return Max date <= NOW */
SELECT yt.id, yt.remote_id, yt.type, yt.date
FROM YourTable yt
INNER JOIN (SELECT remote_id, type, max(date) as maxdate
FROM YourTable
WHERE date <= DATE(NOW())
GROUP BY remote_id, type
HAVING COUNT(*) > 1) sq
ON yt.remote_id = sq.remote_id
AND yt.type = sq.type
AND yt.date = sq.maxdate
The group by clause groups all rows that have identical values of one or more columns together and returns one row in the result set for them. If you use aggregate functions (min, max, sum, avg etc.) that will be applied for each "group".
SELECT id, remote_id, type, max(date)
FROM blah
GROUP BY remote_id, date;
I'm not whore where today's date comes in, but assumed that was part of the complex query that you didn't describe and I assume isn't directly relevant to your question here.
Try this:
SELECT a.*
FROM table a INNER JOIN
(
select remote_id, type, MAX(date) date, COUNT(1) cnt from table
group by remote_id, type
) b
WHERE a.remote_id = b.remote_id,
AND a.type = b.type
AND a.date = b.date
AND ( (b.cnt = 1) OR (b.cnt>1 AND b.date <= DATE(NOW())))
Try this
select id, remote_id, type, MAX(date) from table
group by remote_id, type
Hey Carson! You could try using the "distinct" keyword on those two fields, and in a union you can use Count() along with group by and some operators to pull non-unique (greatest and less-than today) records!